A particle executing SHM along a straight line has zero velocity at points $A$ and $B$ whose distance from $O$ on the same line $O A B$ are $a$ and $b$, respectively. If the velocity at the mid point between $A$ and $B$ is $v$, then its time period is

Question

A particle executing SHM along a straight line has zero velocity at points $A$ and $B$ whose distance from $O$ on the same line $O A B$ are $a$ and $b$, respectively. If the velocity at the mid point between $A$ and $B$ is $v$, then its time period is, $\frac{\pi(b+a)}{v}$, $\pi\left(\frac{b-a}{v}\right)$, $\left(\frac{b+a}{2 v}\right)$, $\left(\frac{b-a}{2 v}\right)$

MARKS App · Accepted Answer

According to the question, $\therefore$ Amplitude $=\frac{\text { Distance travelled by the particles }}{2}(A \text { to } B)$ Amplitude of particles executing simple harmonic motion (SHM) along a straight line $A B$ is $(a)=\frac{b-a}{2}$. Velocity of particle, $v=$ Amplitude $\times$ Oscillation frequency $\begin{aligned} \therefore \quad v & =a \omega \ v & =\left(\frac{b-a}{2}\right) \omega \text { or } \omega=\frac{2 v}{b-a} \quad \ldots (i) \end{aligned}$ $\therefore$ Time period, $T=\frac{2 \pi}{\omega}$ Putting the value of $\omega$ from Eq. (i) to above formula, $\Rightarrow \quad T=\frac{2 \pi}{2 v} \times(b-a) \Rightarrow T=\frac{b-a}{v} \pi$ So, the time period of a particle executing SHM along a straight line from points $A$ to $B$ is, $T=\frac{b-a}{v} \pi$

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