Question

Two identical balls $A$ and $B$ , each of mass $0.1kg$ , 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 $0.06m$ . Each spring has a natural length of $0.06\pi m$ and spring constant $0.1N{m}^{-1}$ , 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, $R=0.06m$

The natural length of spring ${\text{l}}_0=\text{0.06}\pi =\pi \text{R}$ (Half circle) and spring constant,
$k=0.1N{m}^{-1}$
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
$\mathrm{f}=\frac{1}{2\pi }\sqrt{\frac{4\times 0.1}{0.1}}=\frac{1}{\pi }\mathrm{Hz}$