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 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. Identify the wrong statement.

• A. At time , its position lies in between and O.
• B. The phase difference between points A and B must be expressed as
• C. At time , its speed is decreasing
• D. At time it is not going towards .

Answer 1

The correct option is B The phase difference between points A and B must be expressed as

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 decreases. Therefore, the particle lies in between $-x_m$ and O.
At time $t_2$,velocity is positive and its magnitude is less than maximum i.e., it has yet not crossed O. Therefore, it lies in between $x_m$ and O.
Phase of particle at time $t_{1}is(180+{\theta }_1)$
Phase of particle at time $t_{2}is(270+{\theta }_2)$
Phase difference is $90+{\theta }_2-{\theta }_1$
${\theta }_2-{\theta }_1$ can be negative making $\Delta \varphi <{90}^{\circ }$ but cannot be more than ${90}^{\circ }$.