Physics - SHM Question with Solution | TestHub

Given $A=7\text{ mm}=7\times 10^{-3}\text{ m}$ and $v_{\max }=4.4\text{ m/s}$.

The maximum velocity of a particle in S.H.M. is given by $v_{\max }=A\omega$.

Since $\omega =\frac{2\pi }{T}$, we have $v_{\max }=\frac{2\pi A}{T}$.

Rearranging for $T$:

$T=\frac{2\pi A}{v_{\max }}$

Substituting the given values:

$T=\frac{2\times \frac{22}{7}\times 7\times 10^{-3}}{4.4}$

$T=\frac{44\times 10^{-3}}{4.4}$

$T=10\times 10^{-3}\text{ s}=0.01\text{ s}$

Answer: $0.01\mathrm{s}$