STD 11 - 13. oscillations — NEET STD 11 Science — Question

Time period is $12 \,s$ from diagram.

$\omega=\frac{2 \pi}{12}=\frac{\pi}{6}$

Amplitude $A=4$

Initial phase is determined by putting known values in the equation.

$2=4 \sin \left(\frac{\pi}{6} t+\phi\right)$

$\sin ^{-1} \frac{1}{2}=\phi[t=0]$

$\frac{\pi}{6}=\phi$

Hence equation is $x=\left(\frac{\pi}{6} t+\frac{\pi}{6}\right)$