The displacement time graph of a particle executing S.H.M. is given in figure
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 ?

(A) The force is zero $\mathrm{t}=\frac{3 \mathrm{~T}}{4}$

(B) The acceleration is maximum at $\mathrm{t}=\mathrm{T}$

(C) The speed is maximum at $\mathrm{t}=\frac{\mathrm{T}}{4}$

(D) The P.E. is equal to $\mathrm{K}$.E. of the oscillation

at $\mathrm{t}=\frac{\mathrm{T}}{2}$

1. (A), (B) and (D)

2. $(\mathrm{B}),(\mathrm{C})$ and $(\mathrm{D})$

3. (A) and (D)

4. (A), (B) and (C)

Correct Option: , 4

Solution:

(A) $F=m a$

$a=-\omega^{2} X$

at $\frac{3 \mathrm{~T}}{4}$ displacement zero $(\mathrm{x}=0)$, so $\mathrm{a}=0$

$F=0$

(B) at $\mathrm{t}=\mathrm{T}$

displacement $(x)=A$

$\mathrm{x}$ maximum, So acceleration is maximum.

(C) $\mathrm{V}=\omega \sqrt{\mathrm{A}^{2}-\mathrm{x}^{2}}$

$\mathrm{V}_{\max }$ at $\mathrm{x}=0$

$\mathrm{~V}_{\max }=\mathrm{A} \omega$

at $\mathrm{t}=\frac{\mathrm{T}}{4}, \quad \mathrm{x}=0, \quad$ So $\mathrm{V}_{\max }$

(D) $\mathrm{KE}=\mathrm{PE}$

$\therefore$ at $x=\frac{A}{\sqrt{2}}$

at $\mathrm{t}=\frac{\mathrm{T}}{2} \quad \mathrm{x}=-\mathrm{A} \quad$ (So not possible)