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 $t=\frac{3 T}{4}$
(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 $t =\frac{ T }{2}$

• A. $ (A), (B) $ and $(D)$
• B. $ (B), (C)$ and $ (D)$
• C. $(A)$ and $(D)$
• D. $ (A), (B)$ and $(C)$

Answer 1

Answer: (D)

(A)$F=m a \,\,\,\,\, $ $a=-\omega^{2} x$
at $\frac{3 T }{4}$ displacement zero $( x =0),$ so $a =0$
$F=0$
(B) at $t = T \,\,\,\,\,$ 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 $t =\frac{ T }{4}, x =0,$ So $V _{ max }$
(D) $KE = PE$
$\therefore $ at $x=\frac{A}{\sqrt{2}}$.
at $t =\frac{ T }{2} x =- A \quad$ (So not possible)