Wave Mechanics Question 596
Question: A rod of mass M and length L is hinged at its centre of mass so that it can rotate in a vertical plane. Two springs each of stiffness k are connected at its ends, as shown in the figure. The time period of SHM is
Options:
A) $ 2\pi \sqrt{\frac{M}{6k}} $
B) $ 2\pi \sqrt{\frac{M}{3k}} $
C) $ 2\pi \sqrt{\frac{ML}{k}} $
D) $ \pi \sqrt{\frac{M}{6k}} $
Show Answer
Answer:
Correct Answer: A
Solution:
[a] The restoring torque $ ( forsmall\theta ) $
$ {\tau _{rest}}=-[ \frac{kL\theta }{2}\times \frac{L}{2} ]\times 2=\frac{kL^{2}}{2}(-\theta ) $
$ \
Therefore ,\alpha =\frac{{\tau _{rest}}}{I}=\frac{kL^{2}/2}{ML^{2}/12}(-\theta )=\frac{6k}{M}(-\theta ) $
$ \
Therefore T=2\pi \sqrt{\frac{M}{6k}} $ .
BETA
