Question

A particle executes SHM with time period T and amplitude A. The maximum possible average velocity in time ${\displaystyle \frac{\text{T}}{\text{4}}}$ is:
A. ${\displaystyle \frac{\text{2A}}{\text{T}}}$
B. ${\displaystyle \frac{\text{4A}}{\text{T}}}$
C. ${\displaystyle \frac{\text{6A}}{\text{T}}}$
D. ${\displaystyle \frac{\text{4}\sqrt{2}\text{A}}{\text{T}}}$

Answer 1

Hint: To solve this question, we need to use the basic theory of simple harmonic motion (SHM). First suppose, SHM is represented by x=A sin wt and then we directly use some basic formula so that we will get the answer.

Formula used- The maximum possible average velocity = ${\displaystyle \frac{\text{maximum displacement}}{\text{time}}}$

Complete Step-by-Step solution:
As we know, in mechanics and physics, simple harmonic motion is a special type of periodic motion where the restoring force on the moving object is directly proportional to, and opposite of, the object's displacement vector.
Now, in this case
We have, time = ${\displaystyle \frac{\text{T}}{\text{4}}}$
The maximum possible average velocity = ${\displaystyle \frac{\text{maximum displacement}}{\text{time}}}$
That means, ${\text{V}}_{\text{avg}}$= ${\displaystyle \frac{{\text{d}}_{\text{max}}}{{\displaystyle \frac{\text{T}}{\text{4}}}}}$

${\text{V}}_{\text{avg}}$= ${\displaystyle \frac{\text{4}{\text{d}}_{\text{max}}}{\text{T}}}$

here T→2π
${\displaystyle \frac{\text{T}}{\text{4}}}$$\to$${\displaystyle \frac{\pi }{\text{2}}}$
∴θ = ${\displaystyle \frac{\pi }{\text{4}}}$
∴ ${\text{d}}_{\text{max}}$=PQ=2Asinθ
= 2a $\text{sin}\left({\displaystyle \frac{\pi }{\text{4}}}\right)$
∴ ${\text{d}}_{\text{max}}$= ${\displaystyle \frac{\text{2A}}{\sqrt{\text{2}}}}$ = $\sqrt{\text{2}}\text{A}$
Now, as we know:
${\text{V}}_{\text{avg}}$= ${\displaystyle \frac{{\text{d}}_{\text{max}}}{{\displaystyle \frac{\text{T}}{\text{4}}}}}$
        = ${\displaystyle \frac{\sqrt{\text{2}}\text{A}}{{\displaystyle \frac{\text{T}}{\text{4}}}}}$
       = ${\displaystyle \frac{\text{4}\sqrt{\text{2}}\text{A}}{\text{T}}}$
Therefore, a particle executes SHM with time period T and amplitude A. The maximum possible average velocity in time ${\displaystyle \frac{\text{T}}{\text{4}}}$ is ${\displaystyle \frac{\text{4}\sqrt{\text{2}}\text{A}}{\text{T}}}$.
So, option (D) is the correct answer.

Note- An oscillation follows simple harmonic motion if it fulfils the following two rules: Acceleration is always in the opposite direction to the displacement from the equilibrium position. Acceleration is proportional to the displacement from the equilibrium position.