A simple harmonic oscillator has an amplitude A and time period $T$. The time required by it to trav

Question

A simple harmonic oscillator has an amplitude A and time period $T$. The time required by it to travel from $x=A$ to $x=A / 2$ is\

Professor PC Thomas Classes · Answer

(a) : For S.H.M., $x=A \sin \left(\frac{2 \pi}{T} tight)$\ when $x=A, A=A \sin \left(\frac{2 \pi}{T} tight)$\ $	herefore \quad \sin \left(\frac{2 \pi}{T} \cdot tight)=1 \Rightarrow \sin \left(\frac{2 \pi}{T} \cdot tight)=\sin \left(\frac{\pi}{2}ight)$\ $\Rightarrow \quad t=(T / 4)$\ When $x=\frac{A}{2}, \frac{A}{2}=A \sin \left(\frac{2 \pi}{T} \cdot tight)$\ or $\sin \frac{\pi}{6}=\sin \left(\frac{2 \pi}{T} tight)$ or $t=(T / 12)$\ Now, time taken to travel from $x=A$ to $x=A / 2$ $$ =T / 4-T / 12=T / 6 $$