Question

A particle is executing linear SHM of amplitude $A$ and time period $T$. If $v$ refers to the average speed of the particle during any time interval of $\frac{T}{3}$, then the maximum possible value of $v$ in terms of $A$ is

• A. $\frac{3\sqrt{3}}{T}A$
• B. $\frac{\sqrt{3}A}{T}$
• C. $\frac{2\sqrt{3}}{T}A$
• D. $\frac{3A}{T}$

Answer 1

The correct option is A $\frac{3\sqrt{3}}{T}A$

Let us consider the situation shown in the phasor diagram.


When a particle is moving from $\text{A}$ to $\text{B}$, let the change in phase of the particle be $\theta =\omega t.$
Given, Time interval $t=\frac{T}{3}$
$\Rightarrow \theta =\omega t=\frac{2\pi }{T}\times \frac{T}{3}=\frac{2\pi }{3}$
Which means change in phase of the particle in moving from $\text{A}$ to $\text{B}$ is ${120}^{\circ }$ on the reference circle.
From the phasor diagram, we can deduce that, the maximum average velocity for a time interval of $\frac{T}{3}$ would be from $\text{P}$ to $\text{Q}$.
$\therefore$ Maximum average velocity $v=\frac{\text{Displacement between P and Q}}{\text{Time interval}}$
$\Rightarrow v=\frac{\frac{\sqrt{3}A}{2}+\frac{\sqrt{3}A}{2}}{\left(T/3\right)}=\frac{3\sqrt{3}A}{T}$
Thus, option (a) is the correct answer.

Alternate solution:
Let us suppose that the particle starts at the mean position. The equation of SHM is given by
$x=A\sin \omega t......\left(1\right)$
Differentiating $\left(1\right)$ with respect to time, we get
$v=A\omega \cos \omega t$
Since the maximum value of speed is at $x=0$, the maximum average velocity for a time interval $\frac{T}{3}$ would be from $-\frac{T}{6}$ to $\frac{T}{6}$.
Average velocity of particle in a time interval $t$ is given by
$v_{av}=\frac{{\int }_0^{t}vdt}{{\int }_0^{t}dt}$
$\Rightarrow v_{av}=\frac{3}{T}{\int }_{-T/6}^{T/6}A\omega cos\omega tdt$
$\Rightarrow v_{av}=\frac{3A}{T}\left(2\sin \left(\frac{\omega T}{6}\right)\right)$
Since $\omega T=2\pi$,
$\Rightarrow v_{av}=\frac{3\sqrt{3}A}{T}$
Thus, option (a) is the correct answer.