Question

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 a 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 spring constant $0.1 \, N \, m^{- 1}$ , Initially, both the balls are displaced by an angle $\theta \, =\frac{\pi }{6} \, rad$ with respect to the diameter $ \, PQ \, $ of the circle (as shown in the figure) and released from rest.

The frequency of oscillation of balls is

• A. $\frac{1}{2 \pi } \text{ Hz}$
• B. $\frac{1}{3 \pi } \text{ Hz}$
• C. $\frac{1}{\pi } \text{ Hz}$
• D. $\frac{1}{4 \pi } \text{ Hz}$

Answer 1

Answer: (C)

Given, Mass of each block A and B, m = 0.1 kg
Radius of circle, $R=0.06 \, m$

The natural length of spring $\textit{l}_{0} = \text{0.06} \pi = \pi \text{R}$ (Half circle) and spring constant,
$k=0.1 \, N \, m^{- 1}$
In the stretched position elongation in each spring
$\text{x} = \text{R} \theta $
Let us draw FBD of A

Spring in lower side is stretched by 2x and on upper side compressed by 2x. Therefore, a force 2kx on each block would be exerted by each spring.
Hence, a restoring force, F = 4kx will act on A in the direction shown in figure.
Restoring torque of this force about origin
$\tau = - \text{F. R} = - \left(4 \text{kx}\right) \text{R} = - \left(4 \text{kR} \theta \right) \text{R}$
or $\tau = - 4 \text{kR}^{2} \cdot \theta $ ..... (i)
Since, $\tau \propto - \theta $ , each ball executes angular SHM about origin O.
Equation. (i) can be written as
$ \mathrm{I} \alpha=-4 \mathrm{kR}^2 \theta $
or $\left(\mathrm{mR}^2\right) \alpha=-4 \mathrm{kR}^2 \theta$
or $\alpha=-\left(\frac{4 \mathrm{k}}{\mathrm{m}}\right) \theta$
Frequency of oscillation, $\mathrm{f}=\frac{1}{2 \pi} \sqrt{\frac{ { acceleration }}{ { displacement }}}$
$ =\frac{1}{2 \pi} \sqrt{|\frac{\alpha}{\theta}|} \mathrm{f} =\frac{1}{2 \pi} \sqrt{\frac{4 \mathrm{k}}{\mathrm{m}}} $
Substituting the values, we have
$ \mathrm{f}=\frac{1}{2 \pi} \sqrt{\frac{4 \times 0.1}{0.1}}=\frac{1}{\pi} \mathrm{Hz} $