JEE Mains · Physics · STD 11 - 13. oscillations
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
- A \(\frac{4\pi}{3}\)
- B \(\frac{3}{8}\pi\)
- C \(\frac{8\pi}{3}\)
- D \(\frac{7}{3}\pi\)
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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