Question

The velocity time graph of a particle executing SHM is shown in the figure. Choose the incorrect statement

(1) Force is maximum at t = 3T/2
(2) Potential energy of the particle is minimum at t = T/4
(3) Kinetic energy of the particle is maximum at t = T
(4) Acceleration of particle is maximum at t= T/4

Please explain the reason too

Answer 1

above graph is given velocity-time graph and it is cosine graph. so, displacement-time graph must be sine graph.

so, particle starts to move from origin. and achieve extreme position at t = T/4, 5T/4 and mean position at t = T/2, T, 3T/2

(1) Force , F = ma = -$m\omega^2x$ in figure it is clear that at t = 3T/2 , x = 0. so, force = 0 hence, force is not maximum at 3T/2 . it's incorrect statement.

(2) No, it's incorrect statement because at t = T/4 particle is located at extreme position and we know Potential energy at extreme position is maximum.

(3) yes, kinetic energy is maximum only at mean position. at t = T , mean position. hence, it's correct statement.

(4)yes, acceleration of particle at t = T/4, will be maximum as acceleration is directly proportional to position of particle in SHM. hence, it's correct statement.

so, only (1), (2) options are incorrect. and rest options are correct.