question_answer

A horizontal platform with an object placed on it is executing S.H.M. in the vertical direction. The amplitude of oscillation is \[3.92\times {{10}^{-3}}m\]. What must be the least period of these oscillations, so that the object is not detached from the platform                                                        [AIIMS 1999]

A)            0.1256 sec                              

B)            0.1356 sec

C)            0.1456 sec                              

D)            0.1556 sec

Correct Answer: A

Solution :

                   By drawing free body diagram of object during the downward motion at extreme position, for equilibrium of mass \[mg-R=mA\]     (A = Acceleration)                    For critical condition R = 0                    so mg = mA \[\Rightarrow mg=ma{{\omega }^{2}}\]                    \[\Rightarrow \omega =\sqrt{g/a}=\sqrt{\frac{9.8}{3.92\times {{10}^{-3}}}}=50\]                      \[\Rightarrow T=\frac{2\pi }{\omega }=\frac{2\pi }{50}=0.1256sec\]