The (displacement-time) graph of a particle executing SHM is shown in the figure. Then,
     

(a) the force is zero at \(t=\dfrac{3T}{4}\)
(b) the acceleration is maximum at \(t=\dfrac{4T}{4}\) 
(c) the velocity is maximum at \(t=\dfrac{T}{4}\)
(d) the potential energy is equal to the kinetic energy of oscillation at \(t=\dfrac{T}{2}\)
 
Which of the statement/s given above is/are true?
1. (a), (b), (d)
2. (a), (b), (c)
3. (b), (c), (d)
4. (c), (d)
(2) Hint: The velocity is maximum at the mean position and the acceleration is maximum at the extreme position.
Consider the diagram
Step 1: Find the velocity and the acceleration at different positions.
From the given diagram; it is clear that
1. At t=3T4, the displacement of the particle is zero. Hence, the particle executing SHM
will be at mean position i.e., x=0. So, the acceleration is zero and the force is also zero.
2. At t=4T4, the displacement is maximum i.e., extreme position, so acceleration is maximum.
3. At t =T4 corresponds to the mean position, so the velocity will be maximum at this position.
4. At t=2T4=T2, corresponds to the extreme position, so KE=0 and PE =maximum.