Oscillations - Result Question 19

22. A particle starts simple harmonic motion from the mean position. Its amplitude is A and time period is T.What is its displacement when its speed is half of its maximum speed

[1996]

(a) $\frac{\sqrt{2}}{3} A$

(b) $\frac{\sqrt{3}}{2} A$

(c) $\frac{2}{\sqrt{3}} A$

(d) $\frac{A}{\sqrt{2}}$

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Answer:

Correct Answer: 22. (b)

Solution:

  1. (b) $v _{\max }=A \omega$ when $v=\frac{v _{\max }}{2}=\frac{A \omega}{2}$

$v=\omega \sqrt{A^{2}-y^{2}}$

$\Rightarrow \frac{A \omega}{2}=\omega \sqrt{A^{2}-y^{2}} \Rightarrow y= \pm \frac{\sqrt{3}}{2} A$

The displacement at which the speed is $n$ times the maximum speed is given by $y=a \sqrt{1-n^{2}}$