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(D) From the graph,the displacement $x$ follows a cosine function starting from $x = A$ at $t = 0$,so $x(t) = A \cos(\omega t)$.$(A)$ At $t = \frac{3T

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$(A), (B)$ and $(C)$

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(D) From the graph,the displacement $x$ follows a cosine function starting from $x = A$ at $t = 0$,so $x(t) = A \cos(\omega t)$.$(A)$ At $t = \frac{3T}{4}$,the displacement $x = A \cos(\omega \cdot \frac{3T}{4}) = A \cos(\frac{3\pi}{2}) = 0$. Since $F = -m\omega^2 x$,if $x = 0$,then $F = 0$. Thus,$(A)$ is true.$(B)$ At $t = T$,$x = A \cos(\omega T) = A \cos(2\pi) = A$. The acceleration $a = -\omega^2 x = -\omega^2 A$. The magnitude of acceleration is maximum at the extreme positions $(x = \pm A)$. Thus,$(B)$ is true.$(C)$ At $t = \frac{T}{4}$,$x = A \cos(\omega \cdot \frac{T}{4}) = A \cos(\frac{\pi}{2}) = 0$. The speed $v = \omega \sqrt{A^2 - x^2}$ is maximum when $x = 0$. Thus,$(C)$ is true.$(D)$ At $t = \frac{T}{2}$,$x = A \cos(\omega \cdot \frac{T}{2}) = A \cos(\pi) = -A$. At $x = -A$,$P.E.$ is maximum and $K.E. = 0$. Therefore,$P.E. 
eq K.E.$. Thus,$(D)$ is false.Therefore,statements $(A), (B),$ and $(C)$ are true.

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