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JEE Mains · Physics · STD 11 - Oscillations
  • JEE Mains 2023
  • Easy
  • MCQ

The maximum potential energy of a block executing simple harmonic motion is \(25\,J\). A is amplitude of oscillation. At \(A / 2\), the kinetic energy of the block is \(...............\)

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Answer & Solution

Correct Answer

(C) \(18.75\)

Step-by-step Solution

Detailed explanation

\(u _{\max }=\frac{1}{2} m \omega^2 A ^2=25\,J\)

\(KE \text { at } \frac{ A }{2}=\frac{1}{2} mv _1^2=\frac{1}{2} m \omega^2\left( A ^2-\frac{ A ^2}{4}\right)\)

\(KE =\frac{1}{2} m \omega^2 \frac{3 A ^2}{4}=\frac{3}{4}\left(\frac{1}{2} m \omega^2 A ^2\right)\)

\(KE =\frac{3}{4} \times 25=18.75\,J\)

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