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Q.
The displacement-time graph of a particle executing SHM is shown in figure. Which of the following statement is/are incorrect?
Oscillations
Solution:
Consider the figure given below
From this figure, it is clear that
(a) 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, acceleration is zero and force is also zero.
(b) at $t=\frac{4 T}{4}=T$, 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 $t=\frac{2 T}{4}=\frac{T}{2}$, the particle will be at extreme position, so $KE =0$ and $PE =$ maximum.
