Q.Define simple harmonic motion. Show that the motion of (point) projection of a particle performing uniform circular motion on any diameter is simple harmonic.
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Start your 14-day free trial to unlock the full solution →SHM is a to-and-fro motion where acceleration a = -(omega^2)x. The projection (foot of perpendicular) of a particle in uniform circular motion onto a diameter executes x = A cos(omega t), giving a = -(omega^2)x, proving it is SHM.
Definition of SHM: Simple harmonic motion is the oscillatory (to-and-fro) motion of a body in which the acceleration (or restoring force) is directly proportional to the displacement from a fixed mean position and is always directed towards that mean position.
Mathematically: a = -(omega^2) x, where omega is the angular frequency and the minus sign shows the direction is opposite to displacement.
Projection of uniform circular motion:
Consider a particle P moving with uniform speed on a circle of radius A, centre O, with constant angular velocity omega. Let it start on the reference (x) axis and rotate anticlockwise. After time t, the radius OP makes angle theta = omega t with the x-axis.
Let N be the foot of the perpendicular from P onto a diameter (say the x-axis). As P goes round the circle, N moves back and forth along the diameter between the two ends. N is the projection of P.
The displacement of N from the centre O is:
x = A cos(theta) = A cos(omega t)
Velocity of N (differentiate x with respect to t):
v = dx/dt = - A omega sin(omega t)
Acceleration of N (differentiate v):
a = dv/dt = - A omega^2 cos(omega t)
But A cos(omega t) = x, so:
a = - omega^2 x
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