Question

8. A simple harmonic oscillator has an amplitude A and time period T. The time required by it to travel x =√3/2A to x = A is​

Answer 1

Let the displacement of the simple harmonic oscillator be given by,

$\sf{\longrightarrow x=A\sin\left(\dfrac{2\pi}{T}\,t+\phi\right)}$

Let $\sf{t_1}$ and $\sf{t_2}$ be the time taken by the oscillator to travel by a displacement of $\sf{x=\dfrac{\sqrt3}{2}\,A}$ and $\sf{x=A}$ respectively from mean position.

Then we get,

$\sf{\longrightarrow \dfrac{\sqrt3}{2}\,A=A\sin\left(\dfrac{2\pi}{T}\,t_1+\phi\right)}$

$\sf{\longrightarrow\sin\left(\dfrac{2\pi}{T}\,t_1+\phi\right)=\dfrac{\sqrt3}{2}}$

$\sf{\longrightarrow\dfrac{2\pi}{T}\,t_1+\phi=\dfrac{\pi}{3}\quad\quad\dots(1)}$

And,

$\sf{\longrightarrow A=A\sin\left(\dfrac{2\pi}{T}\,t_2+\phi\right)}$

$\sf{\longrightarrow \sin\left(\dfrac{2\pi}{T}\,t_2+\phi\right)=1}$

$\sf{\longrightarrow\dfrac{2\pi}{T}\,t_2+\phi=\dfrac{\pi}{2}\quad\quad\dots(2)}$

Subtracting (1) from (2),

$\sf{\longrightarrow\dfrac{2\pi}{T}\left(t_2-t_1\right)=\dfrac{\pi}{2}-\dfrac{\pi}{3}}$

$\sf{\longrightarrow\dfrac{2\pi}{T}\left(t_2-t_1\right)=\dfrac{\pi}{6}}$

$\sf{\longrightarrow\underline{\underline{t_2-t_1=\dfrac{T}{12}}}}$

This is the time taken by the oscillator to travel from $\sf{x=\dfrac{\sqrt3}{2}\,A}$ to $\sf{x=A.}$