JEE Main · 2022 · 27 Jun 2022 · Shift 2 · Elasticity & SHM

For SHM with amplitude $A=8$ cm, the displacement is:

$x=A\cos (\omega t+\varphi )$

At maximum displacement (say at $t=0$): $x=A$, so $\varphi =0$

At half amplitude: $x=\frac{A}{2}$

$\frac{A}{2}=A\cos (\omega t)$

$\cos (\omega t)=\frac{1}{2}$

$\omega t=\frac{\pi }{3}$ (taking the first occurrence)

Since $\omega =\frac{2\pi }{T}=\frac{2\pi }{6}=\frac{\pi }{3}$ rad/s:

$t=\frac{\pi /3}{\pi /3}=1$ s

Alternatively, the phase change from maximum displacement (phase $0$) to half amplitude (phase $\pi /3$) is $\Delta \varphi =\frac{\pi }{3}$:

$t=\frac{\Delta \varphi }{\omega }=\frac{\pi /3}{\pi /3}=1$ s

The time taken is 1 second.