The displacement time graph of a particle executing SHM is shown in figure. Which of the following statement is/are true?

Options

A. The force is zero at `t= (3T)/(4)`
B. The acceleration is maximum at `t= (4T)/(4)`
C. The velocity is maximum at `t= (T)/(4)`
D. The PE is equal to KE of oscillation `t= (T)/(2)`.

Answer

A, B, C

Explanation

From graph it is clearly seen that at time `t= (3T)/(4)`, the displacement of oscillator is zero and acceleration at mean position is zero and hence force is also zero so, option A is true.

At time `t= (4T)/(4)`, oscillator is at extreme positive point and at that point acceleration is maximum `[a= -omega^(2)|A|]` hence, option B is true.
At time `t= (T)/(4)`, oscillator is at mean position and hence its velocity is maximum `[v= -omega sqrt(A^(2)-x^(2))]`, so option C is true.
At time `t= (T)/(2)`, oscillator is at extreme negative point and at this point kinetic energy is zero `[K= (1)/(2) k(A^(2)-x^(2))]` and hence potential energy is maximum `[U= (1)/(2)kx^(2)]` so, option D is false.