Physics - SHM Question with Solution | TestHub

Path length$=16\text{ cm}$
$\therefore$Amplitude$A=8\text{ cm}$

Period$T=2\pi \sqrt{\frac{l}{g}}$
$=2\pi \sqrt{\frac{l}{{\pi }^2}}=\mathrm{2 }\mathrm{s}$
$=2\pi \times \frac{1}{\pi }=2\mathrm{ s}$
Maximum velocity$V_{\text{max}=}=a\omega$
$=A\times \frac{2\pi }{{\rm T}}$
$=8\times \frac{2\pi }{2}$
$=8\mathrm{\pi }\mathrm{ }\mathrm{c}\mathrm{m} {\mathrm{s}}^{-1}$