Q 11-13-011NEETJEE MainEasy
A particle executes SHM given by $x = 0.05\sin\left(4\pi t + \dfrac{\pi}{6}\right)$ m. Its amplitude, frequency and initial phase are
Answer: (A) $5$ cm, $2$ Hz, $\dfrac{\pi}{6}$
Concept. Compare with $x = A\sin(\omega t + \phi)$; $f = \dfrac{\omega}{2\pi}$.
Step 1. Read off. $A = 0.05$ m $= 5$ cm, $\omega = 4\pi$ rad/s, $\phi = \dfrac{\pi}{6}$.
Step 2. Frequency. $f = 2$ Hz.
Answer. $5$ cm, $2$ Hz, $\dfrac{\pi}{6}$ (Option A).
Solution by Sreeraj P, M.Sc Physics