JEE Mains · Physics · STD 11 - 13. oscillations
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: \(.............cm\).
- A \(4\)
- B \(2\)
- C \(5\)
- D \(3\)
Answer & Solution
Correct Answer
(B) \(2\)
Step-by-step Solution
Detailed explanation
\(K E=P E+\frac{P E}{4}\) \(K E=\frac{5}{4} P E\) \(\frac{1}{2} m \omega^2\left(A^2-x^2\right)=\frac{5}{4} \times \frac{1}{2} m \omega^2 x^2\) \({\left[ v =\omega \sqrt{ A ^2- x ^2}\right]}\) \(A^2-x^2=\frac{5}{4} x^2\) \(\frac{9 x^2}{4}=A^2\) \(x=\frac{2}{3} A\)…
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