Question

A particle is executing SHM between points $-X_m$ and $X_m$, as shown in figure-I. The velocity $V(t)$ of the particle is partially graphed and shown in figure-II. Two points $A$ and $B$ corresponding to time $t_1$ and time $t_2$ respectively are marked on the $V(t)$ curve:

• A. At time $t_1$, it is going towards $X_m$
• B. At time $t_1$, its speed is decreasing
• C. At time $t_2$, its position lies in between $-X_m$ and $O$
• D. The phase difference $\Delta f$ between points $A$ and $B$ must be expressed as $90^{\circ }<\Delta f<180^{\circ }$

Answer 1

Answer: (B), (C)

At time $t_1$, velocity of the particle is negative i.e. going towards $-X_m$.
From the graph, at time $t_1$, its speed is decreasing.
Therefore particle lies in between $-X_m$ and $0$ .
At time $t_2$, velocity is positive and its magnitude is less than maximum i.e, it has yet not crossed $O$.
It lies in between $-X_m$ and $0.$
Phasc of particle at time $t_1$ is $\left(180+{\theta }_1\right)$
Phase of particle at time $t_2$ is $\left(270+{\theta }_2\right)$
Phase difference is $90+\left({\theta }_2-{\theta }_1\right)$
${\theta }_2-{\theta }_1$ can be negative making $\Delta f<90^{\circ }$ but can not be more than $90^{\circ }$.