Q 11-13-150JEE MainJEE Main 2017 (8 Apr)Easy
The ratio of maximum acceleration to maximum velocity in a simple harmonic motion is $10\ \text{s}^{-1}$. At $t = 0$ the displacement is $5$ m. What is the maximum acceleration? The initial phase is $\dfrac\pi4$.
Answer: (D) $500\sqrt2\ \text{m s}^{-2}$
$$\frac{a_{\max}}{v_{\max}} = \frac{\omega^2A}{\omega A} = \omega = 10\ \text{s}^{-1}$$
With $x = A\sin(\omega t + \pi/4)$, at $t = 0$: $5 = A\sin\dfrac\pi4 = \dfrac{A}{\sqrt2}$, so $A = 5\sqrt2$ m.
$$a_{\max} = \omega^2A = 100\times5\sqrt2 = 500\sqrt2\ \text{m s}^{-2}$$
Solution by Sreeraj P, M.Sc Physics