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JEE Mains · Physics · STD 11 - Oscillations
  • JEE Mains 2024
  • Hard
  • Numerical

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

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Answer & Solution

Correct Answer

(B) \(22\)

Step-by-step Solution

Detailed explanation

\(\mathrm{m}=0.5 \mathrm{~kg}\)

\(\mathrm{~F}=-50(\mathrm{x})\)

\(\mathrm{ma}=(-50 \mathrm{x})\)

\(0.5 \mathrm{a}=-50 \mathrm{x}\)

\(\mathrm{a}=(-100 \mathrm{x})\)

\(\mathrm{W}^2=100 \Rightarrow(\mathrm{w}=10)\)

\(\mathrm{T}=\frac{2 \pi}{10}=\left(\frac{\pi}{5}\right)=\frac{22}{7 \times 15}=\left(\frac{22}{35}\right)\)

\(\frac{\pi}{35}=\frac{22}{35} \Rightarrow \mathrm{x}=22\)

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