Show that S.H.M. is a Projection of U.C.M. on Any Diameter - Physics

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Show that S.H.M. is a projection of  U.C.M. on any diameter



Linear S.H.M. is defined as the linear periodic motion of a body, in which the restoring force (or acceleration) is always directed towards the mean position and its magnitude is directly proportional to the displacement from the mean position.
There is basic relation between S.H.M. and U.C.M. that is very useful in understanding S.H.M. For an object performing U.C.M. the projection of its motion along any diameter of its path executes S.H.M.
Consider particle ‘P’ is moving along the circumference of circle of radius ‘a’ with constant angular speed of ω in anticlockwise direction as shown in figure.
Particle P along circumference of circle has its projection particle on diameter AB at point M. Particle P is called reference particle and the circle on which it moves, its  projection moves back and forth along the horizontal diameter, AB.

The x–component of the displacement of P is always same as displacement of M, the x–component of the velocity of P is always same as velocity of M and the x–component of the acceleration of M.
Suppose that particle P starts from initial position with initial phase α (angle between radius OP and the x – axis at the time t = 0) In time t the angle between OP and x - axis is ( ωt + α ) as particle P moving with constant angular velocity (ω) as shown in figure.
cos( ωt + α ) = `x/a`

`therefore x=acos(omegat+alpha)`                       ......(1)

This is the expression for displacement of particle M at time t.
As velocity of the particle is the time rate of change of displacement then we have


`therefore v=-aomegasin(omegat+alpha)`            ......(2)

As acceleration of particle is the time rate of change of velocity, we have

`a=(dv)/dt=d/dt [-aomegasin(omegat+alpha)]`


`therefore a= omega^2x`

It shows that acceleration of particle M is directly proportional to its displacement and its direction is opposite to that of displament. Thus particle M performs simple harmonic motion but M is projection of particle performing U.C.M. hence S.H.M. is projection of U.C.M. along a diameter, of circle.

Concept: Linear Simple Harmonic Motion (S.H.M.)
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Chapter 5: Oscillations - Exercises [Page 129]


Balbharati Physics 12th Standard HSC Maharashtra State Board
Chapter 5 Oscillations
Exercises | Q 4 | Page 129


Define linear simple harmonic motion.

Choose the correct option:

A body of mass 1 kg is performing linear S.H.M. Its displacement x (cm) at t(second) is given by x = 6 sin `(100t + π/4)`. Maximum kinetic energy of the body is ______.

Two parallel S.H.M.s represented by `"x"_1 = 5 sin(4π"t" + π/3)` cm and `"x"_2 = 3sin (4π"t" + π/4)` cm are superposed on a particle. Determine the amplitude and epoch of the resultant S.H.M.

What does the phase of π/2 indicate in linear S.H.M.? 

Define linear S.H.M.

At extreme positions of a particle executing simple harmonic motion, ______

A particle is performing a linear simple harmonic motion of amplitude 'A'. When it is midway between its mean and extreme position, the magnitudes of its velocity and acceleration are equal. What is the periodic time of the motion?

A particle executes simple harmonic motion and is located at x = a, b, and c at times t0, 2t0, and 3t0 respectively. The frequency of the oscillation is ______.

For a particle executing simple harmonic motion, which of the following statements is NOT correct?

The velocities of a particle performing linear S.H.M are 0.13 m/s and 0.12 m/s, when it is at 0.12 m and 0.13 m from the mean position respectively. If the body starts from mean position, the equation of motion is ____________.

The equation of a particle executing simple harmonic motion is given by x = sin π `("t" + 1/3)` m. At t = 1s, the speed of particle will be ______. (Given π = 3.14)

Two simple harmonic motions are represented by the equations y1 = 0.1 sin `(100pi"t"+pi/3)` and y1 = 0.1 cos πt.

The phase difference of the velocity of particle 1 with respect to the velocity of particle 2 is ______.

A particle is executing simple harmonic motion with amplitude A. When the ratio of its kinetic energy to the potential energy is `1/4`, its displacement from its mean position is ______.

For a particle executing SHM the displacement x is given by x = A cos ωt. Identify the graph which represents the variation of potential energy (P.E.) as a function of time t and displacement x.


A light rod of length 2m suspended from the ceiling horizontally by means of two vertical wires of equal length. A weight W is hung from a light rod as shown in figure.

The rod hung by means of a steel wire of cross-sectional area A1 = 0.1 cm2 and brass wire of cross-sectional area A2 = 0.2 cm2. To have equal stress in both wires, T1/T2 = ______.

What do you know about restoring force?

The displacement of a particle performing simple harmonic motion is `1/3` rd of its amplitude. What fraction of total energy will be its kinetic energy?


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