Oscillations Ques 17
17. 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: 17.(b)
Solution:
- (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 $y=a \sqrt{1-n^{2}}$
BETA
