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 $\frac{2 \mathrm{~A}}{3}$. The new amplitude of motion is $\frac{\mathrm{nA}}{3}$. The value of $\mathrm{n}$ is
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Answer: (7)
Solution:
$v=\omega \sqrt{A^{2}-x^{2}}$
at $\mathrm{x}=\frac{2 \mathrm{~A}}{3}$
$v=\omega \sqrt{\mathrm{A}^{2}-\left(\frac{2 \mathrm{~A}}{3}\right)^{2}}=\frac{\sqrt{ } 5 \mathrm{~A} \omega}{3}$
New amplitude $=\mathrm{A}^{\prime}$
$\mathrm{v}^{\prime}=3 \mathrm{v}=\sqrt{5} \mathrm{~A} \omega=\omega \sqrt{\left(\mathrm{A}^{\prime}\right)^{2}-\left(\frac{2 \mathrm{~A}}{3}\right)^{2}}$
$\mathrm{A}^{\prime}=\frac{7 \mathrm{~A}}{3}$
