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: \(.............cm\).
Pick an option to check your answer.
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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