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