question_answer

A particle is executing simple harmonic motion with a time period T. AT time t = 0, it is at its position of equilibrium. The kinetic energy-time graph of the particle will look like                                              [JEE Main 2017]

A)  

B)  

C)  

D)  

Correct Answer: B

Solution :

(b)

Time taken to reach the extreme position from equilibrium position is\[\frac{T}{4}\]. Velocity is maximum at equilibrium position and zero at extreme position.

\[V=A\,\omega \,\cos \omega t\]

\[K.E=\frac{1}{2}m{{v}^{2}}\]

(m is the mass of particle and v is the velocity of particle

\[K.E=\frac{1}{2}m{{A}^{2}}{{\omega }^{2}}{{\cos }^{2}}\omega t\]

Hence graph of K.E. v/s time is square cos function