Q 11-13-139JEE MainJEE Main 2019 (11 Jan, Shift 1)Easy
A particle undergoing simple harmonic motion has time dependent displacement given by $x(t) = A\sin\dfrac{\pi t}{90}$. The ratio of kinetic to potential energy of this particle at $t = 210$ s will be
Answer: (D) $\dfrac13$
At $t = 210$ s the phase is $\dfrac{210\pi}{90} = \dfrac{7\pi}{3}$, so $\sin\theta = \dfrac{\sqrt3}{2}$ and $\cos\theta = \dfrac12$.
$$\frac{K}{U} = \frac{\cos^2\theta}{\sin^2\theta} = \frac{1/4}{3/4} = \frac13$$
Solution by Sreeraj P, M.Sc Physics