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
- A \(\pi \mathrm{s}\)
- B \(2 \pi \mathrm{s}\)
- C \(3 \pi \mathrm{s}\)
- D \(4 \pi \mathrm{s}\)
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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