Q: One end of a spring of negligible unstretched length and spring constant k is fixed at the origin (0,0) . A point particle of mass m carrying a positive charge **q** is attached at its other end . The entire system is kept on a smooth horizontal surface . When a point dipole **p** pointing towards the charge **q** is fixed at origin , the spring gets stretched to length **l** and attains a new equilibrium position . If the point mass is now displaced slightly by Δl << l from its equilibrium position and released , it is found to oscillate at frequency $\frac{1}{\delta} \sqrt{\frac{k}{m}}$ . The value of δ is ———-

Ans: (3.14)

Solution: Electrostatic force = Spring force

$F_e = F_{sp}$

$ \frac{1}{4\pi \epsilon_0}\frac{2p q}{l^3} = k l$

Now , the mass m is displaced by Δl = x from the mean position

F_{net} = F_{sp} – F_{e}

$F_{net} = k(l+x) – \frac{1}{4\pi \epsilon_0}\frac{2p q}{(l+x)^3} $

$F_{net} = k(l+x) – \frac{1}{4\pi \epsilon_0}\frac{2p q}{l^3(1+x/l)^3} $

$F_{net} = kl + kx – \frac{1}{4\pi \epsilon_0}\frac{2p q}{l^3}(1-\frac{3 x}{l}) $

Substituting $ \frac{1}{4\pi \epsilon_0}\frac{2p q}{l^3} = k l$

$F_{net} = kx + kl (\frac{3x}{l}) = 4 k x $ ; This force is restoring in nature .

Therefore , k_{eq} = 4 k

Time period , $\displaystyle T = 2\pi \sqrt{\frac{m}{4k}}$

$\displaystyle T = \pi \sqrt{\frac{m}{k}}$

Frequency , $\displaystyle \nu = \frac{1}{\pi} \sqrt{\frac{k}{m}}$