Q.A simple pendulum performs S.H.M of period 4 seconds. How much time after crossing the mean position, will the displacement of the bob be one third of its amplitude?

[!TLDR] $\omega=2\pi/T=\pi/2$ rad/s. Setting $A/3=A\sin(\omega t)$ gives $\omega t=\sin^{-1}(1/3)\approx0.3398$ rad, s …

With period $T=4$ s, angular frequency is $\omega=\frac{2\pi}{T}=\frac{2\pi}{4}=\frac{\pi}{2}\approx1.5708$ rad/s. Since the bob crosses the mean position at t = 0, its displacement is $x=A\sin(\omega t)$. We want the time at which $x=\frac{A}{3}$: $\frac{A}{3}=A\sin(\omega t)\Rightarrow\sin(\omega t)=\frac13$. So $\omega t= …

Write $x=A\sin(\omega t)$ (starts at mean position), set $x=A/3$, solve for $\omega t=\sin^{-1} …

Working $\sin^{-1}(1/3)$ in degrees but then dividing by $\omega$ in rad/s without c …