Astronomy

How do I calculate moon phase orientation at any Coordinates given all the necessary data?

How do I calculate moon phase orientation at any Coordinates given all the necessary data?


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I'm having a problem for a couple of days where I'm trying to accurately calculate the phase angle of the moon at any given Coordinates (lon,lat).If I know my current Coordinates, Moon and sun topocentric right ascension, declination, altitude and zenith Coordinates.

Given all of that data, how do I achieve that ? The result I'm looking for preferable in degrees where 0 degrees means the crescent moon pointing directly towards me, 180 degrees means the crescent moon pointing at the other direction and so on…

I have tried a formula where I measured the angle between the sun zenith and a certain coordinate in relation to the moon zenith, on paper this could work, the results were decent but not to the accuracy I'm looking for. Any help would be much appreciated.


Find Terminal Coordinates, Given a Bearing and a Distance

This function will calculate the end coordinates, in degrees, minutes and seconds, given an initial set of coordinates, a bearing or azimuth (referenced to True North or 0 degrees), and a distance.

The function uses the Great Circle method of calculating distances between two points on the Earth. The shortest distance between two points on the surface of a sphere is an arc, not a straight line. (Try this using a string on a globe surface.) Because of the curved surface, the angle from Point A looking to Point B generally will not be converse of the angle looking from Point B to Point A. The function Distance and Azimuths Between Two Sets of Coordinates, shows this more clearly.

Bearings or azimuths start with 0 degrees toward true north, 90 degrees east, 180 degrees south, and 270 degrees west (clockwise rotation).

Applicants will find this program helpful in determining compliance with the minimum spacing table in 47 CFR 73.207 for FM stations or 47 CFR 73.610 for television stations. DXers (long distance listeners and viewers) can use this function to find the distance and bearings to, and antenna orientation to best receive, distant stations. Station coordinates may be found through the AM Query, FM Query, or the TV Query.

Questions on Find Terminal Coordinates may be directed to Dale Bickel, [email protected]


Information about AM and FM broadcast radio stations is available at the Audio Division on the FCC's website, and at Broadcast Radio Links.

Information about television stations is available at the Video Division.


How do astronomers determine the position of other celestial objects?

For very distant objects you would just measure the angular position in the sky, and estimate the distance from the redshift. The angular position is typically expressed using equatorial coordinates (right ascension and declination). This subtracts the effect of the Earth's daily rotation, but the coordinates would still change slowly with time due to precession, so the coordinates are qualified by an epoch (eg. J2000.0). Other coordinate systems are available. The precise origin of the coordinate system is immaterial for such objects: somewhere in the vicinity of the solar system is accurate enough.

Objects within the solar system are usually in orbit around something, and often moving quite quickly, so rather than just measure their position at an instant in time what you usually want to do is calculate their orbital elements (the parameters of the orbit). Depending on how quickly the orbit is being perturbed by other objects, this allows you to calculate their location at any time within a given time window. There are six orbital elements because there are six degrees of freedom, but several ways in which they can be expressed.

Regarding how you derive the orbital elements, there are a number of tools at your disposal. Obviously you can measure angular position at particular moments in time. Distance and radial velocity can sometimes be measured by RADAR, or in the case of the moon, LIDAR (using the retroreflectors landed there). One particularly accurate type of measurement is to record the time at which one object eclipses or transits another. It is also possible to make use of information from tracking the orbits of space probes. Given a sufficient number of such measurements it is possible to obtain a unique solution for the orbital elements.

Intermediate distances can be more difficult to measure accurately. Parallax can be used to measure distance with reasonable accuracy up to a few hundred parsecs. Other methods include using type 1a supernovae as 'standard candles' - if the apparent magnitude is observed then the approximate distance can be inferred.

Coordinates are typically geocentric (relative to the Earth), heliocentric (to the Sun), or barycentric (to the centre of mass of the orbiting objects). In cases where it material and not implied by the context, it ought to be stated which.


3. Error and Uncertainty

Positions are the products of measurements. All measurements contain some degree of error. Errors are introduced in the original act of measuring locations on the Earth surface. Errors are also introduced when second- and third-generation data is produced, say, by scanning or digitizing a paper map.

In general, there are three sources of error in measurement: human beings, the environment in which they work, and the measurement instruments they use.

Human errors include mistakes, such as reading an instrument incorrectly, and judgments. Judgment becomes a factor when the phenomenon that is being measured is not directly observable (like an aquifer), or has ambiguous boundaries (like a soil unit).

Environmental characteristics, such as variations in temperature, gravity, and magnetic declination, also result in measurement errors.

Instrument errors follow from the fact that space is continuous. There is no limit to how precisely a position can be specified. Measurements, however, can be only so precise. No matter what instrument, there is always a limit to how small a difference is detectable. That limit is called resolution.

Figure 5.4.1, below, shows the same position (the point in the center of the bullseye) measured by two instruments. The two grid patterns represent the smallest objects that can be detected by the instruments. The pattern at left represents a higher-resolution instrument.

The resolution of an instrument affects the precision of measurements taken with it. In the illustration below, the measurement at left, which was taken with the higher-resolution instrument, is more precise than the measurement at right. In digital form, the more precise measurement would be represented with additional decimal places. For example, a position specified with the UTM coordinates 500,000. meters East and 5,000,000. meters North is actually an area 1 meter square. A more precise specification would be 500,000.001 meters East and 5,000,000.001 meters North, which locates the position within an area 1 millimeter square. You can think of the area as a zone of uncertainty within which, somewhere, the theoretically infinitesimal point location exists. Uncertainty is inherent in geospatial data.

Precision takes on a slightly different meaning when it is used to refer to a number of repeated measurements. In the Figure 5.4.3, below, there is less variance among the nine measurements at left than there is among the nine measurements at right. The set of measurements at left is said to be more precise.

Hopefully, you have noticed that resolution and precision are independent from accuracy. As shown below, accuracy simply means how closely a measurement corresponds to an actual value.

I mentioned the U.S. Geological Survey's National Map Accuracy Standard in Chapter 2. In regard to topographic maps, the Standard warrants that 90 percent of well-defined points tested will be within a certain tolerance of their actual positions. Another way to specify the accuracy of an entire spatial database is to calculate the average difference between many measured positions and actual positions. The statistic is called the root mean square error (RMSE) of a data set.


3 Things to Know About Coordinate Rotation

1. Is rotation a transformation?

Rotation of coordinates to a new location is considered a type of transformation of those points, but transformations are not always a rotation. They can and often are much more complex than rotating points about an axis.

2. Are rotations clockwise or counterclockwise?

Rotations can be both clockwise and counterclockwise, however, the calculator above solves for clockwise rotation.

3. Is rotation and revolution the same?

The terms revolution and rotation are synonomous. One revolution is equal to a rotation of 360 degrees.


How to Calculate Current on a 3-phase, 208V Rack PDU (Power Strip)

In recent years, extending 3-phase power distribution all the way to server cabinets and racks has become extremely popular in new data center builds—for many good reasons. Principally, for cabinet power capacities above 5kVA, utilizing 3-phase rack power strips can significantly reduce the copper required to supply such dense loads.

But unfortunately, many users (rightly) find it cumbersome to provision and calculate current (amperage) for 3-phase power in the rack—for example, a typical question would be:

If I plug in a 250W power supply (about 1.2 amps) on this receptacle right here, how will that affect the current that flows through each of the three phases of this rack power strip? Which one will be closest to tripping a circuit breaker?

In North America, where 3-phase, 208V power distribution is wired “line-to-line”, the answer to this question is particularly counter-intuitive. I was astonished to find that the almighty Internet offers virtually no good tools to help answer this question, and so created one together with my colleagues at Raritan (link follows).

