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 radis 0.06 m. Each sping has a natural lenght of $0.006\pi m$ and force constant of $0.1N/m$. 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. equal to frequency of oscillation of ball A

Answer 1

Answer: (B), (D)

The correct options are
B $\frac{1}{\pi }Hz$
D equal to frequency of oscillation of ball A

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, $m_r=\frac{m_1{m}_2}{m_1+m_2}=\frac{m}{2}\Rightarrow K_{eff}=K_1+K_2=2K$
$\therefore n=\frac{1}{2\pi }\sqrt{\frac{K_{eff}}{m_r}}=\frac{1}{2\pi }\sqrt{\frac{2K}{m}\times 2}n=\frac{1}{\pi }\sqrt{\frac{0.1}{0.1}}=\frac{1}{\pi }Hz$.