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JEE Mains · Physics · STD 11 - Oscillations
  • JEE Mains 2023
  • Medium
  • Numerical

The amplitude of a particle executing \(SHM\) is \(3\,cm\). The displacement at which its kinetic energy will be \(25 \%\) more than the potential energy is: \(.............cm\).

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Answer & Solution

Correct Answer

(B) \(2\)

Step-by-step Solution

Detailed explanation

\(K E=P E+\frac{P E}{4}\)

\(K E=\frac{5}{4} P E\)

\(\frac{1}{2} m \omega^2\left(A^2-x^2\right)=\frac{5}{4} \times \frac{1}{2} m \omega^2 x^2\) \({\left[ v =\omega \sqrt{ A ^2- x ^2}\right]}\)

\(A^2-x^2=\frac{5}{4} x^2\)

\(\frac{9 x^2}{4}=A^2\)

\(x=\frac{2}{3} A\)

\(\therefore x=\frac{2}{3} \times 3\,cm\)

\(x=2\,cm\)

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