Why 3-Phase (208V) Power Strip Loading Is Difficult

With single-phase power strips, loading and provisioning are straightforward: if you add a device to the rack that draws 10 amps—in turn, 10 amps of additional load is drawn from the input line of the power strip.

But with 3-phase power strips, when you add the same 10 amp device to a server cabinet, it is unclear what results. From which of the three lines will additional current flow? How much of the 10 amps flows from which of the three lines? etc.

As we shall see, the answer is not always obvious.

1. In North America, a 208v, 3-phase power strip is divided, well, into 3 sections:

  • L1/L2: outlets that are wired to draw current from lines L1 and L2 (aka, XY)
  • L2/L3: outlets that are wired to draw current from lines L2 and L3 (aka, YZ)
  • L3/L1: outlets that are wired to draw current from lines L1 and L3 (aka, XZ)

2. One would intuitively expect the following to occur when a load is applied:

  • Start with a completely empty rack/power strip
  • Add a 10A load onto an outlet that is supplied from L1/L2
  • In turn, 10A of current flows from L1 and 10A of current flows from L2.

This is intuitive, and indeed, it does happen exactly as you would expect.

3. But next, let us apply another load:

  • Start with the above cabinet (where 10A is already flowing on both L1 and L2)
  • Now, add another (second) 10A load to the cabinet, but plug it into an outlet that is supplied from L2/L3
  • You would expect an additional 10A load to flow on L2 and 10A on L3

In fact, this does not happen!

  • Indeed, 10A flows from L1 (no change)
  • As you would expect, 10A flows from L3—a result of the new load
  • But on L2, the current becomes 17.3A—NOT 20A (10A + 10A).

On the surface, 17.3A on L2 appears to be a completely random number. There are two 10A loads that are wired to L2, but the current is not 10A, not 20A, not 15A, but… 17.3A .

I asked one of our senior power engineers to explain this in laymen’s terms. He sent me the following diagram:

Absolutely true story. Obviously, the moral of the story here is: don’t ask an engineer to explain things in laymen’s terms.

In all seriousness, the essential truth illustrated by the diagram is that the amount of current on a given line (L1, L2, and L3) depends on the amount of current on the other two lines. Calculating each line value requires vector addition (a.k.a. “complex” arithmetic), which really cannot be done in your head—or even with simple Excel formulae.

These calculations simply require more than straightforward arithmetic. And therefore, are neither intuitive nor simple to execute “back-of-the-envelope” calculations. This issue comes up so often with my clients that, for your convenience and reference, I have created the following…

3-Phase Rack Power Strip Current + Power Capacity Calculation Tool

This 3-phase current and power capacity tool, which you can download by clicking on the button above, is completely vendor agnostic. Regardless of the manufacturer of your rack power strip, this tool will be applicable.

Generally-speaking, 3-phase rack power strips can be purchased most commonly in the following 208V electrical configurations:

  • 30A (24A derated) : NEMA L21-30P or NEMA L15-30P input plugs
  • 50A (35A derated) : Hubbell CS8365C “California-style” input plug, but with only three branch circuit breakers [or, less desirably, fuses]on the unit.
  • 50A (40A derated) : Hubbell CS8365C “California-style” input plug, but with six branch circuit breakers[or, less desirably, fuses] on the unit.
  • 60A (45A derated) : IEC 60309 60A, 3-pole / 4-wire, “pin-and-sleeve” input plug, but with smaller, 6# AWG gauge input conductors. Check your manufacturer’s spec sheet closely. If it states that: (a) the rated current is 45A or (b) the rated power capacity is 16.2kVA [not 17.3kVA], then you have this type of power strip.
  • 60A (48A derated) : IEC 60309 60A, 3-pole / 4-wire, “pin-and-sleeve” input plug, but with larger, 4# AWG gauge input conductors. Check your manufacturer’s spec sheet closely. If you have this type of power strip, the spec sheet will explicitly state either: (a) the rated current is 48A or (b) the rated power capacity is 17.3kVA [not 16.2kVA].

Simply download the 3-phase Capacity Planning Tool, and select the correct worksheet among the five tabs along the bottom. Enter in the expected amperages per outlet, being careful to enter loads underneath the correct corresponding circuit breaker heading. (That is, look at your power strip physically [or on engineering submittal drawings], to determine whether a given outlet is on the L1/L2, L2/L3, or L3/L1 branches—also known as XY, YZ, and XZ).

The tool will automatically calculate the resulting current (in amps) on each of the three lines, and highlight any possible error. If you do not see any red warnings (“OVERLOAD” or “ERROR”) on your spreadsheet, then you are good to go!

Note that, if you have a fully-redundant environment (Tier IV), wherein each server cabinet uses two rack power strips originating from two redundant UPS feeds, you should enter loads into this tool assuming that one of the two rack power strips is offline. This ensures that you have enough power capacity on the rack power strip, in case such a power loss does occur.

Fine Print Required By Law

  • If you find this tool useful, please consider linking to this blog post
  • If you find this tool useful, please consider Raritan’s extensive line of 3-phase rack power strips—among the most comprehensive in the market
  • This tool is provided only for your guidance and convenience. Its calculations are not warranted by me or by Raritan (i.e., check your work with actual measurements).

Finally, as it is relevant: allow me to plug the real-time energy metering provided by Raritan’s portfolio of rack power distribution units. Our devices provide billing-grade accurate power information (amps, kWh, voltage, power factor, kVA, kW) for every cabinet—and in many models, for every power supply in every cabinet.

Such information perfectly complements the provisioning task which this 3-phase capacity tool enables. Also, your purchase helps fund my children’s future college educations.

Hope you find this helpful. Comments welcome, as always.

Learn about Raritan Power or request a call from a Data Center Power expert.


How to Calculate Power Factor Correction

wikiHow is a “wiki,” similar to Wikipedia, which means that many of our articles are co-written by multiple authors. To create this article, 27 people, some anonymous, worked to edit and improve it over time.

This article has been viewed 391,868 times.

Power factor correction allows you to calculate apparent power, true power, reactive power and your phase angle. [1] X Research source Consider the equation of a right triangle. So to calculate the angle you need to know your Cosine, Sine and Tangent laws. You also will need to know The Pythagorean Theorem ( c² = a² + b² ) for calculating the magnitudes of the sides of the triangle. You will also need to know what units each type of power is in. Apparent power is measured in Volt-Amps. True power is measured in Watts and your Reactive power is measured in the units called Volt-Amp-Reactive (VAR’s). There are several equations to calculate these and all will be covered in the article. You now have the basis of what you are trying to calculate.


Positive: 1 to 180

If you mod any positive number between 1 and 180 by 360, you will get the exact same number you put in. Mod here just ensures these positive numbers are returned as the same value.

Negative: -180 to -1

Using mod here will return values in the range of 180 and 359 degrees.

Special cases: 0 and 360

Using mod means that 0 is returned, making this a safe 0-359 degrees solution.


The simple approach of just averaging them has weird edge cases with angles when they wrap from 359' back to 0'.

A much earlier question on SO asked about finding the average of a set of compass angles.

An expansion of the approach recommended there for spherical coordinates would be:

  • Convert each lat/long pair into a unit-length 3D vector.
  • Sum each of those vectors
  • Normalise the resulting vector
  • Convert back to spherical coordinates

Thanks! Here is a C# version of OP's solutions using degrees. It utilises the System.Device.Location.GeoCoordinate class

I found this post very useful so here is the solution in PHP. I've been using this successfully and just wanted to save another dev some time.

Very useful post! I've implemented this in JavaScript, hereby my code. I've used this successfully.

Javascript version of the original function

In the interest of possibly saving someone a minute or two, here is the solution that was used in Objective-C instead of python. This version takes an NSArray of NSValues that contain MKMapCoordinates, which was called for in my implementation:

very nice solutions, just what i needed for my swift project, so here's a swift port. thanks & here's also a playground project: https://github.com/ppoh71/playgounds/tree/master/centerLocationPoint.playground

Java Version if anyone needs it. Constants defined static to not calculate them twice.

