Question

Two identical balls A and B each of mass 0.1 kg are attached to two identical mass less 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 centers of the balls can move in a circle of radius 0.06 m. Each spring has a natural length of 0.06 π m and force constant 0.1 N/m. Initially, both the balls are displaced by angle θ = π/6 radian with respect to the diameter PQ of the circle and released from rest. The frequency of oscillation of the ball B is

Answer 1

The frequency of oscillation of the ball B -

$\bigstar\;\boxed{\bf \dfrac{1}{\pi}}$

Explanation :

Given :

• Two identical balls A and B each of mass 0.1 kg are attached to two identical mass less springs.
• The pipe is fixed in a horizontal plane.
• The centers of the balls can move in a circle of radius 0.06 m.
• Each spring has a natural length of 0.06 π m and force constant 0.1 N/m.
• Both the balls are displaced by angle θ = $\bf \dfrac{\pi }{6}$ radiant with respect to the diameter P Q of the circle and released from rest.

To Find :

The frequency of oscillation of the ball B.

Solution :

We will use Reduced Mass concept.

We know that :

$\bigstar\;\boxed{\mu=\dfrac{M_1\;M-2}{M_1+M_2}}}$

So,

$\implies \sf \mu=\dfrac{M_1\;M-2}{M_1+M_2}\\ \\ \\\implies \sf \dfrac{m \times m}{2m}\\ \\ \underline{\underline{\textbf{Both are of same masses So,}}}\\ \\ \\\implies \sf \dfrac{m}{2}$

The value of m is given - 0.1 kg.

So,

$\implies \sf \dfrac{0.1}{2}\\ \\ \\\implies \sf 0.05\;Kg$

$\rule{300}{1.5}$

We can see in the figure that the springs are parallel.

Therefore,

The value of Keff -

$\implies \sf K +K\\ \\ \\\implies \sf 2k$

So,

The value of k in the question is  0.1 N/m.

Therefore,

$\implies \sf 2\times0.1\\ \\ \\\implies \sf 0.2\;N/m$

$\rule{300}{1.5}$

Now, The time period -

$\implies \sf T = 2\pi \sqrt{\dfrac{\mu}{K}}\\ \\ \\\implies \sf 2\pi \sqrt{\dfrac{0.05}{0.2}}\\ \\ \\\implies \sf 2\pi \sqrt{\dfrac{1}{4}}}\\ \\ \\\therefore T = \bf \pi\;seconds.$

$\rule{300}{1.5}$

We have to find the frequency.

We know that :

$\bigstar\;\boxed{\bf Frequency = \dfrac{1}{T}}}$

The value of T - π

∴ The frequency of oscillation of the ball B -

$\bigstar\;\boxed{\bf \dfrac{1}{\pi}}$

$\rule{300}{1.5}$

• Reduced Mass Concept :

When two bodies in relative motion are acted upon by a central force including Newton's law then the system can be replaced by a single mass called the reduced mass.

Answer 2

Answer:

Given:

Identical balls of mass m = 0.1 kg are attached to mass-less springs. The spring mass system is made to move inside a circular pipe in the horizontal plane.

Radius of the circle = 0.06 m .

Natural length of spring = 0.06 π m

Force constant = 0.1 N/m

Initial angular displacement = π/6 rad.

To find:

Frequency of Oscillation of ball B

Diagram:

A simplified diagram can be shown like the one attached .

Concept:

First of all , majority of the data has been provided to confuse the Answerer as they are not involved in calculation of Frequency. Those data can be used for calculation of Total Energy of system or velocity of balls.

We just need the mass , force constant data only.

Also since the circular pipe is in horizontal plane , we have to ignore gravity

Always remember this :

For binary mass and spring system which are not attached to any fixed boundary , it's always recommended to use reduced mass concept of SHM.

Calculation:

Reduced mass is denoted as μ

$\mu = \dfrac{(m1)(m2)}{(m1 + m2)}$

Since m1 = m2 = m

$= > \mu = \dfrac{(m)(m)}{(m + m)}$

$= > \mu = \dfrac{ {m}^{2} }{(2m)}$

$= > \mu = \dfrac{m}{2}$

$= > \mu = \dfrac{(0.1)}{2}$

$= > \mu = 0.05 \: kg$

Now , we need to calculate the effective spring constant . We can easily understand that the springs are in parallel combination. Hence the effective spring constant will be algebraic sum of individual spring constants.

$K_{eff} = K_{1} + K_{2}$

$= > K_{eff} = K + K$

$= > K_{eff} = 2K$

$= > K_{eff} = 2(0.01)$

$= > K_{eff} = 0.02 \: N {m}^{ - 1}$

Now , Time period will be :

$T = 2\pi \sqrt{ \dfrac{ \mu}{K_{eff}} }$

$= > T = 2\pi \sqrt{ \dfrac{ 0.05}{0.2} }$

$= > T = 2\pi \sqrt{ \dfrac{1}{4} }$

$= > T = \: \dfrac{2\pi}{2}$

$= > T = \: \pi \: sec$

We know frequency is reciprocal of time period :

$\bigstar \: \: freq. = \dfrac{1}{\pi} \: hz$