JEE Mains · Physics · STD 11 - 13. oscillations
A particle performs simple harmonic motion with amplitude \(A\). Its speed is trebled at the instant that it is at a distance \(\frac{{2A}}{3}\) from equilibrium position. The new amplitude of the motion is
- A \(A\)\(\sqrt 3 \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\)
- B \(\;\frac{{7A}}{3}\)
- C \(\;\frac{A}{3}\sqrt {41} \)
- D \(3A\)
Answer & Solution
Correct Answer
(B) \(\;\frac{{7A}}{3}\)
Step-by-step Solution
Detailed explanation
Let new amplitude is \(A'\) initial velocity \(v^{2}=\omega^{2}\left(A^{2}-\left(\frac{2 A}{3}\right)^{2}\right)\) \(...(1)\) Where \(A\) is initial amplitude \(\&\, \omega\) is angular frequency. Final velocity…
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