If you are interested in obtaining a very simplified 'center' of the points (for example, to simply center a map to the center of your gmaps polygon), then here's a basic approach that worked for me.

This returns the middle lat/lng coordinate for the center of a polygon.

In Django this is trivial (and actually works, I had issues with a number of the solutions not correctly returning negatives for latitude).

For instance, let's say you are using django-geopostcodes (of which I am the author).

point is a Django Point instance that can then be used to do things such as retrieve all objects that are within 10km of that centre point

Changing this to raw Python is trivial

Here is the Android version based on @Yodacheese's C# answer using Google Maps api:

Here is the python Version for finding center point. The lat1 and lon1 are latitude and longitude lists. it will retuen the latitude and longitude of center point.

Dart Implementation for Flutter to find Center point for Multiple Latitude, Longitude.

Latitude and Longitude List

This is is the same as a weighted average problem where all the weights are the same, and there are two dimensions.

Find the average of all latitudes for your center latitude and the average of all longitudes for the center longitude.

Caveat Emptor: This is a close distance approximation and the error will become unruly when the deviations from the mean are more than a few miles due to the curvature of the Earth. Remember that latitudes and longitudes are degrees (not really a grid).

The document contains lots of other useful stuff

Out of object in PHP. Given array of coordinate pairs, returns center.

I did this task in javascript like below

Dart/Flutter Calculate the center point of multiple latitude/longitude coordinate pairs

If you want all points to be visible in the image, you'd want the extrema in latitude and longitude and make sure that your view includes those values with whatever border you want.

(From Alnitak's answer, how you calculate the extrema may be a little problematic, but if they're a few degrees on either side of the longitude that wraps around, then you'll call the shot and take the right range.)

If you don't want to distort whatever map that these points are on, then adjust the bounding box's aspect ratio so that it fits whatever pixels you've allocated to the view but still includes the extrema.

To keep the points centered at some arbitrary zooming level, calculate the center of the bounding box that "just fits" the points as above, and keep that point as the center point.


How to Calculate Slope and Intercepts of a Line

This article was co-authored by Grace Imson, MA. Grace Imson is a math teacher with over 40 years of teaching experience. Grace is currently a math instructor at the City College of San Francisco and was previously in the Math Department at Saint Louis University. She has taught math at the elementary, middle, high school, and college levels. She has an MA in Education, specializing in Administration and Supervision from Saint Louis University.

There are 8 references cited in this article, which can be found at the bottom of the page.

This article has been viewed 314,806 times.

The slope of a line measures how steep the line is. [1] X Research source You could also say it is the rise over the run that is, how much the line rises vertically compared with how much it runs horizontally. Being able to find the slope of a line, or using the slope to find points on the line, is an important skill used in economics, [2] X Research source geoscience, [3] X Research source accounting/finance and other fields.


Important Questions for CBSE Class 11 Physics Chapter 14 - Oscillations

Free PDF download of Important Questions with solutions for CBSE Class 11 Physics Chapter 14 - Oscillations prepared by expert Physics teachers from latest edition of CBSE(NCERT) books. CoolGyan.Org to score more marks in your Examination.

1 Marks Questions

1.The girl sitting on a swing stands up. What will be the effect on periodic time of swing?

Ans.The periodic time T is directly proportional to the square root of effective length of pendulum (l). When the girl starts up, the effective length of pendulum (i.e. Swing) decreases, so the Time period (T) also decreases.

2.At what distance from the mean position, is the kinetic energy in a simple harmonic oscillator equal to potential energy?

Ans.Let the displacement of particle executing S.H.M = Y

Amplitude of particle executing S.H.M = a

The kinetic energy =

3.The soldiers marching on a suspended bridge are advised to go out of steps. Why?

Ans.The soldiers marching on a suspended bridge are advised to go out of steps because in such a case the frequency of marching steps matches with natural frequency of the suspended bridge and hence resonance takes place, as a result amplitude of oscillation increases enormously which may lead to the collapsing of bridge.

4.Is the motion of a simple pendulum strictly simple harmonic?

Ans.It is not strictly simple harmonic because we make the assumption that Sinθ =θ, which is nearly valid only if θ is very small.

5.Can a simple pendulum experiment be done inside a satellite?

Ans.Since time period of a simple pendulum is :-

Since, inside a satellite, effective value of ‘g’ = O

So, when g = O, T = α. Therefore, inside the satellite, the pendulum does not oscillate at all. So, it can not be preformed inside a satellite.

6.Give some practical examples of S. H. M?

Ans.Some practical examples of S. H. M. are :-

1) Motion of piston in a gas – filled cylinder.

2) Atoms vibrating in a crystal lattice.

3) Motion of helical spring.

7.What is the relation between uniform circular motion and S.H.N?

Ans.Uniform form circular motion can be thought of as two simple harmonic motion operating at right angle to each other.

8.What is the minimum condition for a system to execute S.H.N? .N? H i[p

Ans.The minimum condition for a body to posses S.H.N. is that it must have elasticity and inertia.

9. The piston in the cylinder head of a locomotive has a stroke (twice the amplitude) of 1.0 m. If the piston moves with simple harmonic motion with an angular frequency of 200 rad/min, what is its maximum speed?

Ans.Angular frequency of the piston, = 200 rad/ min.

Amplitude,

The maximum speed () of the piston is give by the relation:

One Mark Question

  1. How is the time period effected, if the amplitude of a simple pendulum is in Creased?
    Ans. No effect on time period when amplitude of pendulum is increased or decreased.
  2. Define force constant of a spring.
    Ans. The spring constant of a spring is the change in the force it exerts, divided by the change in deflection of the spring.
  3. At what distance from the mean position, is the kinetic energy in simple harmonic oscillator equal to potential energy?
    Ans. Not at the mid-point, between mean and extreme position. it will be at
  4. How is the frequency of oscillation related with the frequency of change in the of K. E and PE of the body in S.H.M.?
    Ans. P.E. or K.E. completes two vibrations in a time during which S.H.M completes one vibration or the frequency of P.E. or K.E. is double than that of S.H.M
  5. What is the frequency of total energy of a particle in S.H.M.?
    Ans. The frequency of total energy of particle is S.H.M is zero because it retain constant.
  6. How is the length of seconds pendulum related with acceleration due gravity of any planet?
    Ans. Length of the seconds pendulum proportional to acceleration due to gravity)
  7. If the bob of a simple pendulum is made to oscillate in some fluid of density greater than the density of air (density of the bob density of the fluid), then time period of the pendulum increased or decrease.
    Ans. Increased
  8. How is the time period of the pendulum effected when pendulum is taken to hills Or in mines?
    Ans. As T will increase.
  9. Define angular frequency. Give its S.I. unit.
    Ans. It is the angle covered per unit time or it is the quantity obtained by multiplying frequency by a factor of 2.
    = 2π v, S.I. unit is rads s -1
  10. Does the direction of acceleration at various points during the oscillation of a simple pendulum remain towards mean position?
    Ans. No, the resultant of Tension in the string and weight of bob is not always towards the mean position.
  11. What is the time period for the function f(t) = sin tot +cos dot may represent the simple harmonic motion?
    Ans.
  12. When is the swinging of simple pendulum considered approximately SHMT?
    Ans. Swinging through small angles.
  13. Can the motion of an artificial satellite around the earth be taken as SHMT?
    Ans. No, it is a circular and periodic motion but not SHM.
  14. What is the phase relationship between displacement, velocity and acceleration in SHM?
    Ans. In SHM, -The velocity leads the displacement by a phase II.2 radians and acceleration leads the velocity by a phase /2 radians.
  15. What forces keep the simple pendulum in motion?
    Ans. The Component of Weight (mg sin)
  16. How Will the time period of a simple pendulum change when its length is doubled?
    Ans.
  17. If the motion of revolving particle is periodic in nature, give the nature of motion or projection of the revolving particle along the diameter.
    Ans. S.H.M
  18. In a forced oscillation of a particle, the amplitude is maximum for a frequency (1) of the force, while the energy is maximum for a frequency () of the force. What is the relation between and ?
    Ans. Both amplitude and energy of the particle can be maximum only in the case of resonance, for resonance to occur .
  19. When will the motion of a simple pendulum be simple harmonic?
    Ans. When the displacement of bob from the mean position is so small that Sin .
  20. A simple harmonic motion of acceleration 'a' and displacement 'x' is represented by . What is the time period of S.H.M?
    Ans.
  21. What is the main difference between forced oscillations and resonance?
    Ans. The frequency of external periodic force is different from the natural frequency of the oscillator in case of forced oscillation but in resonance two frequencies are equal.
  22. Define amplitude of S.H.M.
    Ans. The maximum displacement of oscillating particle on either side of its mean position is called its amplitude.
  23. What is the condition to be satisfied by a mathematical relation between time and displacement to describe a periodic motion?
    Ans. A periodic motion repeats after a definite time interval T. So, y(t) = y(t + T) = y(t + 2T) etc.
  24. Why the pitch of an organ pipe on a hot summer day is higher?
    Ans. On a hot day, the velocity of sound will be more since (frequency proportional to Velocity) the frequency of sound increases and hence its pitch increases.
  25. If any liquid of density higher than the density of water is used in a resonance tube, how will the frequency change?
    Ans. The frequency of vibration depends on the length of the air column and not on reflecting media, hence frequency does not change.

