Question

The displacement time graph of a particle executing S.H.M. is given in figure : (sketch is schematic and not to scale)

Which of the following statements is/are true for this motion?
$\left(a\right)$The force is zero at $t=\frac{3T}{4}$
$\left(b\right)$The acceleration is maximum at $t=T$
$\left(c\right)$The speed is maximum at $t=\frac{T}{4}$
$\left(d\right)$The P.E. is equal to K.E. of the oscillation at $t=\frac{T}{2}$

• A. $\left(a\right),\left(b\right)$ and $\left(d\right)$
• B. $\left(b\right),\left(c\right)$ and $\left(d\right)$
• C. $\left(a\right),\left(b\right)$ and $\left(c\right)$
• D. $\left(a\right)$ and $\left(d\right)$

Answer 1

The correct option is C $\left(a\right),\left(b\right)$ and $\left(c\right)$

From graph equation of SHM $X=A\cos \omega t$
(a) At $t=\frac{3T}{4},$ particle is at mean position.
$\therefore$ Acceleration = 0, Force = 0

(b)At $t=T,$ particle is again at extreme position. So acceleration is maximum

(c)At $t=\frac{T}{4}$, particle is at mean position. So velocity is maximum.
Acceleration $=0$

(d) When $KE=PE$
$\Rightarrow \frac{1}{2}k(A^2-x^2)=\frac{1}{2}k{x}^2$
Here, A = amplitude of SHM
x = displacement from mean position
$\Rightarrow A^2=2x^2\Rightarrow x=\frac{\pm A}{\sqrt{2}}$
But at $t=\frac{T}{2},x=–A$
$\therefore \left(a\right),\left(b\right)\text{and}\left(c\right)$ are correct.