Question

The Kinetic energy of a particle in SHM is E when the particle passes through the mean position. If its mass is m and amplitude A, the time period is:

Answer 1

In case of linear SHM,

Kinetic energy at mean position is maximum. So,

$E=\frac{m}{2}(vmax{)}^2$

Also, $v_{max}=A\omega$. So,

$E=\frac{m}{2}(A\omega {)}^2$

$\frac{2E}{m{A}^2}={\omega }^2$

$\Rightarrow \omega =A\sqrt{\frac{2E}{m}}$

As $\omega =2\pi f$

$\Rightarrow \omega =\frac{2\pi }{T}$. So,

$\Rightarrow \frac{2\pi }{T}=\left(\sqrt{\frac{2E}{m}}\right)\frac{1}{A}$

$\Rightarrow \sqrt{2}\frac{\left(\pi A\right)}{\sqrt{m}}(E{)}^{1/2}=T$

$T={\left(\frac{2E}{m}\right)}^{1/2}\pi A$.