Physics - SHM Question with Solution | TestHub

The kinetic energy of a particle executing simple harmonic motion is maximum at the mean position, which is given by:

$K_{\text{max}}=\frac{1}{2}m{\omega }^2{A}^2$

Substituting the given values $K_{\text{max}}=16\text{ J}$, $m=5.12\text{ kg}$, and $A=25\text{ cm}=\frac{1}{4}\text{ m}$:

$16=\frac{1}{2}\times 5.12\times {\omega }^2\times {\left(\frac{1}{4}\right)}^2$

$16=\frac{1}{2}\times 5.12\times {\omega }^2\times \frac{1}{16}$

$16=\frac{5.12}{32}{\omega }^2$

$16=0.16{\omega }^2$

${\omega }^2=\frac{16}{0.16}=100$

$\omega =10\text{ rad/s}$

The period of oscillation is given by $T=\frac{2\pi }{\omega }$:

$T=\frac{2\pi }{10}=\frac{\pi }{5}\text{ s}$

Answer: $\frac{\pi }{5}\mathrm{s}$