Q 11-13-013NEETJEE MainEasy
The maximum speed and maximum acceleration of a particle in SHM are $4$ m/s and $16\ \text{m/s}^2$. Its amplitude is
Answer: (C) $1$ m
Concept. $v_{\max} = A\omega$ and $a_{\max} = A\omega^2$, so $\omega = \dfrac{a_{\max}}{v_{\max}}$.
Step 1. Angular frequency. $\omega = \dfrac{16}{4} = 4$ rad/s.
Step 2. Amplitude. $A = \dfrac{4}{4} = 1$ m.
Answer. $1$ m (Option C).
Solution by Sreeraj P, M.Sc Physics