TS EAMCET · Physics · Oscillations
A particle executes simple harmonic motion according to the equation \(x(t)=A \sin ^2(\alpha t)\). If the time period of the SHM is \(0.2 \mathrm{~s}\), then the value of \(\alpha\) (in units of \(\mathrm{rad} / \mathrm{s}\) ) is
- A \(2 \pi\)
- B \(10 \pi\)
- C \(5 \pi\)
- D \(2.5 \pi\)
Answer & Solution
Correct Answer
(C) \(5 \pi\)
Step-by-step Solution
Detailed explanation
Given that, the equation of SHM \( \begin{aligned} x(t) & =A \sin ^2(\alpha t) \\ & =A\left[\frac{1-\cos 2 \alpha t}{2}\right]=\frac{A}{2}-\frac{A}{2} \cos 2 \alpha t \end{aligned} \) We know that, general equation of SHM is \(x=a_1+a_2 \cos \omega t\)...(i)…
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