question_answer

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

A) \[0.1556\text{ }sec\]       

B)        \[0.1456\text{ }sec\]     

C)        \[0.1356\text{ }sec\]     

D)        \[0.1256\text{ }sec.\]

Correct Answer: D

Solution :

                As the platform is executing SHM, its time period will be minimum when it has the maximum acceleration. We know that in SHM, the maximum acceleration is given by     \[{{a}_{\max }}={{\omega }^{2}}A\]                    Now if the body is not to be detached from the platform, \[{{a}_{\max }}\] should be less than the acceleration due to gravity. In the limiting case, \[{{a}_{\max }}=g\] \[\Rightarrow \]  \[{{\omega }^{2}}A=g\] \[\Rightarrow \,\,\frac{4{{\pi }^{2}}A}{{{T}^{2}}}=g\] \[\Rightarrow \]  \[T=2\pi \sqrt{\frac{A}{g}}=2\pi \times \sqrt{\frac{3.92\times {{10}^{-4}}}{9.8}}\] \[\therefore \]  \[{{T}_{\min }}=0.1256\sec .\]