2 Marks Questions

1.A simple harmonic oscillator is represented by the equation : Y = 0.40 Sin (440t+0.61)

Y is in metres

t is in seconds

Find the values of 1) Amplitude 2) Angular frequency 3) Frequency of oscillation 4) Time period of oscillation, 5) Initial phase.

Ans.The given equation is:- Y = 0.40 Sin (440 + 0.61)

Comparing it with equation of S.H.M. Y = a Sin (w t +∅0 )

2) Angular frequency = w = 440Hz

3) Frequency of oscillation,

4) Time period of oscillations, T =

5) Initial phase, ∅ = 0.61 rad.

2.The springs of spring factor k, 2k, k respectively are connected in parallel to a mass m. If the mass = 0.08kg m and k = 2N|m, then find the new time period?

Ans. Total spring constant, K 1 = K1 + K2 + K3 (In parallel)

3.The bob of a vibrating simple pendulum is made of ice. How will the period of swing will change when the ice starts melting?

Ans .The period of swing of simple pendulum will remain unchanged till the location of centre of gravity of the bob left after melting of the ice remains at the fixed position from the point of suspension. If centre of gravity of ice bob after melting is raised upwards, then effective length of pendulum decreases and hence time period of swing decreases. Similarly, if centre of gravity shifts downward, time period increases.

4.An 8 kg body performs S.H.M. of amplitude 30 cm. The restoring force is 60N, when the displacement is 30cm. Find: - a) Time period b) the acceleration c) potential and kinetic energy when the displacement is 12cm?

Ans.Here m = 8 kg

a) f = 60 N, Y = displacement = 0.30m

K = =

As, Angular velocity = w =

Time period, T =

P.E. = Potential energy =

Kinetic energy = K.E =

=

Kinetic energy = K. E. = 7.56J

5.A particle executing SH.M has a maximum displacement of 4 cm and its acceleration at a distance of 1 cm from its mean position is 3 cm/s 2 . What will be its velocity when it is at a distance of 2cm from its mean position?

Ans.The acceleration of a particle executing S.H.M is –

w = Angular frequency Y = Displacement

Given A = 3cm / s 2 Y = 1cm

w =

The velocity of a particle executing S.H.M is :-

6.What is ratio of frequencies of the vertical oscillations when two springs of spring constant K are connected in series and then in parallel?

Ans .If two spring of spring constant K are connected in parallel, then effective resistance in parallel = KP = K + K = 2K

Let fP = frequency in parallel combination.

In Series combination, effective spring constant for 2 sprigs of spring constant K is :-

Let fS = frequency in series combination

7.The kinetic energy of a particle executing S.H.M. is 16J when it is in its mean position. If the amplitude of oscillations is 25cm and the mass of the particle is 5.12kg. Calculate the time period of oscillations?

Ans.K. E. = Kinetic energy = 16J

a = amplitude = 25cm or 0.25m

The Maximum value of K. E. is at mean position which is =

So,

Now, T = Time Period =

8.The time period of a body suspended by a spring is T. What will be the new time period if the spring is cut into two equal parts and 1) the body is suspended by one part. 2) Suspended by both parts in parallel?

Ans.Since time period of oscillation, a body of mass ‘m’ suspended from a spring with force constant ‘k’ are:-

1) On cutting the spring in two equal parts, the length of one part is halved and the force constant of each part will be doubled (2k). Therefore, the new time period is :→

So,

2) If the body is suspended from both parts in parallel, then the effective force constant of parallel combination = 2k + 2k = 4k. Therefore, time period is:→

9.A simple pendulum is executing Simple harmonic motion with a time T. If the length of the pendulum is increased by 21 %. Find the increase in its time period?

Ans.Now, time period of simple pendulum,

l = length of simple pendulum

g = acceleration due b gorily

So,

Since 2π and g are constant, so

If

So,

Therefore percentage increase in time period =

10.A particle is executing S H M of amplitude 4 cm and T = 4 sec. find the time taken by it to move from positive extreme position to half of its amplitude?

Ans.If Y = displacement

Given Y =

So,

Now, T = Time Period, w =

Let WA is work done by spring A & kA = Spring Constant

WB is work done by spring B& kB = Spring Constant

11.Two linear simple harmonic motions of equal amplitudes and angular frequency w and 2 w are impressed on a particle along axis X and Y respectively. If the initial phase difference between them is , find the resultant path followed by the particle?

Ans.Let s = amplitude of each S.H.M.

Then give simple harmonic motions may be represented by:

Now,

x → Displacement along X – axis

y → Displacement along y – axis.

Now, Cos 2 θ = 1 – 2 Sin 2 θ

For equation 1) Sin w t = Putting the value

12.The acceleration due to gravity on the surface of moon is 1.7 m/s 2 . What is the time period of simple pendulum on moon if its time period on the earth is 3.5s?

Ans.If l = length of simple pendulum,

g = Acceleration due to gravity.

Then, on earth

On Moon

g1 = Acceleration due to gravity on moon = 1.7 m | S 2

g = Acceleration due to gravity on earth = 9.8m | s 2

13. Using the correspondence of S. H. M. and uniform circular motion, find displacement, velocity, amplitude, time period and frequency of a particle executing SH.M?

Ans. If initially at t = 0

Then Drop a perpendicular From P on AB,

If the displacement OM = Y

Ratios of circle of reference = Amplitude = a

then In ∆ O P M :→ Angle POD = Angle OPM ( Alternate Angles)

Sin

Now, w = Angular speed

Then

So,

2) Velocity,

So, V = a w

Form equation of displacement :→

So, V = a w

3) Acceleration :→ A =

The acceleration is proportional to negative of displacement is the characteristics of S. H. M.

14.A particle executing S.H.N. along a straight line has a velocity of um/s when its displacement from mean position is 3 m and 3 m / s when displacement is 4m. Find the time taken to travel 2.5 m from the positive extremity of its oscillation?

Ans.Velocity = v1 = u m/s

then, displacement = 3m let y1 = displacement = 3m

For second case, v2 = 3m / s and y2 = displacement = 4m

When the particle is 2.5m from the positive extreme position, its displacement from the mean position, y = 5 – 2.5 = 2.5m. Since the time is measured when the particle is at extreme position:→

15.Springs is spring constant K, 2K, 4K, K ----- are connected in series. A mass M Kg is attached to the lower end of the last spring and system is allowed to vibrate. What is the time period of oscillation?

