Q84 / 90 · JEE Main 2024 (08 Apr Shift 2) · Physics
An object of mass $0.2 \mathrm{~kg}$ executes simple harmonic motion along $\mathrm{x}$ axis with frequency of $\left(\frac{25}{\pi}\right) \mathrm{Hz}$. At the position $x=0.04 \mathrm{~m}$ the object has kinetic energy $0.5 \mathrm{~J}$ and potential energy $0.4 \mathrm{~J}$. The amplitude of oscillation is _____$\mathrm{cm}$.
Correct answer: 6
Solution
$\begin{aligned} & \text { Total energy }=\text { K.E. }+ \text { P.E. } \\ & \text { at } x=0.04 \mathrm{~m}, \text { T.E. }=0.5+0.4=0.9 \mathrm{~J} \\ & \begin{aligned} \text { T.E } & =1 \mathrm{~m} \omega^2 \mathrm{~A}^2=0.9 \\ & =\frac{1}{2} \times 0.2\left(2 \pi \times \frac{25}{\pi}\right)^2 \times \mathrm{A}^2=0.9 \\ \Rightarrow \mathrm{A} & =0.06 \mathrm{~m} \\ \mathrm{~A} & =6 \mathrm{~cm}\end{aligned}\end{aligned}$
Hence, the correct answer is 6.