Wave Mechanics Question 547
Question: A particle starts with S.H.M. from the mean position as shown in figure below. Its amplitude is A and its time period is T. At one time, its speed is half that of the maximum speed. What is the displacement at that time?
Options:
A) $ \frac{\sqrt{2}A}{3} $
B) $ \frac{\sqrt{3}A}{2} $
C) $ \frac{2A}{\sqrt{3}} $
D) $ \frac{3A}{\sqrt{2}} $
Show Answer
Answer:
Correct Answer: B
Solution:
[b] $ v=\omega {{[A^{2}-x^{2}]}^{1/2}}\Rightarrow x={{[ A^{2}-\frac{V^{2}}{{{\omega }^{2}}} ]}^{1/2}} $ Given that $ v=\frac{{v _{\max }}}{2}=\frac{A\omega }{2} $ . $ so,{{[ A^{2}-\frac{A^{2}{{\omega }^{2}}}{4{{\omega }^{2}}} ]}^{1/2}}=\frac{\sqrt{3}}{2}.A $
BETA
