# A particle is executing simple harmonic motion with a time period T. At time $t=0$, it is at its position of quilibrium. The kinetic energy - time graph of the particle will look like

Solution :
Time taken to reach the extreme position from equilibrium position is $\large\frac{T}{4}$
Velocity is maximum at equilibrium position and zero at extreme position.
$v= A \omega \cos \omega mv^2$
$K.E = \large\frac{1}{2} $$mv^2 (m is the mass of particle and v is the velocity of particle) K.E= \large\frac{1}{2}$$mA^2 \omega ^2 \cos ^2 \omega t$
Hence graph of K.E. v/s time is square cos function