Physics - SHM Question with Solution | TestHub

Angular frequency of the SHM is $\omega =2\pi f=2\pi \times 5=10\pi$ 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 of the SHM is equal to the extension at the equilibrium position. Thus, $A=x_0=\frac{g}{{\omega }^2}$.

The maximum speed of the mass is given by $v_{\text{max}}=A\omega$.

Substituting the value of $A$, we get $v_{\text{max}}=\left(\frac{g}{{\omega }^2}\right)\omega =\frac{g}{\omega }$.

Using $g=10$ m/s${}^2$ and $\omega =10\pi$ rad/s, we get $v_{\text{max}}=\frac{10}{10\pi }=\frac{1}{\pi }$ m/s.

Answer: $1/\pi$ m/s