Ncert Exercises & Solutions

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 = \frac{3 T}{4}

, 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

= \frac{4 T}{4}

, the displacement is maximum i.e., extreme position, so acceleration is maximum.

3. At t

= \frac{T}{4}

corresponds to the mean position, so the velocity will be maximum at this position.

4. At

t = \frac{2 T}{4} = \frac{T}{2},

corresponds to the extreme position, so KE=0 and PE =maximum.