Ans.For effective resistance of spring of individual spring constant k1, k2, ------ kn

16.A particle is moving with SHN in a straight line. When the distance of the particle from mean position has values x1 and x2 the corresponding values of velocities are v1 and v2. Show that the time period of oscillation is given by:

Ans.If a = amplitude y = displacement w = angular frequency

For first case. u1 2 = w 2 (a 2 – x1 2 ) →1) ( velocity = u1 Displacement = x1)

For second case, u2 2 = w 2 (a 2 – x2 2 ) →2) (velocity = u2 Displacement = x2)

Subtracting equation 2) from equation 1)

17. A circular disc of mass 10 kg is suspended by a wire attached to its centre. The wire is twisted by rotating the disc and released. The period of torsional oscillations is found to be 1.5 s. The radius of the disc is 15 cm. Determine the torsional spring constant of the wire. (Torsional spring constant α is defined by the relation J = -α />¸, where J is the restoring couple and />¸ the angle of twist).

Ans. Mass of the circular disc, m = 10 kg

Radius of the disc, r = 15 cm = 0.15 m

The torsional oscillations of the disc has a time period, T = 1.5 s

The moment of inertia of the disc is:

I

=

= 0.1125 kg

Time period,

α is the torsional constant.

Hence, the torsional spring constant of the wire is 1.972 Nm rad –1 .

2 MARKS QUESTIONS

  1. Which of the following condition is not sufficient for simple harmonic motion and why?
    (i) acceleration and displacement(ii) restoring force and displacement
    Ans. Condition (i) is not sufficient, because direction of acceleration is not mentioned. In SHM, the acceleration is always in a direction opposite to that of the displacement.
  2. The formula for time period T for a loaded spring, Does-the time period depend on length of the spring?
    Ans. Although length of the spring does not appear in the expression for the time period, yet the time period depends on the length of the spring. It is because, force constant of the spring depends on the length of the spring.
  3. Water in a U-tube executes S.H.M. Will the time period for mercury filled up to the same height in the tube be lesser of greater than that in case of Water?
    Ans. The time period of the liquid in a U-tube executing S.H.M. does not depend upon density of the liquid, therefore time period will be same, when the mercury is filled up to the same height in place of water in the U-tube.
  4. There are two springs, one delicate and another hard or stout one. For which spring, the frequency of the oscillator will be more?
    Ans. We have,
    So, when a hard spring is loaded with a mass m. The extension I will be lesser W.r.t. delicate one. So frequency of the oscillation of the hard spring Will be more and if time period is asked it will be lesser.
  5. Time period of a particle in S.H.M. depends on the force constant K and mass m of the particle . A simple pendulum for small angular displacement executes S.H.M approximately. Why then is the time period of a pendulum independent of the mass of the pendulum?
    Ans. Restoring force in case of simple pendulum is given by

    So force constant itself proportional to m as the value of k is substituted in the formula, m is canceled out.
  6. What is the frequency of oscillation of a simple pendulum mounted in a cabin that is falling freely?
    Ans. The pendulum is in a state of Weight less ness i.e. g. = 0. The frequency of pendulum
  7. Even after the breakup of one prong of tunning fork it produces a round of same frequency, then what is the use of having a tunning fork with two prongs?
    Ans. Two prongs of a tunning fork set each other in resonant Vibrations and help to maintain the vibrations for a longer time.
  8. The displacement of a particle in S.H.M may be given byy = a sin(+ o)show that if the time t is increased by 2 Io, the value of y remains the same.
    Ans. The displacement at any time t is y = a sin()
    ∴ displacement at any time will be


    Hence, the displacement at time t and () are same.
  9. Differentiate between closed pipe and open pipe at both ends of same length for frequency of fundamental note and harmonics.
    Ans.
    (i) in a pipe open at both ends, the frequency of fundamental note produced is twice as that produced by a closed pipe of same length.
    (ii) An open pipe produces all the harmonics, while in a closed pipe, the even harmonics are absent.
  10. How does the frequency of a tuning fork change, when the temperature is increased?
    Ans. As the temperature increases, the length of the prong of the tunning fork increases. This increases the wavelength of the stationary waves set up in the tuning fork. As frequency, . so frequency of the tunning fork decreases.

3 Marks Questions

1.A mass = m suspend separately from two springs of spring constant k1 and k2 gives time period t1 and t2 respectively. If the same mass is connected to both the springs as shown in figure. Calculate the time period ‘t’ of the combined system?

Ans. If T = Time Period of simple pendulum

then,

For first spring :

For second spring :

When springs is connected in parallel, effective spring constant, k = k = k1 + k2

Or

2.Show that the total energy of a body executing SHN is independent of time?

Ans.Let y = displacement at any time‘t’

So,

Now, kinetic energy = K. E. =

Potential energy =

Thus total mechanical energy is always constant is equal to . The total energy is independent to time. The potential energy oscillates with time and has a maximum value of . Similarly K. E. oscillates with time and has a maximum value of . At any instant = constant = . The K. E or P.E. oscillates at double the frequency of S.H.M.

3.A particles moves such that its acceleration ‘a’ is given by a = -b x where x = displacement from equilibrium position and b is a constant. Find the period of oscillation? [2]

Ans.Given that a = -b x, Since a x and is directed apposite to x, the particle do moves in S. H. M.

4.A particle is S.H.N. is described by the displacement function:

If the initial (t = 0) position of the particle is 1 cm and its initial velocity is π cm | s, What are its amplitude and phase angle?

Ans.

5.Determine the time period of a simple pendulum of length = l when mass of bob = m Kg? [3]

Ans.It consist of a heavy point mass body suspended by a weightless inextensible and perfectly flexible string from a rigid support which is free to oscillate.

The distance between point of suspension and point of oscillation is effective length of pendulum.

l = length of simple pendulum

Let the bob is displaced through a small angle θ the forces acting on it:-

1) weight = Mg acting vertically downwards.

2) Tension = T acting upwards.

Divide Mg into its components → Mg Cos θ & Mg Sin θ

- ve sign shows force is divested towards the ocean positions. If θ = Small,

Sin θ

i.e.1.) Time period depends on length of pendulum and ‘g’ of place where experiment is done.

2) T is independent of amplitude of vibration provided and it is small and also of the mass of bob.

6. Which of the following examples represent periodic motion?

(a) A swimmer completing one (return) trip from one bank of a river to the other and back.

(b) A freely suspended bar magnet displaced from its N-S direction and released.

(c) A hydrogen molecule rotating about its center of mass.

(d) An arrow released from a bow.

Ans.(b) and (c)

(a) The swimmer's motion is not periodic. The motion of the swimmer between the banks of a river is back and forth. However, it does not have a definite period. This is because the time taken by the swimmer during his back and forth journey may not be the same.

(b) The motion of a freely-suspended magnet, if displaced from its N-S direction and released, is periodic. This is because the magnet oscillates about its position with a definite period of time.

(c) When a hydrogen molecule rotates about its centre of mass, it comes to the same position again and again after an equal interval of time. Such motion is periodic.

(d) An arrow released from a bow moves only in the forward direction. It does not come backward. Hence, this motion is not a periodic.

7. Figure 14.27 depicts four x-t plots for linear motion of a particle. Which of the plots represent periodic motion? What is the period of motion (in case of periodic motion)?

Ans.(b) and (d) are periodic

(a) It is not a periodic motion. This represents a unidirectional, linear uniform motion. There is no repetition of motion in this case.

(b) In this case, the motion of the particle repeats itself after 2 s. Hence, it is a periodic motion, having a period of 2 s.

