One end of a spring of negligible unstretched length and spring constant is fixed at the origin . A point particle of mass carrying a positive charge is attached at its other end. [2010]

Question

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 $\vec{p}$ pointing towards the charge $q$ is fixed at the origin, the spring gets stretched to a length $l$ and attains a new equilibrium position (see figure below). If the point mass is now displaced slightly by $Δ l ≪ l$ from its equilibrium position and released, it is found to oscillate at frequency $\frac{1}{δ} \sqrt{\frac{k}{m}} .$ The value of $δ$ is ________. [2010]

Smart Achievers · Accepted Answer

(3.14)

$Original frequency, f = \frac{2}{δ} \sqrt{\frac{K}{m}}$

$If dipole appears at equilibrium,$

$q$

$When displaced towards right by length x_{0}$

$f_{net} = \frac{2 K P}{{(ℓ + x_{0} + x)}^{3}} \cdot q - K (x_{0} + x)$

$m a = \frac{2 K P q}{{(ℓ + x_{0})}^{3}} {[1 + \frac{x}{ℓ - x_{0}}]}^{3} - K (x_{0} + x)$

$= \frac{2 K P q}{{(ℓ + x_{0})}^{3}} {[1 - \frac{3 x}{ℓ + x_{0}}]}^{3} - K (x_{0} + x)$

$= \frac{6 K P q x}{{(ℓ + x_{0})}^{4}} k x = \frac{3 x}{ℓ + x_{0}} K x_{0} - K x = - K x [\frac{3 x_{0}}{ℓ + x_{0}} + 1]$

$As' ℓ' is negative, m a = - 4 K x$ $\Rightarrow a = \frac{- 4 K}{m} x$

$New frequency, f' = \frac{1}{2 π} \sqrt{\frac{4 K}{m}} = 2 f = \frac{1}{π} \sqrt{\frac{K}{m}}$

$∴$ $δ = π = 3.14$

Solution

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