Q 11-13-143JEE MainJEE Main 2019 (12 Jan, Shift 2)Easy
A simple harmonic motion is represented by: $$y = 5\left(\sin3\pi t + \sqrt3\cos3\pi t\right)\ \text{cm}$$ The amplitude and time period of the motion are:
Answer: (B) $10$ cm, $\dfrac23$ s
$\sin\theta + \sqrt3\cos\theta = 2\sin\left(\theta + \dfrac\pi3\right)$, so
$$y = 10\sin\left(3\pi t + \frac\pi3\right)\ \text{cm}$$
Amplitude $10$ cm; $\omega = 3\pi$ gives $T = \dfrac{2\pi}{3\pi} = \dfrac23$ s.
Solution by Sreeraj P, M.Sc Physics