Q 11-13-061JEE MainJEE Main 2024 (8 Apr, Shift 2)Easy
An object of mass $0.2\ \text{kg}$ executes simple harmonic motion along the $x$ axis with frequency of $\left(\dfrac{25}{\pi}\right)\ \text{Hz}$. At the position $x = 0.04\ \text{m}$ the object has kinetic energy $0.5\ \text{J}$ and potential energy $0.4\ \text{J}$. The amplitude of oscillation is ______ cm.
Numerical value type. Enter your answer.
Answer: 6
$\omega = 2\pi f = 50\ \text{rad s}^{-1}$. Total energy $E = 0.5 + 0.4 = 0.9\ \text{J}$:
$$\frac12m\omega^2A^2 = 0.9 \;\Rightarrow\; A^2 = \frac{2\times0.9}{0.2\times2500} = 0.0036 \;\Rightarrow\; A = 0.06\ \text{m} = 6\ \text{cm}$$
Solution by Sreeraj P, M.Sc Physics