WBJEE · Physics · Oscillations
The velocity of a particle executing a simple harmonic motion is \(13 \mathrm{ms}^{-1}\), when its distance from the equilibrium position \((Q)\) is \(3 \mathrm{m}\) and its velocity is \(12 \mathrm{ms}^{-1},\) when it is \(5 \mathrm{m}\) away from \(Q .\) The frequency of the simple harmonic motion is
- A \(\frac{5 \pi}{8}\)
- B \(\frac{5}{8 \pi}\)
- C \(\frac{8 \pi}{5}\)
- D \(\frac{8}{5 \pi}\)
Answer & Solution
Correct Answer
(B) \(\frac{5}{8 \pi}\)
Step-by-step Solution
Detailed explanation
The speed of a particle executing simple harmonic motion is \(v=\omega \sqrt{a^{2}-x^{2}}\) where, \(a=\) Amplitude \(\omega=\) Angular frequency \(x=\) Displacement or \[ v^{2}=\omega^{2}\left(a^{2}-x^{2}\right) \] According to the question, Here…
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