Question

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

• A. The force is zero at $t=\frac{3T}{4}$.
• B. The acceleration is maximum at $t=\frac{4T}{4}$.
• C. The velocity is maximum at $t=\frac{T}{4}$.
• D. The PE is equal to $KE$ of oscillation at $t=\frac{T}{2}$.

Answer 1

Answer: (D)

Consider the figure given below
From this figure, it is clear that
(a) at $t=\frac{3T}{4}$, the displacement of the particle is zero. Hence, the particle executing SHM will be at mean position, i.e. $x=0$. So, acceleration is zero and force is also zero.
(b) at $t=\frac{4T}{4}=T$, displacement is maximum, i.e. the particle is at extreme position, so acceleration is maximum.
(c) Similarly, at $t=\frac{T}{4}$, the particle will be at to mean position, so velocity will be maximum at this position.
(d) at $t=\frac{2T}{4}=\frac{T}{2}$, the particle will be at extreme position, so $KE=0$ and $PE=$ maximum.