Question

Average velocity of a particle performing $SHM$ in one time period is
A. $Zero$
B. $\dfrac{{A\omega }}{2}$
C. $\dfrac{{A\omega }}{{2\pi }}$
D. $\dfrac{{2A\omega }}{\pi }$

Answer 1

In the question, they’ve asked us for the average velocity of a particle performing SHM in one time period. The question is solely based on one’s understanding of the period of a sin or cos function. First, you need to derive an equation for the velocity of a particle executing SHM. Then, you can solve for average over a time period.

Formula used:
$v = A\omega \cos \left( {\omega t + \theta } \right)$

Complete step-by-step answer:
The equation for a particle performing simple harmonic motion is given by
$x = A\sin \left( {\omega t + \theta } \right)$
Where,
$x$ is the displacement of the particle
$A$ is the amplitude
$\omega$ is the angular velocity
$t$ is the time
If we differentiate the above equation with respect to time on both the sides, we have
\eqalign{ & x = A\sin \left( {\omega t + \theta } \right) \cr & \Rightarrow \dfrac{{dx}}{{dt}} = A\dfrac{{d\left( {\sin \left( {\omega t + \theta } \right)} \right)}}{{dt}} \cr & \Rightarrow \dfrac{{dx}}{{dt}} = A\omega \cos \left( {\omega t + \theta } \right) \cr}
The rate of change of displacement with respect to time is velocity. So, we have the velocity of the particle as
$v = A\omega \cos \left( {\omega t + \theta } \right)$
Where ‘$v$’ is the velocity of the particle
Now, average velocity over one time period will be the average of the cos function over $2\pi$ radians, which is zero.
Thus, the average velocity of the particle executing SHM is zero.

So, the correct answer is “Option A”.

Note: Notice that, we’re representing the displacement of the particle by ‘x’ not ‘y’. The variable ‘x’ indicates the displacement of the particle, while ‘y’ indicates the displacement of the wave. In the solution we’ve considered $2\pi$ radians for one time period as any particle that performs SHM will complete $2\pi$ radians of phase angle in one cycle i.e. one time period.