A) \[\frac{1}{2\pi }\sqrt{\frac{1}{g}}\]
B) \[v\,\sin \,\frac{\theta }{2}\]
C) \[\sqrt{mg}\]
D) zero
Correct Answer: D
Solution :
Key Idea: Work done is the product of force and displacement. If a body in periodic motion moves along the same path to and fro about a definite point (equilibrium position), then the motion of body is oscillatory. As the body goes to one side of its equilibrium position, comes back to that position goes to the other side, and again returns to the same position, it is said to complete one vibration or one oscillation, i.e., displacement is zero. Work done = force \[\times \] displacement Since, displacement is zero, work done is also zero.
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