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Simple Harmonic Motion question

2024 · 9 Apr · Shift 2 · Q86
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  5. /2024 · 9 Apr · Shift 2 · Q86

Simple Harmonic Motion question

2024 · 9 Apr · Shift 2 · Q86

JEE MainPhysicsSimple Harmonic MotionNumerical+4 / −1
A particle of mass 0.50 kg0.50 \mathrm{~kg}0.50 kg executes simple harmonic motion under force F=−50(Nm−1)xF=-50(\mathrm{Nm}^{-1}) xF=−50(Nm−1)x. The time period of oscillation is x35s\frac{x}{35} s35x​s. The value of xxx is ‾\underline{\hspace{2cm}}​. (Given π=227\pi=\frac{22}{7}π=722​)
Numerical answer
View written solutionFree

Correct answer: 22

  1. Given data
  • Mass of particle: m=0.50 kgm = 0.50\,\text{kg}m=0.50kg
  • Restoring force: F=−50xF = -50xF=−50x

For SHM, the restoring force has the form F=−kxF = -kxF=−kx So, comparing, k=50 N m−1k = 50\,\text{N m}^{-1}k=50N m−1

  1. Formula for time period of SHM

The time period is T=2πmkT = 2\pi\sqrt{\frac{m}{k}}T=2πkm​​

Substitute the values: T=2π0.5050T = 2\pi\sqrt{\frac{0.50}{50}}T=2π500.50​​

T=2π0.01T = 2\pi\sqrt{0.01}T=2π0.01​

T=2π(0.1)=0.2πT = 2\pi(0.1) = 0.2\piT=2π(0.1)=0.2π

  1. Use the given value of π\piπ

Given, π=227\pi = \frac{22}{7}π=722​

Therefore, T=0.2×227=2235 sT = 0.2 \times \frac{22}{7} = \frac{22}{35}\,\text{s}T=0.2×722​=3522​s

  1. Compare with the given form

The question says the time period is x35 s\frac{x}{35}\,\text{s}35x​s

So, x35=2235\frac{x}{35} = \frac{22}{35}35x​=3522​

Hence, x=22x = 22x=22

  1. Final answer

22\boxed{22}22​

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