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 $t=\frac{3 T}{4}$
$(B)$ 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 $t=\frac{T}{2}$
JEE MAIN 2020, Diffcult
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Sol. $(A)$ $F =$ ma $\quad a =-\omega^{2} x$
at $\frac{3 T }{4}$ displacement zero $( x =0),$ so $a =0$
$F=0$
$(B)$ at $t=T \quad$ displacement $(x)=A$ $x$ maximum, So acceleration is maximum.
$(C)$ $V =\omega \sqrt{ A ^{2}- x ^{2}}$
$V _{\max }$ at $x =0$
$V _{\max }= A \omega$
at $t =\frac{ T }{4}, x =0, \quad$ So $V _{ max }$
$(D)$ $KE = PE$
$\therefore$ at $x=\frac{A}{\sqrt{2}}$
at $t=\frac{T}{2} \quad x=-A \quad$ (So not possible)
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