### JEE Main Solved Paper 2017 Question 1

##### Question: 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 Solved Paper-2017]

#### Options:

A)

B)

C)

D)

## Show Answer

#### Answer:

Correct Answer: B

#### Solution:

- 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}mv^{2} $ (m is the mass of particle and v is the velocity of particle $ K.E=\frac{1}{2}mA^{2}{{\omega }^{2}}{{\cos }^{2}}\omega t $ Hence graph of K.E. v/s time is square cos function