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JEE Mains · Physics · STD 11 - Oscillations
  • JEE Mains 2021
  • Medium
  • MCQ

A particle starts simple harmonic motion from the mean position. Its amplitude is \(a\) and total energy \(E\). At one instant its kinetic energy is \(3E/4.\) Its displacement at that instant is

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Answer & Solution

Correct Answer

(B) \(y=\frac{a}{2}\)

Step-by-step Solution

Detailed explanation

(b) \(\frac{K}{E} = \frac{{\frac{1}{2}m{\omega ^2}({a^2} - {y^2})}}{{\frac{1}{2}m{\omega ^2}{a^2}}}\)

\(= \frac{{{a^2} - {y^2}}}{{{a^2}}} \)

\(= 1 - \frac{{{y^2}}}{{{a^2}}}\)

So, \(\frac{{\left( {\frac{{3E}}{4}} \right)}}{E} = 1 - \frac{{{y^2}}}{{{a^2}}}\)

\(\Rightarrow \frac{{{y^2}}}{{{a^2}}} = 1 - \frac{3}{4} = \frac{1}{4}\)

\( \Rightarrow y = \frac{a}{2}\).

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