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
Concept. Compare with $x = A\sin(\omega t + \phi)$; $T = \dfrac{2\pi}{\omega}$.
Step 1. Read off. $A = 5$ m, $\omega = \pi$ rad/s.
Step 2. Period. $T = \dfrac{2\pi}{\pi} = 2$ s.
Answer. $5$ m, $2$ s (Option D).
Solution by Sreeraj P, M.Sc Physics