JEE Mains · Physics · STD 11 - 13. oscillations
A particle executes \(S.H.M.\) with amplitude \('a'\) and time period \(V\). The displacement of the particle when its speed is half of maximum speed is \(\frac{\sqrt{ x } a }{2} .\) The value of \(x\) is \(\ldots \ldots \ldots\)
- A \(1\)
- B \(5\)
- C \(2\)
- D \(3\)
Answer & Solution
Correct Answer
(D) \(3\)
Step-by-step Solution
Detailed explanation
\(V =\omega \sqrt{ A ^{2}- x ^{2}} \quad V _{\max }= A\omega\) \(\frac{ A\omega }{2}=\omega \sqrt{ A ^{2}- x ^{2}}\) \(\frac{ A ^{2}}{4}= A ^{2}- x ^{2}\) \(x ^{2}=\frac{3 A ^{2}}{4}\) \(x =\frac{\sqrt{3}}{2} \,A\)
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