Oscillations Question 7
Question 7 - 2024 (31 Jan Shift 1)
A particle performs simple harmonic motion with amplitude A. Its speed is increased to three times at an instant when its displacement is $\frac{2 A}{3}$. The new amplitude of motion is $\frac{nA}{3}$. The value of $n$ is
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Answer: (7)
Solution:
$v=\omega \sqrt{A^{2}-x^{2}}$
at $x=\frac{2 A}{3}$
$v=\omega \sqrt{A^{2}-\left(\frac{2 A}{3}\right)^{2}}=\frac{\sqrt{ } 5 A \omega}{3}$
New amplitude $=A^{\prime}$
$v^{\prime}=3 v=\sqrt{5} A \omega=\omega \sqrt{\left(A^{\prime}\right)^{2}-\left(\frac{2 A}{3}\right)^{2}}$
$A^{\prime}=\frac{7 A}{3}$
