A particle of mass 0.50 kg executes simple harmonic motion under force F =-50 ( Nm -1 ) x . The time period of oscillation is x/35 s . The value of x is (Given π=22/7 )

Physics · Simple Harmonic Motion

JEE Main 2024 — 9 April, Shift 2 — Question 47

A particle of mass 0.50 kg executes simple harmonic motion under force $\mathrm{F}=-50\left(\mathrm{Nm}^{-1}\right) \mathrm{x}$ . The time period of oscillation is $\frac{x}{35} \mathrm{~s}$ . The value of x is (Given $\pi=\frac{22}{7}$ )

Answer: 22

Numerical answer — enter this value.

Step-by-step solution

Given restoring force:

F=-kx \Rightarrow k=50\,\text{N m}^{-1}

and mass $m=0.50\,\text{kg}$ .

Time period of SHM:

T=2\pi\sqrt{\frac{m}{k}} =2\pi\sqrt{\frac{0.5}{50}} =2\pi\sqrt{\frac{1}{100}} =\frac{2\pi}{10}=\frac{\pi}{5}\,\text{s}

Given:

T=\frac{x}{35}\ \text{s}

\frac{x}{35}=\frac{\pi}{5} \Rightarrow x=7\pi

Using $\pi=\frac{22}{7}$ :

x=22

Answer key and solution verified before publishing.

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Exam: JEE Main 2024
Paper: 9 April, Shift 2
Subject: Physics
Chapter: Simple Harmonic Motion
Topic: Kinematics of SHM, Phase and Energy in SHM