JEE Mains · Physics · STD 11 - Oscillations
- JEE Mains 2019
- Medium
- MCQ
A particle executes simple harmonic motion with an amplitude of \(5\, cm\). When the particle is at \(4\, cm\) from the mean position, the magnitude of its velocity is \(SI\,units\) is equal to that of its acceleration. Then, its periodic time in seconds is
Pick an option to check your answer.
Answer & Solution
Correct Answer
(C) \(\frac{8\pi}{3}\)
Step-by-step Solution
Detailed explanation
\(\left|v_{4}\right|=\left|a_{4}\right|\)
\(\Rightarrow \quad(w \sqrt{A^{2}-x^{2}})_{4}=\left(w^{2} x\right)_{4}\)
\(\Rightarrow \quad w \sqrt{25-16}=w^{2} \times 4\)
\(\Rightarrow \quad w=\frac{3}{4}\)
\(T=\frac{2 \pi}{w}=2 \pi \frac{4}{3}=\frac{8 \pi}{3}\)
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