NEET Solved Paper 2017 Question 28
Question: A particle executes linear simple harmonic motion with an amplitude of 3 cm. When the particle is at 2 cm from the mean position, the magnitude of its velocity is equal to that of its acceleration. Then its time period in seconds is
Options:
A) $ \frac{2\pi }{\sqrt{3}} $
B) $ \frac{\sqrt{5}}{\pi } $
C) $ \frac{\sqrt{5}}{2\pi } $
D) $ \frac{4\pi }{\sqrt{5}} $
Show Answer
Answer:
Correct Answer: D
Solution:
- $ v=\omega \sqrt{A^{2}-x^{2}} $ $ a=x{{\omega }^{2}} $ $ v=a $ $ \omega \sqrt{A^{2}-x^{2}}=x{{\omega }^{2}} $ $ \sqrt{{{(3)}^{2}}-{{(2)}^{2}}}=2( \frac{2\pi }{T} ) $ $ \sqrt{5}=\frac{\pi }{T} $
$ \Rightarrow $ ${\space}$ $ T=\frac{4\pi }{\sqrt{5}} $
BETA
