Physics - SHM Question with Solution | TestHub

Frequency of the particle, $f=10\text{ Hz}$

Angular frequency, $\omega =2\pi f=20\pi \text{ rad/s}$

At the equilibrium position, the extension in the spring is $x_0=\frac{mg}{k}=\frac{g}{{\omega }^2}$

Since the spring is unstretched at the highest point of oscillation, the amplitude $A$ of the SHM is equal to the equilibrium extension $x_0$.

$A=\frac{g}{{\omega }^2}$

Maximum speed of the particle is given by $v_{\text{max}}=A\omega$

$v_{\text{max}}=\left(\frac{g}{{\omega }^2}\right)\omega =\frac{g}{\omega }$

Substituting the given values, $v_{\text{max}}=\frac{10}{20\pi }=\frac{1}{2\pi }\text{ m/s}$

Answer: $\frac{1}{2\pi }\mathrm{m}/\mathrm{s}$