An object of mass 0.2 kg executes simple harmonic motion along x axis with frequency of (25/π ) Hz. At the position x=0.04 m the object has kinetic energy 0.5 J and potential energy 0.4 J . The amplitude of oscillation is cm .

Physics · Simple Harmonic Motion

JEE Main 2024 — 8 April, Shift 2 — Question 54

An object of mass 0.2 kg executes simple harmonic motion along x axis with frequency of $\left(\frac{25}{\pi}\right)$ Hz. At the position $x=0.04 \mathrm{~m}$ the object has kinetic energy 0.5 J and potential energy 0.4 J . The amplitude of oscillation is $\qquad$ cm .

Answer: 6

Numerical answer — enter this value.

Step-by-step solution

Total energy $=$ K.E. + P.E.

at $\mathrm{x}=0.04 \mathrm{~m}$ , T.E. $=0.5+0.4=0.9 \mathrm{~J}$

T.E $=1 \mathrm{~m} \omega^{2} \mathrm{~A}^{2}=0.9$

$=\frac{1}{2} \times 0.2\left(2 \pi \times \frac{25}{\pi}\right)^{2} \times \mathrm{A}^{2}=0.9$ $\Rightarrow \mathrm{A}=0.06 \mathrm{~m}$

$\mathrm{A}=6 \mathrm{~cm}$

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Exam: JEE Main 2024
Paper: 8 April, Shift 2
Subject: Physics
Chapter: Simple Harmonic Motion
Topic: Kinematics of SHM, Phase and Energy in SHM