Simple Harmonic Motion Ques 5
- A particle performs simple harmonic motion with amplitude $A$. Its speed is trebled at the instant that it is at a distance $\frac{2}{3} A$ from equilibrium position. The new amplitude of the motion is
(2016 Main)
(a) $\frac{A}{3} \sqrt{41}$
(b) $3 \mathrm{~A}$
(c) $A \sqrt{3}$
(d) $\frac{7}{3} A$
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Answer:
Correct Answer: 5.( d )
Solution:
$ \text { 8. } \begin{aligned} v & =\omega \sqrt{A^2-x^2} A \mathrm{t}, x=\frac{2 A}{3} \\ v & =\omega \sqrt{A^2-\left(\frac{2 A}{3}\right)^2}=\frac{\sqrt{5}}{3} \omega A \end{aligned} $
As, velocity is trebled, hence $v^{\prime}=\sqrt{5} A \omega$
This leads to new amplitude $A^{\prime}$
$ \begin{aligned} & \therefore \quad \omega \sqrt{A^{\prime 2}-\left(\frac{2 A}{3}\right)^2}=\sqrt{5} A \omega \\ & \Rightarrow \quad \omega^2\left[A^{\prime 2}-\frac{4 A^2}{9}\right]=5 A^2 \omega^2 \\ & \Rightarrow \quad A^{\prime 2}=5 A^2+\frac{4}{9} A^2=\frac{49}{9} A^2 \\ & A^{\prime}=\frac{7}{3} A \end{aligned} $
BETA
