Question

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

1. The force is zero at `t = (T)/4`.
2. The acceleration is maximum at `t = (4T)/4`.
3. The velocity is maximum at `t = T/4`.
4. The P.E. is equal to K.E. of oscillation at `t = T/2`.

Answer 1

a, b and c

Explanation:

Consider the diagram

From the given diagram; it is clear that

1. At `t = (3T)/4`, the displacement of the particle is zero. Hence. the particle executing SHM will be at the mean position ie., x = 0. So, acceleration is zero and force is also zero.
2. At t = `(4T)/4`, displacement is maximum i.e., extreme position, so acceleration is maximum.
3. At t = `T/4`, corresponds to the mean position. so velocity will be maximum at this position.
4. At t = `(2T)/4 = T/2`, corresponds to extreme position, so KE = 0 and PE = maximum.