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JEE Mains · Physics · STD 11 - Oscillations
  • JEE Mains 2021
  • Medium
  • Numerical

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 \(\ldots \ldots \ldots\)

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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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