Question

The particle is executing $\text{SHM}$ on a line $4\text{cm}$ long. If its velocity at mean position is $12\text{cm/s}$, its frequency in $\text{Hertz}$ will be :

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

Answer 1

The correct option is D $\frac{3}{\pi }$

Let $A$ be the amplitude of $\text{SHM}$.
Length of $\text{SHM}$ path $\left(2A\right)$ is $4\text{cm}$
$\Rightarrow 2A=4\Rightarrow A=2\text{cm}$
From formula
$v_{mean}=A\omega \Rightarrow 12=2\left(2\pi f\right)$
$\Rightarrow f=\frac{12}{4\pi }\Rightarrow f=\frac{3}{\pi }\text{Hz}$.
Hence, option $\left(d\right)$ is the correct answer.