A particle is performing S.H.M. about its mean position with an amplitude 'a' and periodic time ' T '. The…
A particle is performing S.H.M. about its mean position with an amplitude 'a' and periodic time ' T '. The speed of the particle when. its displacement from mean position' is $\frac{\mathrm{a}}{3}$ will be
$\frac{2 \pi \mathrm{a}}{\mathrm{T}}$
$\frac{4 \sqrt{2} \pi \mathrm{a}}{3 \mathrm{~T}}$
$\frac{4 \pi^2 a}{3 T}$
$\frac{\sqrt{3} \pi^2 a}{2 T}$
Solution
The speed of a particle performing SHM at displacement ' $x$ ' from the mean position is given by, $V=\omega \sqrt{a^2-x^2}$
$\begin{aligned}
& V=\omega \sqrt{a^2-\left(\frac{a}{3}\right)^2} \\
& V=\omega \sqrt{\frac{8 a^2}{9}} \\
& V=\frac{2 \pi}{T} \times \frac{2 \sqrt{2}}{3} \times a \\
& V=\frac{4 \sqrt{2} \pi a}{3 T}
\end{aligned}$