NEET 2020 Physics Question 145 Solution

Consider a particle executing simple harmonic motion about the origin. Its displacement as a function of time is given by: $$x(t) = A \sin(\omega t)$$ where $$A$$ is the amplitude and $$\omega$$ is the angular frequency. The initial phase is $$\phi_x = 0$$.

Velocity is the first time derivative of displacement: $$v(t) = \frac{dx}{dt} = A \omega \cos(\omega t) = A \omega \sin\left(\omega t + \frac{\pi}{2}\right)$$ Acceleration is the second time derivative of displacement: $$a(t) = \frac{dv}{dt} = -A \omega^2 \sin(\omega t)$$

Using the trigonometric identity $$-\sin(\theta) = \sin(\theta + \pi)$$: $$a(t) = A \omega^2 \sin(\omega t + \pi)$$ Here, the phase of acceleration is $$\phi_a = \omega t + \pi$$.

The phase difference between acceleration and displacement is: $$\Delta \phi = \phi_a - \phi_x = (\omega t + \pi) - (\omega t) = \pi\text{ rad} = 180^\circ$$ Physically, when displacement is at its maximum positive value ($$+A$$), acceleration is directed toward the center at its maximum negative value ($$-\omega^2 A$$).