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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

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