A) \[2\pi \sqrt{\frac{m}{\rho gA}}\]
B) \[2\pi \sqrt{\frac{m}{2\rho gA}}\]
C) \[2\pi \sqrt{\frac{2m}{\rho gA}}\]
D) None of these
Correct Answer: A
Solution :
[a] The restoring force \[F=-(PA)\] \[=-[\rho g(2y\sin {{30}^{o}})]A\] \[=\rho gA(-y)\] \[\therefore \,\,\,\,a=\frac{F}{m}=\frac{\rho gA}{m}\] Thus \[T=2\pi \sqrt{\frac{m}{\rho gA}}\]You need to login to perform this action.
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