MHT CET · Physics · Oscillations
A particle executes S.H.M. starting from the mean position. Its amplitude is 'a' and its periodic time is 'T'. At a certain instant, its speed ' \(u\) ' is half that of maximum speed \(V_{\text {max }}\). The displacement of the particle at that instant is
- A \(\frac{2 a}{\sqrt{3}}\)
- B \(\frac{\sqrt{2} a}{3}\)
- C \(\frac{3 \mathrm{a}}{\sqrt{2}}\)
- D \(\frac{\sqrt{3} a}{2}\)
Answer & Solution
Correct Answer
(D) \(\frac{\sqrt{3} a}{2}\)
Step-by-step Solution
Detailed explanation
\(u = \frac{1}{2} V_{\text{max}}\) \(\omega \sqrt{a^2 - x^2} = \frac{1}{2} (a\omega)\)
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