(c) It is not a periodic motion. This is because the particle repeats the motion in one position only. For a periodic motion, the entire motion of the particle must be repeated in equal intervals of time.

(d) In this case, the motion of the particle repeats itself after 2 s. Hence, it is a periodic motion, having a period of 2 s.

8. Which of the following relationships between the acceleration a and the displacement x of a particle involve simple harmonic motion?

(b) a = -200

(d) a = 100

Ans.(c) A motion represents simple harmonic motion if it is governed by the force law:

m is the mass (a constant for a body)

Among the given equations, only equation a = –10 x is written in the above form with Hence, this relation represents SHM.

9. The acceleration due to gravity on the surface of moon is 1.7 />. What is the time period of a simple pendulum on the surface of moon if its time period on the surface of earth is 3.5 s? (g on the surface of earth is 9.8 />)

Ans.Acceleration due to gravity on the surface of moon,

Acceleration due to gravity on the surface of earth, g = 9.8 m

Time period of a simple pendulum on earth, T = 3.5 s

l is the length of the pendulum

The length of the pendulum remains constant.

On moon's surface, time period,

Hence, the time period of the simple pendulum on the surface of moon is 8.4 s.

10. A simple pendulum of length l and having a bob of mass M is suspended in a car. The car is moving on a circular track of radius R with a uniform speed v. If the pendulum makes small oscillations in a radial direction about its equilibrium position, what will be its time period?

Ans.The bob of the simple pendulum will experience the acceleration due to gravity and the centripetal acceleration provided by the circular motion of the car.

Acceleration due to gravity = g

Centripetal acceleration

v is the uniform speed of the car

R is the radius of the track

Effective acceleration () is given as:

Time period,

Where,l is the length of the pendulum

∴Time period, T

4 Marks Questions

1. Which of the following examples represent (nearly) simple harmonic motion and which represent periodic but not simple harmonic motion?

(a) the rotation of earth about its axis.

(b) motion of an oscillating mercury column in a U-tube.

(c) motion of a ball bearing inside a smooth curved bowl, when released from a point slightly above the lower most point.

(d) general vibrations of a polyatomic molecule about its equilibrium position.

Ans.(b) and (c) are SHMs

(a) and (d) are periodic, but not SHMs

(a) During its rotation about its axis, earth comes to the same position again and again in equal intervals of time. Hence, it is a periodic motion. However, this motion is not simple harmonic. This is because earth does not have a to and fro motion about its axis.

(b) An oscillating mercury column in a U-tube is simple harmonic. This is because the mercury moves to and fro on the same path, about the fixed position, with a certain period of time.

(c) The ball moves to and fro about the lowermost point of the bowl when released. Also, the ball comes back to its initial position in the same period of time, again and again. Hence, its motion is periodic as well as simple harmonic.

(d) A polyatomic molecule has many natural frequencies of oscillation. Its vibration is the superposition of individual simple harmonic motions of a number of different molecules. Hence, it is not simple harmonic, but periodic.

2. Which of the following functions of time represent (a) simple harmonic, (b) periodic but not simple harmonic, and (c) non-periodic motion? Give period for each case of periodic motion ( is any positive constant):

(a)

(b)

(c) 3 cos (π/4 - 2t)

(d) cos t + cos 3t + cos 5t

(e) exp

(f) 1 + t +

This function represents SHM as it can be written in the form:

Its period is:

(b) Periodic, but not SHM

3. A spring balance has a scale that reads from 0 to 50 kg. The length of the scale is 20 cm. A body suspended from this balance, when displaced and released, oscillates with a period of 0.6 s. What is the weight of the body?

Ans. Maximum mass that the scale can read, M = 50 kg

Maximum displacement of the spring = Length of the scale, l = 20 cm = 0.2 m

Maximum force exerted on the spring, F = Mg

g = acceleration due to gravity = 9.8 m/

F = 509.8 = 490

∴Spring constant,

Mass m, is suspended from the balance.

Time period,

∴Weight of the body = mg = 22.36 9.8 = 219.167 N

Hence, the weight of the body is about 219 N.

4. Figures 14.29 correspond to two circular motions. The radius of the circle, the period of revolution, the initial position, and the sense of revolution (i.e. clockwise or anti-clockwise) are indicated on each figure.

Obtain the corresponding simple harmonic motions of the x-projection of the radius vector of the revolving particle P, in each case.

Ans.(a) Time period, T = 2 s

At time, t = 0, the radius vector OP makes an angle with the positive x-axis, i.e., phase angle

Therefore, the equation of simple harmonic motion for thex-projection of OP, at time t, is given by the displacement equation:

(b) Time period, T = 4 s

At time t = 0, OP makes an angle π with the x-axis, in the anticlockwise direction. Hence, phase angle, Φ = + π

Therefore, the equation of simple harmonic motion for the x-projection of OP, at time t, is given as:

5. Cylindrical piece of cork of density of base area A and height h floats in a liquid of density. The cork is depressed slightly and then released. Show that the cork oscillates up and down simple harmonically with a period

Where is the density of cork. (Ignore damping due to viscosity of the liquid).

Ans.Base area of the cork = A

Density of the liquid =

Density of the cork =

Weight of the cork = Weight of the liquid displaced by the floating cork

Let the cork be depressed slightly by x. As a result, some extra water of a certain volume is displaced. Hence, an extra up-thrust acts upward and provides the restoring force to the cork.

Up-thrust = Restoring force, F = Weight of the extra water displaced

F = –(Volume />Density />g)

Volume = Area Distance through which the cork is depressed

F = – A g … (i)

According to the force law:

…….(ii)

The time period of the oscillations of the cork:

……….(iii)

= Volume of the cork Density

= Base area of the cork /> Height of the cork /> Density of the cork

= Ah

Hence, the expression for the time period becomes:

6. A mass attached to a spring is free to oscillate, with angular velocity , in a horizontal plane without friction or damping. It is pulled to a distance x0and pushed towards the centre with a velocity v0at time t = 0. Determine the amplitude of the resulting oscillations in terms of the parameters Ô°, x0and v0. [Hint: Start with the equation x = a cos () and note that the initial velocity is negative.]

Ans.The displacement equation for an oscillating mass is given by:

x=

Velocity,

At t= 0,

… (i)

And,

… (ii)

Squaring and adding equations (i) and (ii), we get:

Hence, the amplitude of the resulting oscillation is.

7. A spring having with a spring constant 1200 N is mounted on a horizontal table as shown in Fig. A mass of 3 kg is attached to the free end of the spring. The mass is then pulled sideways to a distance of 2.0 cm and released.

Determine (i) the frequency of oscillations, (ii) maximum acceleration of the mass, and (iii) the maximum speed of the mass.

Ans.Spring constant, k = 1200 N

Displacement, A = 2.0 cm = 0.02 cm

(i) Frequency of oscillation v, is given by the relation:

Where, T is the time period

Hence, the frequency of oscillations is 3.18 cycles per second.

(ii) Maximum acceleration (a) is given by the relation:

a =

= Angular frequency =

Hence, the maximum acceleration of the mass is 8.0.

(iii) Maximum velocity,

Hence, the maximum velocity of the mass is 0.4 m/s.

8. Answer the following questions:

(a) Time period of a particle in SHM depends on the force constant k and mass m of the particle:

. A simple pendulum executes SHM approximately. Why then is the time period of a pendulum independent of the mass of the pendulum?

(b) The motion of a simple pendulum is approximately simple harmonic for small angle oscillations. For larger angles of oscillation, a more involved analysis shows that T is greater than. Think of a qualitative argument to appreciate this result.

(c) A man with a wristwatch on his hand falls from the top of a tower. Does the watch give correct time during the free fall?

(d) What is the frequency of oscillation of a simple pendulum mounted in a cabin that is freely falling under gravity?

