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 \(\ldots \ldots \ldots\)
Pick an option to check your answer.
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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