Class 11 Physics Oscillations | SHM and Uniform Circular Motion |

__SHM & Uniform Circular Motion__

Uniform Circular motion can be interpreted as a SHM.

To explain above statement:- Consider a ball tied to a thread and moving in a circular path such a way it appears to be in circular motion for a person who is observing it from top view or who is standing in the same plane as we are. But it appears to be SHM if somebody is standing in the same plane of motion.

Case 1: It is executing circular motion.

Case 2: Circular motion observed by a person when standing on the same plane.

Case 3: It is appears to be SHM for a person who is standing in the same line of sight.

Mathematically:-

Consider any particle moving in a circular path whose radius is R

Angular velocity = w

Angular position =∫θ dt

=wt + φ

Consider the projection of particle on x-axis be P’.

Displacement = Acosθ

**x=A cos(wt+ φ)**

the above equation same as the equation of simple harmonic motion

As the particle is moving in the same way the projections are also moving.

- When the particle is moving in the upper part of circle then the projections start moving towards left.
- When the particle is moving in the lower part of the circle then the projections are moving towards right.
- We can conclude that the particle is swinging from left to right and again from right to left.
- This to and fro motion is SHM.

In the above figure we can see that a reference point P ' moving with uniform circular motion in a reference circle of radius A. Its projection P on the x-axis executes simple harmonic motion.

__Conclusion: -__

SHM is the projection of uniform circular motion on the diameter of the circle in which the SHM takes place.

** Problem:-**In the given figuresit corresponds 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.

** Answer**: - (a) Time period, t = 2 s

Amplitude, A = 3 cm

At time, t = 0, the radius vector OP makes an angle π/2 with the positive x-axis, phase angle Φ = +π/2

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

A= cos [(2 πt/T) + Φ]

=3cos (2 πt/2 + π/2) = -3sin (2πt/2)

=-3sinπt cm.

(b) Time Period, t = 4 s

Amplitude, a = 2 m

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 the time t, is given as__:__

=a cos [(2 πt/T) + Φ]

=2 cos [(2 πt/T) + π]

x=-2 cos (π/2 t) m

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