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  • A circular spring of natural length  is cut and welded with two beads of masses is cut and welded with two beads of masses <img alt="m_{1}" src

A circular spring of natural length l_{0} is cut and welded with two beads of masses is cut and welded with two beads of masses m_{1} and m_{2} each such that the ratio of the original spring is k then find the frequency of oscillation of the heads in a smooth horizontal rigid tube. Assume m_{1}=m and m_{2}=3 m^{\prime}.

   

                         

Option: 1

25 \sqrt{\frac{k}{3m}}


Option: 2

5 \sqrt{\frac{k}{3m}}


Option: 3

5 \sqrt{\frac{k}{m}}


Option: 4

\frac{5}{3} \sqrt{\frac{k}{m}}


Answers (1)

best_answer

When m_1  is displaced x, each spring will be deformed by same amount. Hence, the springs are connected in parallel. The equivalent spring constant is

                                k_{e q}=k_{1}+k_{2}

 

                

\\ \text{If the spring is cut, the force constant of spring} \propto \frac{1}{l} \\ \Rightarrow k_{1} l_{1}=k_{2} l_{2}=k l \\ \text{Substituting} \ l_{1}=l / 5 \ and \ l_{2}=4 / 5, \text{we have} \ \ k_{1}=5 k \ and \\ k_{2}=\frac{5}{4} k \\ Then \ \ k_{e q}=\frac{25}{4} k \\ \text{Now we have two partical of masses} \ m_{1} \ and \ m_{2} \ \\ \text{and one spring of stiffness} \ k_{e q}=\frac{25}{4} k \\ \text{The reduced mass is } \mu=\frac{m_{1} m_{2}}{\left(m_{1}+m_{2}\right)} \\ where \ m_{1}=m \ and \ m_{2}=3 m \\ \text{This gives} \ \ \mu=3 / 4 m \\ \ Substituting \ \mu=3 / 4 m \ and \ k_{e q}=25 / 4 k \ in \ the \ formula

\omega=\sqrt{\frac{k_{e q}}{\mu}} \Rightarrow \omega=\sqrt{\frac{\frac{25}{4} k}{\frac{3}{4} m}}=\sqrt{\frac{25 k}{3 m}} = 5 \sqrt{\frac{k}{3m}}

Posted by

vishal kumar

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