Physics - SHM Question with Solution | TestHub

At the equilibrium position, the spring force balances the weight of the mass:

$k{x}_0=mg\Rightarrow x_0=\frac{mg}{k}$

Since the spring is unstretched at the highest point of oscillation, the amplitude of the simple harmonic motion is equal to the equilibrium extension:

$A=x_0=\frac{mg}{k}$

The angular frequency of the system is $\omega =\sqrt{\frac{k}{m}}$, which gives $\frac{m}{k}=\frac{1}{{\omega }^2}$.

Therefore, the amplitude can be written as:

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

The maximum speed of the mass in SHM is:

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

Given the frequency $f=10$ Hz, the angular frequency is:

$\omega =2\pi f=20\pi$ rad/s

Substituting $g=10$ m/s${}^2$ and $\omega =20\pi$ rad/s:

$v_{\text{max}}=\frac{10}{20\pi }=\frac{1}{2\pi }$ m/s

Answer: $\left(\frac{1}{2\pi }\right)$ m/s