Question

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

• A. At time $t_1$, its position lies at $O$.
• B. At time $t_2$, its position lies at $O$.
• C. At time $t_1$, its position lies between $-X_m$ and $0$.
• D. At time $t_2$, its position lies between $-X_m$ and $0$.

Answer 1

The correct option is C At time $t_1$, its position lies between $-X_m$ and $0$.

We know that acceleration of a $SHM$ is given by, $a=-{\omega }^{2}x$ ...........(1)
where $\omega$ is angular frequency and $x$ is displacement from mean position.

From above diagram, we can see that, acceleration ( slope of v-t graph ) at time $t_1$ is positive and at time $t_2$ is negative.
Therefore, from (1), displacement at time $t_1$ is negative and at time $t_2$ is positive. Hence, particle is between $0$ and $-X_m$ at time $t_1$.