One end of a spring of negligible unstretched length and spring...

Original frequency $f = \dfrac{1}{2\pi}\sqrt{\dfrac{k}{m}}$

If dipole appears,

At equilibrium

$\dfrac{2Kp}{{\left(ℓ + {x}_{0}\right)}^{3}} \cdot q = k{x}_{0} . . . \left(i\right)$

Now disturbed

$\dfrac{2Kp}{{\left(ℓ + {x}_{0} + x\right)}^{3}} \cdot q - k\left({x}_{0} + x\right) = ma$

$⇒ ma = \dfrac{2Kpq}{{\left(ℓ + {x}_{0}\right)}^{3}}{\left[1 + \dfrac{x}{ℓ + {x}_{0}}\right]}^{- 3} - k\left({x}_{0} + x\right)$

$⇒ ma = \dfrac{2Kpq}{{\left(ℓ + {x}_{0}\right)}^{3}}{\left[1 - \dfrac{3x}{ℓ + {x}_{0}}\right]}^{- 3} - k\left({x}_{0} + x\right) = - \dfrac{6Kpqx}{{\left(ℓ + {x}_{0}\right)}^{4}} - kx$

$= - \dfrac{3x}{ℓ + {x}_{0}}k{x}_{0} - kx = - kx\left[\dfrac{3{x}_{0}}{ℓ + {x}_{0}} + 1\right]$

As$ℓ$ is negative,   

$ma = - 4kx$;   $⇒ a = \dfrac{4k}{m}x$

New frequency :

${f}^{'} = \dfrac{1}{2\pi}\sqrt{\dfrac{4k}{m}} = 2f$ $= \dfrac{1}{\delta}\sqrt{\dfrac{k}{m}}$

$∴ \delta = \pi = 3 . 14$