Question

The kinetic energy of a particle executing S.H.M. is $ 16J $ when it is at its mean position if the amplitude of oscillation is $ 25cm $ , and the mass of the particle is $ 5.12kg $ then time period of the oscillation is:
(A) $ 20\pi \sec $
(B) $ 2\pi \sec $
(C) $ \dfrac{\pi }{5}\sec $
(D) $ 5\pi \sec $

Answer 1

Hint :Here, we have the example of the S.H.M. and its kinetic energy is given so we have to use the formula for kinetic energy at the mean position is
 $ K.E. = \dfrac{1}{2}m{\omega ^2}({A^2} - {x^2}) $ , and time period of oscillation is $ T = \dfrac{{2\pi }}{\omega } $

Complete Step By Step Answer:
Here, the given date is as follows:
Kinetic energy, $ K.E. = 16J $
Amplitude, $ A = 25 \times {10^{ - 2}}m $
Mass , $ m = 5.12kg $
Displacement, $ x = 0 $ …(mean position)
Thus, using this information let us consider the kinetic energy for S.H.M. as below
 $ K.E. = \dfrac{1}{2}m{\omega ^2}({A^2} - {x^2}) $ …… $ (1) $
Now, put all the required values in the equation $ (1) $ , we get
 $ \Rightarrow 16 = \dfrac{1}{2} \times 5.12 \times {\omega ^2}{\left( {25 \times {{10}^{ - 2}}} \right)^2} $
After solving the above equation, we get
 $ \Rightarrow {\omega ^2} = 100 $
 $ \Rightarrow \omega = 10rad/\sec $
The time period is given by
 $ T = \dfrac{{2\pi }}{\omega } $ …… $ (2) $
Thus, from above equations we can put the requires values in equation $ (2) $
 $ \Rightarrow T = \dfrac{{2\pi }}{{10}} = \dfrac{\pi }{5}\sec $
Thus the time period of an oscillation is obtained as $ \dfrac{\pi }{5}\sec $
The correct answer is option C.

Note :
Simple harmonic motion is a periodic motion where the restoring force on the moving object is directly proportional to the object's displacement magnitude and acts towards the equilibrium position of the object. Kinetic energy of the S.H.M. depends on the amplitude, mass and displacement of the object as well as the angular frequency of the object. As we discussed above the time period is inversely proportional to the angular frequency of the object and proportionality constant is $ 2\pi $ .