Question

The maximum speed of a particle executing SHM is $1{ms}^{-1}$ and maximum acceleration is $1.57{ms}^{-2}$. Its frequency is:

• A. $0.25s^{-1}$
• B. $2s^{-1}$
• C. $1.57s^{-1}$
• D. $2.57s^{-1}$

Answer 1

The correct option is A $0.25s^{-1}$

Let the angular frequency of the oscillation be $\omega$

Maximum acceleration $a={\omega }^{2}A$ where $A$ is the amplitude of oscillation

Maximum velocity $v=\omega A$

$⟹$ $\omega =\frac{a}{v}$

$\therefore$ $\omega =\frac{1.57}{1}=1.57$ $rad/s$

$\therefore$ Frequency of oscillation $\nu =\frac{\omega }{2\pi }=\frac{1.57}{2\times 3.14}=0.25$ $s^{-1}$