question_answer

A sphere of radius r is kept on a concave mirror of radius of curvature R. The arrangement is kept on a horizontal table (the surface of concave mirror is frictionless and sliding not rolling). If the sphere is displaced from its equilibrium position and left, then it executes S.H.M. The period of oscillation will be

A)            \[2\pi \sqrt{\left( \frac{\left( R-r \right)1.4}{g} \right)}\]    

B)            \[2\pi \sqrt{\left( \frac{R-r}{g} \right)}\]

C)            \[2\pi \sqrt{\left( \frac{rR}{a} \right)}\]                       

D)            \[2\pi \sqrt{\left( \frac{R}{gr} \right)}\]

Correct Answer: B

Solution :

                   Tangential acceleration, \[{{a}_{t}}=-g\sin \theta =-g\theta \]                     \[{{a}_{t}}=-g\frac{x}{(R-r)}\]             Motion is S.H.M., with time period                    \[T=2\pi \sqrt{\frac{\text{displacement}}{\text{acceleration}}}\]\[=2\pi \sqrt{\frac{x}{\frac{gx}{(R-t)}}}=2\pi \sqrt{\frac{R-r}{g}}\]