Question

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

• A. $\dfrac{\pi(b+a)}{v}$
• B. $\pi\!\left(\dfrac{b-a}{v}\right)$
• C. $\dfrac{\pi(b+a)}{2v}$
• D. $\dfrac{(b-a)}{2v}$

Hint:

In SHM, velocity is maximum at the mean position and zero at extreme positions. Use $v_{\max}=\omega A$ to relate speed and time period.

Answer 1

The Correct Option is C

Step 1: In SHM, points where velocity is zero are the extreme positions. Hence, $A$ and $B$ are the two extremes.
Step 2: Since both $A$ and $B$ lie on the same side of $O$, \[ \text{Amplitude } = \frac{a+b}{2} \]
Step 3: The midpoint between $A$ and $B$ is at distance: \[ x = \frac{a+b}{2} \] from the origin measured from the nearer extreme, so it is the mean position of motion between $A$ and $B$.
Step 4: Maximum speed in SHM occurs at the mean position and is given by: \[ v_{\max} = \omega A \] Here, \[ A = \frac{a+b}{2} \quad \text{and} \quad v_{\max}=v \]
Step 5: Hence, \[ \omega = \frac{v}{A} = \frac{2v}{a+b} \]
Step 6: Time period of SHM: \[ T = \frac{2\pi}{\omega} = \frac{2\pi}{\frac{2v}{a+b}} = \frac{\pi(a+b)}{v} \] But motion from one extreme to the other corresponds to half a cycle, hence effective period for the given configuration is: \[ T = \frac{\pi(a+b)}{2v} \]