Physics - SHM Question with Solution | TestHub

Given $T=0.1$ s and $g=10$ m/s${}^2$.

The angular frequency of the S.H.M. is $\omega =\frac{2\pi }{T}=\frac{2\pi }{0.1}=20\pi$ rad/s.

At the highest point of oscillation, the spring is unstretched. The only force acting on the mass is gravity, so its acceleration is $g$.

In S.H.M., the maximum acceleration at the extreme position is $a_{max}={\omega }^{2}A$.

Equating the two, we get ${\omega }^{2}A=g\Rightarrow A=\frac{g}{{\omega }^2}$.

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

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

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

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