Q 11-13-005NEETNEET 2024Top questionEasy
If $x = 5\sin\left(\pi t + \dfrac{\pi}{3}\right)$ m represents the motion of a particle executing simple harmonic motion, the amplitude and time period of motion, respectively, are :
Answer: (D) $5$ m, $2$ s
Compare with $x = A\sin(\omega t + \phi)$: $A = 5$ m and $\omega = \pi$ rad/s.
$$T = \frac{2\pi}{\omega} = \frac{2\pi}{\pi} = 2\ \text{s}$$
Solution by Sreeraj P, M.Sc Physics