The maximum potential energy of a block executing simple harmonic motion is $25 \mathrm{~J}$. A is amplitude of oscillation. At $\mathrm{A / 2}$, the kinetic energy of the block is
Solution
$\mathrm{U}_{\max }=\frac{1}{2} \mathrm{~m} \omega^{2} \mathrm{~A}^{2}=25 \mathrm{~J}$
<br/><br/>$\mathrm{KE}$ at $\frac{\mathrm{A}}{2}=\frac{1}{2} m v_{1}^{2}=\frac{1}{2} m \omega^{2}\left(A^{2}-\frac{A^{2}}{4}\right)$
<br/><br/>$=\frac{1}{2} \mathrm{~m} \omega^{2} \frac{3 \mathrm{~A}^{2}}{4}=\frac{3}{4}\left(\frac{1}{2} \mathrm{~m} \omega^{2} \mathrm{~A}^{2}\right)$
<br/><br/>$\mathrm{KE}=\frac{3}{4} \times 25=18.75 \mathrm{~J}$
About this question
Subject: Physics · Chapter: Oscillations · Topic: Simple Harmonic Motion
This question is part of PrepWiser's free JEE Main question bank. 88 more solved questions on Oscillations are available — start with the harder ones if your accuracy is >70%.