Ans.(a) The time period of a simple pendulum,

For a simple pendulum, k is expressed in terms of mass m, as:

k m

= Constant

Hence, the time period T, of a simple pendulum is independent of the mass of the bob.

(b) In the case of a simple pendulum, the restoring force acting on the bob of the pendulum is given as:

g = Acceleration due to gravity

For small θ, sinθ

For large θ, sinθ is greater than θ.

This decreases the effective value of g.

Hence, the time period increases as:

T=

Where, l is the length of the simple pendulum

(c) The time shown by the wristwatch of a man falling from the top of a tower is not affected by the fall. Since a wristwatch does not work on the principle of a simple pendulum, it is not affected by the acceleration due to gravity during free fall. Its working depends on spring action.

(d) When a simple pendulum mounted in a cabin falls freely under gravity, its acceleration is zero. Hence the frequency of oscillation of this simple pendulum is zero.

5 Marks Questions

1.What is Simple pendulum? Find an expression for the time period and frequency of a simple pendulum?

Ans.A simple pendulum is the most common example of the body executing S.H.M, it consist of heavy point mass body suspended by a weightless inextensible and perfectly flexible string from a rigid support, which is free to oscillate.

Let O is the equilibrium position, OP = X

Let θ = small angle through which the bob is displaced.

The forces acting on the bob are:-

1) The weight = M g acting vertically downwards.

2) The tension = T in string acting along Ps.

Resolving Mg into 2 components as Mg Cos θ and Mg Sin θ,

Restoring force F = - Mg Sin θ

- ve sign shows force is directed towards mean position.

Let θ = Small, so Sin θ ≈ θ =

F = - mg

Now, In S.H.M, F = k x →4) k = Spring constant

Equating equation3) & 4) for F

- k x = - m g

Spring factor = k =

Inertia factor = Mass of bob = m

=

=

2. A particle is in linear simple harmonic motion between two points, A and B, 10 cm apart. Take the direction from A to B as the positive direction and give the signs of velocity, acceleration and force on the particle when it is

(a) at the end A,

(b) at the end B,

(c) at the mid-point of AB going towards A,

(d) at 2 cm away from B going towards A,

(e) at 3 cm away from A going towards B, and

(f) at 4 cm away from B going towards A.

Ans.(a) Zero, Positive, Positive

(b) Zero, Negative, Negative

(c) Negative, Zero, Zero

(d) Negative, Negative, Negative

(e) Zero, Positive, Positive

(f) Negative, Negative, Negative

Explanation:
The given situation is shown in the following figure. Points A and B are the two end points, with AB = 10 cm. O is the midpoint of the path.

A particle is in linear simple harmonic motion between the end points

(a) At the extreme point A, the particle is at rest momentarily. Hence, its velocity is zero at this point.

Its acceleration is positive as it is directed along AO.

Force is also positive in this case as the particle is directed rightward.

(b) At the extreme point B, the particle is at rest momentarily. Hence, its velocity is zero at this point.

Its acceleration is negative as it is directed along B.

Force is also negative in this case as the particle is directed leftward.

The particle is executing a simple harmonic motion. O is the mean position of the particle. Its velocity at the mean position O is the maximum. The value for velocity is negative as the particle is directed leftward. The acceleration and force of a particle executing SHM is zero at the mean position.

The particle is moving toward point O from the end B. This direction of motion is opposite to the conventional positive direction, which is from A to B. Hence, the particle's velocity and acceleration, and the force on it are all negative.

The particle is moving toward point O from the end A. This direction of motion is from A to B, which is the conventional positive direction. Hence, the values for velocity, acceleration, and force are all positive.

This case is similar to the one given in (d).

3. The motion of a particle executing simple harmonic motion is described by the displacement function,

x (t) = A cos ( />t + />).

If the initial (t = 0) position of the particle is 1 cm and its initial velocity is />cm/s, what are its amplitude and initial phase angle? The angular frequency of the particle is π s-1. If instead of the cosine function, we choose the sine function to describe the SHM: x = B sin ( />t + α), what are the amplitude and initial phase of the particle with the above initial conditions.

Ans.Initially, at t = 0:

Initial velocity, v = cm/sec.

Angular frequency, = π rad/

…..(i)

Velocity,

……..(ii)

Squaring and adding equations (i) and (ii), we get:

Dividing equation (ii) by equation (i), we get:

Putting the given values in this equation, we get:

….(iii)

Velocity,

Substituting the given values, we get:

……..(iv)

Squaring and adding equations (iii) and (iv), we get:

Dividing equation (iii) by equation (iv), we get:

4. In Exercise 14.9, let us take the position of mass when the spring is unstreched as x = 0, and the direction from left to right as the positive direction of x-axis. Give x as a function of time t for the oscillating mass if at the moment we start the stopwatch (t = 0), the mass is

(a) at the mean position,

(b) at the maximum stretched position, and

(c) at the maximum compressed position.

In what way do these functions for SHM differ from each other, in frequency, in amplitude or the initial phase?

Ans.(a) x = 2sin 20t

(b) x = 2cos 20t

(c) x = –2cos 20t

The functions have the same frequency and amplitude, but different initial phases.

Distance travelled by the mass sideways, A = 2.0 cm

Force constant of the spring, k = 1200 N

Angular frequency of oscillation:

(a) When the mass is at the mean position, initial phase is 0.

Displacement, x = Asin t

(b) At the maximum stretched position, the mass is toward the extreme right. Hence, the initial phase is.

Displacement,

(c) At the maximum compressed position, the mass is toward the extreme left. Hence, the initial phase is.

Displacement,

The functions have the same frequency and amplitude (2 cm), but different initial phases.

5. Plot the corresponding reference circle for each of the following simple harmonic motions. Indicate the initial (t = 0) position of the particle, the radius of the circle, and the angular speed of the rotating particle. For simplicity, the sense of rotation may be fixed to be anticlockwise in every case: (x is in cm and t is in s).

(a) x = -2 sin (3t + π/3)

(c) x = 3 sin (2πt + π/4)

(d) x = 2 cos πt

Ans.(a)

If this equation is compared with the standard SHM equation, then we get:

Phase angle,

Angular velocity,

The motion of the particle can be plotted as shown in the following figure.

(b)

If this equation is compared with the standard SHM equation, then we get:

Phase angle,

Angular velocity,

The motion of the particle can be plotted as shown in the following figure.

(c)

If this equation is compared with the standard SHM equation, then we get:

Phase angle, = 135°

Angular velocity,

The motion of the particle can be plotted as shown in the following figure.

(d) x = 2 cos πt

If this equation is compared with the standard SHM equation, then we get:

Phase angle, = 0

Angular velocity, = π rad/s

The motion of the particle can be plotted as shown in the following figure.

6. Figure 14.30 (a) shows a spring of force constant k clamped rigidly at one end and a mass m attached to its free end. A force F applied at the free end stretches the spring. Figure 14.30 (b) shows the same spring with both ends free and attached to a mass m at either end. Each end of the spring in Fig. 14.30(b) is stretched by the same force F.

(a) What is the maximum extension of the spring in the two cases?

(b) If the mass in Fig. (a) and the two masses in Fig. (b) are released, what is the period of oscillation in each case?

Ans.(a) For the one block system:

When a force F, is applied to the free end of the spring, an extension l, is produced. For the maximum extension, it can be written as:

Where, k is the spring constant

Hence, the maximum extension produced in the spring,

The displacement (x) produced in this case is:

Net force, F = +2 kx

(b) For the one block system:

For mass (m) of the block, force is written as:

Where, x is the displacement of the block in time t

It is negative because the direction of elastic force is opposite to the direction of displacement.

Where,

is angular frequency of the oscillation

∴Time period of the oscillation,

It is negative because the direction of elastic force is opposite to the direction of displacement.

