Question

The maximum velocity in SHM is $v_m$ . The average velocity during motion from one extreme point to the other extreme point is
A. ${\displaystyle \frac{\pi }{2}}{v}_m$
B. ${\displaystyle \frac{2}{\pi }}{v}_m$
C. ${\displaystyle \frac{4}{\pi }}{v}_m$
D. ${\displaystyle \frac{\pi }{4}}{v}_m$

Answer 1

Hint: In this question we shall use the fundamentals of SHM. The maximum velocity is given by $v_m=A\omega$ and the time period T is related to angular frequency by the relation $\omega ={\displaystyle \frac{2\pi }{T}}$. So, we shall express the maximum velocity in terms of time period using both the expressions. Further we will calculate the average velocity given as $v_{avg}={\displaystyle \frac{s}{t}}$ . If we consider the journey of the SHM from one extreme point to another, the total distance will be equal to twice the amplitude of SHM and the total time taken would be half the time period of the SHM. Using these, we will calculate the value of the average velocity.

Complete step by step answer:
The maximum velocity is given by $v_m=A\omega$, where $A$ is the amplitude of the SHM and $\omega$ is the angular frequency.
The time period $T$ is related to angular frequency by the relation,
$\omega ={\displaystyle \frac{2\pi }{T}}$
Substituting in the equation for the maximum velocity we have,
$v_m={\displaystyle \frac{2\pi A}{T}}$
This can be rewritten as
${\displaystyle \frac{A}{T}}={\displaystyle \frac{v_m}{2\pi }}........(1)$
The average velocity is defined for the complete journey and is given as,
$v_{avg}={\displaystyle \frac{s}{t}}$
where $s$ is the total displacement and T is the total time taken.

If we consider the journey of the SHM from one extreme point to another, the total distance will be equal to twice the amplitude of SHM.Hence, we can say that $s=2A$ where $A$ is the amplitude of the SHM.Total time taken would be half the time period of the SHM.Hence, we can say that $t={\displaystyle \frac{T}{2}}$ where $T$ is the time period of the SHM. Substituting in the formula of the average velocity we get,
$v_{avg}={\displaystyle \frac{2A}{{\displaystyle \frac{T}{2}}}}$
$\Rightarrow v_{avg}={\displaystyle \frac{4A}{T}}$
Using the first relation we have,
$v_{avg}=4\times {\displaystyle \frac{v_m}{2\pi }}$
$\therefore v_{avg}={\displaystyle \frac{2v_m}{\pi }}$

Hence option B is the correct answer.

Note:The time period of the SHM is defined as the time taken by the body in SHM to come back to its position from where it started. Hence, the journey from one extreme point to another would take time equal to half the time period of the SHM as in this question.Also, the time period does not depend on the initial point of motion. Be it the extreme points or the mean point, it always remains the same.