# 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 statement is/are true for this motion? (A)    The force is zero at $t=\frac{3T}{4}$ (B)    The acceleration is maximum at t=T (C)    The speed is maximum at $t=\frac{T}{4}$ (D)    The P.E. is equal to K.E. of the oscillation at $t=\frac{T}{2}$ Option: 1 (A) and (D) Option: 2 (A), (B) and (C) Option: 3 (B), (C) and (D) Option: 4 (A), (B) and (D)

From graph equation of SHM

$X=ACos \omega t$

A)

At  $t=\frac{3T}{4}$, the particle is at the mean position.

So  The force is zero at $t=\frac{3T}{4}$

B)

at T particle is at the extreme position so acceleration is maximum

C)

The speed is maximum at $t=\frac{T}{4}$

because at $t=\frac{T}{4}$ particle is at the mean position.

D)

$\begin{array}{l} \mathrm{KE}=\mathrm{PE} \\ \frac{1}{2} \mathrm{k}\left(\mathrm{A}^{2}-\mathrm{x}^{2}\right)=\frac{1}{2} \mathrm{k} \mathrm{x}^{2} \\ \mathrm{~A}^{2}=2 \mathrm{x}^{2} \\ \mathrm{x}=\frac{+\mathrm{A}}{\sqrt{2}} \\ Using \ \frac{\mathrm{A}}{\sqrt{2}}=\mathrm{A} \text { cosot } \\ \mathrm{t}=\mathrm{T} / 8 \end{array}$

So A, B and C are correct

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