Angular frequency,

∴Time period,

7. One end of a U-tube containing mercury is connected to a suction pump and the other end to atmosphere. A small pressure difference is maintained between the two columns. Show that, when the suction pump is removed, the column of mercury in the U-tube executes simple harmonic motion.

Ans.Area of cross-section of the U-tube = A

Density of the mercury column =

Acceleration due to gravity = g

Restoring force, F = Weight of the mercury column of a certain height

F = –(Volume /> Density /> g)

F = –(A 2hg) = –2gh = –k Displacement in one of the arms (h)

2h is the height of the mercury column in the two arms

k is a constant, given by

Time period,

m is the mass of the mercury column

Let l be the length of the total mercury in the U-tube.

Mass of mercury, m = Volume of mercury Density of mercury

= Al

Hence, the mercury column executes simple harmonic motion with time period.

8. An air chamber of volume V has a neck area of cross section a into which a ball of mass m just fits and can move up and down without any friction (Fig.14.33). Show that when the ball is pressed down a little and released, it executes SHM. Obtain an expression for the time period of oscillations assuming pressure-volume variations of air to be isothermal [see Fig. 14.33].

Ans.Volume of the air chamber = V

Area of cross-section of the neck = a

The pressure inside the chamber is equal to the atmospheric pressure.

Let the ball be depressed by x units. As a result of this depression, there would be a decrease in the volume and an increase in the pressure inside the chamber.

Decrease in the volume of the air chamber, ΔV = ax

Volumetric strain

Bulk Modulus of air,

In this case, stress is the increase in pressure. The negative sign indicates that pressure increases with a decrease in volume.

The restoring force acting on the ball, F = p a

In simple harmonic motion, the equation for restoring force is:

Where, k is the spring constant

Comparing equations (i) and (ii), we get:

Time period,

9. You are riding in an automobile of mass 3000 kg. Assuming that you are examining the oscillation characteristics of its suspension system. The suspension sags 15 cm when the entire automobile is placed on it. Also, the amplitude of oscillation decreases by 50% during one complete oscillation. Estimate the values of (a) the spring constant k and (b) the damping constant b for the spring and shock absorber system of one wheel, assuming that each wheel supports 750 kg.

Ans.(a) Mass of the automobile, m = 3000 kg

Displacement in the suspension system, x = 15 cm = 0.15 m

There are 4 springs in parallel to the support of the mass of the automobile.

The equation for the restoring force for the system:

Where, k is the spring constant of the suspension system

Time period,

And

Spring constant, k =

(b) Each wheel supports a mass, M = = 750 kg

For damping factor b, the equation for displacement is written as:

The amplitude of oscillation decreases by 50%.

Time period, = 0.7691 s

Therefore, the damping constant of the spring is 1351.58 kg/s.

10. Show that for a particle in linear SHM the average kinetic energy over a period of oscillation equals the average potential energy over the same period.

Ans.The equation of displacement of a particle executing SHM at an instant t is given as:

A= Amplitude of oscillation

= Angular frequency

The velocity of the particle is:

The kinetic energy of the particle is:

The potential energy of the particle is:

For time period T, the average kinetic energy over a single cycle is given as:

……….(i)

And, average potential energy over one cycle is given as:

…….(ii)

It can be inferred from equations (i) and (ii) that the average kinetic energy for a given time period is equal to the average potential energy for the same time period.

11. A body describes simple harmonic motion with amplitude of 5 cm and a period of 0.2 s. Find the acceleration and velocity of the body when the displacement is (a) 5 cm, (b) 3 cm, (c) 0 cm.

Ans.Amplitude, A= 5 cm = 0.05 m

(a) For displacement, x= 5 cm = 0.05 m

a= –

v

When the displacement of the body is 5 cm, its acceleration is
and velocity is 0.

(b) For displacement, x= 3 cm = 0.03 m

a =

v

When the displacement of the body is 3 cm, its acceleration is and velocity is 0.4πm/s.

(c) For displacement, x= 0

a == 0

V=

When the displacement of the body is 0, its acceleration is 0 and velocity is 0.5 πm/s.

  1. The time period of a body executing S.H.M is 1s. After how much time will its displacement be of its amplitude.
    Ans. Soln:y = sin
    Here and T = 1s
  2. A point describes SHM in a line 6 cm long. Its velocity, when passing through the centre of line is 18 cm S. Find the time period.
    Ans. Soln: Here amplitude r = 6/2 = 3cm When y = 0, v = 18 cms -1
    Now
    Or
    We know
  3. Find the period of vibrating particle (SHM), which has acceleration of 45 cms -2 , when displacement from mean position is 5 cm.
    Ans. Soln: Here y = 4 cm and acceleration a = 45 cms -2
    We know

    And
  4. A 40 gm mass produces on extension of 4 cm in a vertical spring. A mass of 200gm is suspended at its bottom and left pulling down. Calculate the frequency of its vibration.
    Ans. Here mg’ = 40g = 40 × 980 dyne: I = 4cm.
    Say K is the force constant of spring, then
    mg = kl or k = mg/

    When the spring is loaded with mass m = 200 g
  5. The acceleration due to gravity on the surface of the moon is 1.7 ms. What is the time period of a simple pendulum on the moon, if its time period on the earth is 3.5 s? (g = 9.8 ms -2 ].
    Ans. Sol: Here on earth, T = 3.5 s g = 9.8 ms -2
    For simple pendulum

    On moon, g’=1.7 ms -2 and if T’ is time period
    then
    Dividing eq(ii) by eq. (i), we get
    or
  6. Calculate the energy possessed by stone of mass 200 g executing S.H.M of amplitude 1 cm and time period 4s.
    Ans.

  7. particle executes S.H.M. of amplitude 25 cm and time period 3s. What is the minimum time required for the particle to move between two points 12.5 cm on either side of the mean position?
    Ans. Given, r = 25 cm: T = 3s: y = 12.5 cm
    The displacement y =
    or or
    The minimum time taken by the particle 2t = 0.5 s
  8. The Vertical motion of a huge piston in a machine is approximately S.H.M with a frequency of 0.5 S". A block of 10kg is placed on the piston. What is the maximum amplitude of the piston's S.H.M. for the block and piston to remain together?
    Ans. Sol: Given, v = 0.5s -1 g = 9.8ms -1

    amax at the extreme position i.e. r = y
    amax = and amax = g to remain in contact.
    Or
  9. A spring balance has a scale that reads from 0 to 50 kg. The length of the scale is 20 cm. A body suspended from this spring, when displaced and released, oscillates with a period of 0.60s. What is the weight of the body?
    Ans. Here m = 50 kg, I = 0.2 m
    We know mg = kl or
    T = 0.60 s and M is the mass of the body, then using

    Weight of body Mg = 22.34 × 9.8 = 218.93 N.
  10. You are riding in an automobile of mass 3000 kg. Assuming that you are examining the oscillation characteristics of its suspension system. The suspension sags 15 cm when the entire automobile is placed on it. Also, the amplitude of oscillation decreases by 50% during one complete oscillation, Estimate the values of (a) the spring constant and (b) the damping constant 'b' for the spring and shock absorber system of one wheel assuming that each wheel supports 750 kg.
    Ans. Soln:
    (a) Here m = 3000 kg. x = 0.15m
    If k is the spring constant of each spring, then spring constant for four spring connected in parallel will be 4k.


    (b) (a) Here m = 3000 kg, x = 0.15m
    If K is the spring constant of each spring, then spring constant for four spring connected in parallel will be 4K.


    (b) As A’ e -bt/2m
    Or or
    But
    Hence
  11. Find the frequency of note emitted (fundamental note) by a string 1m long and stretched by a load of 20 kg, if this string weighs 4.9 g. Given, g = 980 cms -2 ?
    Ans. Solu, L = 100 cm T = 20kg = 20 × 1000 × 980 dyne

    Now the frequency of fundamental note produced,


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