Physics - SHM Question with Solution | TestHub

For a particle performing S.H.M. starting from the extreme position, the displacement is given by $x=A\cos (\omega t)$.

The acceleration is the second derivative of displacement with respect to time, $a=\frac{d^{2}x}{d{t}^2}=-A{\omega }^2\cos (\omega t)$.

This can be rewritten as $a=A{\omega }^2\cos (\omega t+\pi )$.

Comparing the phases of displacement and acceleration, the phase difference is $\pi \mathrm{rad}$.

Alternatively, since $a=-{\omega }^{2}x$, acceleration and displacement are always in opposite directions, which corresponds to a phase difference of $\pi \mathrm{rad}$.

Answer: $\pi \mathrm{rad}$