A point performs simple harmonic oscillation of period $T$ and the equation of motion is given by $x

Question

A point performs simple harmonic oscillation of period $T$ and the equation of motion is given by $x=a \sin (\omega t+\pi / 6)$. After the elapse of what fraction of the time period, the velocity of the point will be equal to half of its maximum velocity?\

Professor PC Thomas Classes · Answer

(b) : $x=a \sin (\omega t+\pi / 6)$ $$ \frac{d x}{d t}=a \omega \cos (\omega t+\pi / 6) $$ Max. velocity $=a \omega$\ $	herefore \quad \frac{a \omega}{2}=a \omega \cos (\omega t+\pi / 6) \quad$ or, $\cos (\omega t+\pi / 6)=\frac{1}{2}$\ $\Rightarrow \frac{2 \pi}{6}=\frac{2 \pi}{T} \cdot t+\frac{\pi}{6} \Rightarrow \frac{2 \pi}{T} \cdot t=\frac{2 \pi}{6}-\frac{\pi}{6}=+\frac{\pi}{6}$\ $	herefore \quad t=+\frac{\pi}{6} 	imes \frac{T}{2 \pi}=\left|+\frac{T}{12}ight|$\