Physics - SHM Question with Solution | TestHub

The velocity of a particle in simple harmonic motion at a distance $x$ from the mean position is given by

$v=\omega \sqrt{a^2-x^2}$

where $\omega$ is the angular frequency and $a$ is the amplitude.

Given $x=\frac{a}{3}$ and $\omega =\frac{2\pi }{T}$, substituting these values gives

$v=\frac{2\pi }{T}\sqrt{a^2-{\left(\frac{a}{3}\right)}^2}$

$v=\frac{2\pi }{T}\sqrt{a^2-\frac{a^2}{9}}$

$v=\frac{2\pi }{T}\sqrt{\frac{8a^2}{9}}$

$v=\frac{2\pi }{T}\left(\frac{2\sqrt{2}a}{3}\right)$

$v=\frac{4\sqrt{2}\pi a}{3T}$

Answer: $\frac{4\sqrt{2}\pi a}{3T}$