Question

The velocity of a particle executing a simple harmonic motion is $13m{s}^{-1}$, when its distance from the equilibrium position $\left(Q\right)$ is $3m$ and its velocity is $12m{s}^{-1}$, when it is $5m$ away from $Q$. The frequency of the simple harmonic motion is:

• A. $\frac{5\pi }{8}$
• B. $\frac{5}{8\pi }$
• C. $\frac{8\pi }{5}$
• D. $\frac{8}{5\pi }$

Answer 1

The correct option is B $\frac{5}{8\pi }$

For SHM,

$v=\omega \sqrt{A^2-x^2}$

Given:

$13=\omega \sqrt{A^2-3^2}..........\left(i\right)$

$12=\omega \sqrt{A^2-5^2)}.............\left(ii\right)$

Eliminating $A$, we get:

$\frac{{13}^2}{{\omega }^2}+3^2=\frac{{12}^2}{{\omega }^2}+5^2$
$\omega =\frac{5}{4}$

$f=\frac{5}{8\pi }Hz$