In the given figure ABC is a non-conducting semi-circular wire of radius ‘a’ carrying a total charge Q uniformly distributed on it and a point charge q is at its centre. Ends of the wire are attached to two separate springs each having spring constant k as shown in the figure. In the given position system is in equilibrium. Now the point charge is suddenly removed. The amplitude of the resulting oscillation is $\frac{x q Q}{8 π^{2} ε_{0} a^{2} k}$ . Ignoring any force other than spring force and electrostatic force find the value of ‘x’?

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answer is 2.

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Detailed Solution

$\begin{array}{l} F = \int d F \cos θ \\ = \int \frac{1}{4 π ε_{0}} \frac{q λ R (d θ)}{R^{2}} \cos θ \\ = \frac{1}{4 π ε_{0}} \frac{q λ}{R} (\sin θ)_{- π / 2}^{π / 2} \\ = \frac{2 q λ}{4 π ε_{0} R} \\ \Rightarrow 2 k A = \frac{2 q λ}{4 π ε_{0} R} \Rightarrow k A = \frac{q}{4 π ε_{0} R} \frac{Q}{π R} \\ \Rightarrow A = \frac{Q q}{4 π^{2} ε_{0} a^{2} k} (∵ R = a) \\ \Rightarrow A = \frac{2 Q q}{8 π^{2} ε_{0} a^{2} k} \\ \Rightarrow x = 2 \end{array}$