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TS EAMCET · Physics · Oscillations

A particle is exhibiting simple harmonic motion has its displacement \(x\) and velocity \(v\) related as \(4 v^2=25-x^2\). The time period of SHM is

  1. A \(\pi \mathrm{s}\)
  2. B \(2 \pi \mathrm{s}\)
  3. C \(3 \pi \mathrm{s}\)
  4. D \(4 \pi \mathrm{s}\)
Verified Solution

Answer & Solution

Correct Answer

(D) \(4 \pi \mathrm{s}\)

Step-by-step Solution

Detailed explanation

Given that, \(4 v^2=25-x^2 \Rightarrow v^2=\frac{1}{4}\left(25-x^2\right)\) \(\Rightarrow v=\sqrt{\frac{1}{4}\left(25-x^2\right)}=\frac{1}{2} \sqrt{25-x^2}\) \(\begin{equation*}=\frac{1}{2} \sqrt{5^2-x^2} ...\tag{i}\end{equation*}\) And we know that, \(v=\omega \sqrt{A^2-x^2}\)…
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