Q. Two identical balls $A$ and $B$ each of mass $0.1\ kg$ are attached to two identical massless springs. The spring mass system is constrained to move inside a rigid smooth pipe bent in the form of circle as shown in the figure. The pipe is fixed in a horizontal plane. The centres of the balls can move in a circle of radius $0.06\ m.$ Each spring has a natural length of $0.06\pi\ m$ and force constant $0.1N/m.$ Initially both the balls are displaced by an angle $\theta = \pi/6$ radian with respect to the diameter $PQ$ of the circle and released from rest. The frequency of oscillation of the ball $B$ is
A) $\pi\ Hz$
B) $\frac{1}{\pi}Hz$
C) $2\pi\ Hz$
D) $\frac{1}{2\pi}Hz$
Explanation
(b) As here two masses are connected by two springs, this problem is equivalent to the oscillation of a reduced mass $m_{r}$ of a spring of effective spring constant
$T = 2\pi\sqrt{\frac{m_{r}}{K_{eff.}}}$
Here$\ m_{r} = \frac{m_{1}m_{2}}{m_{1} + m_{2}} = \frac{m}{2} \Rightarrow K_{eff.} = K_{1} + K_{2} = 2K$
$\therefore n = \frac{1}{2\pi}\sqrt{\frac{K_{eff.}}{m_{r}}} = \frac{1}{2\pi}\sqrt{\frac{2K}{m} \times 2} = \frac{1}{\pi}\sqrt{\frac{K}{m}} = \frac{1}{\pi}\sqrt{\frac{0.1}{0.1}} = \frac{1}{\pi}Hz$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.