Physics · Simple Harmonic Motion

JEE Main 2024 — 29 January, Shift 1 — Question 56

When the displacement of a simple harmonic oscillator is one third of its amplitude, the ratio of total energy to the kinetic energy is $\frac{x}{8}$ , where $x=$ $\qquad$ .

Answer: 9

Numerical answer — enter this value.

Step-by-step solution

Let total energy $=\mathrm{E}=\frac{1}{2} \mathrm{KA}^{2}$

$U=\frac{1}{2} K\left(\frac{A}{3}\right)^{2}=\frac{K A^{2}}{2 \times 9}=\frac{E}{9}$

$K E=E-\frac{E}{9}=\frac{8 E}{9}$

Ratio $\frac{\text { Total }}{\mathrm{KE}}=\frac{\mathrm{E}}{\frac{8 \mathrm{E}}{9}}=\frac{9}{8}$ $x=9$

Answer key and solution verified before publishing.

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Exam: JEE Main 2024
Paper: 29 January, Shift 1
Subject: Physics
Chapter: Simple Harmonic Motion
Topic: Kinematics of SHM, Phase and Energy in SHM