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

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

Answer 1

Answer: (D)

(A)$F=ma$ $a=-{\omega }^{2}x$
at $\frac{3T}{4}$ 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 $t=\frac{T}{2}x=-A$ (So not possible)