One end of each of two identical springs, natural length 9 cm and force constant$K = 45 Nm$ is attached with a small particle of mass$m = 30 gm.$ Other end of right spring is fixed with a wall and other end of left spring is attached with a fixed block having a positive charge $q = 1\mu C$ as shown in Fig. The particle rests over a smooth horizontal plane and springs are non-deformed.

Deformation of springs when a positive charge  $q = 1\mu C$ is given to the particle i and equilibrium is attained is d.  aAso, frequency of small longitudinal oscillations of the particle is f .Then

When charge is even to the particle, it experiences electrical repulsion. Therefore it starts to move rightwards. Due to its motion right spring is compressed while left spring elongates. Magnitude of deformation of each spring is equal to displacement of the particle. Let this be x. Then distance of particle from fixed charge will be$(0.09 + x)$ metre.

Electrical force,$F=9\times {10}^9\frac{q^2}{(0.09 + x)}.........\left(1\right)$

Considering free body diagram of the particle, (Fig.)

For its equilibrium,$F=2Kx.....\left(2\right)$

$9\times {10}^9\frac{q^2}{(0.09 + x)}=2Kx$

$x=0.01 m or 1 cm$ 

 Let the particle be slightly displaced rightwards by $dr$. Then force in each spring becomes $K (x + dx)$ and increase d $F$ in electrical force $F$ can be calculated by differentiating equation$\left(1\right)$
$dF=9\times {10}^2{q}^2\frac{\left(-2\right)}{{(0.09 + x)}^3}=dx...\left(2\right)$

$=-18dx$

When the particle is released, it accelerates leftwards. Let this acceleration be a. Considering free body diagram of the particle (Fig.)

$m\alpha = 2K (x + dx)-(F+dF).....\left(3\right)$

From equations (2) and (3),

$m\alpha = 2Kdx-dF=108\alpha x$

or acceleration,$a=\frac{108}{m}dx$

$a=\frac{108}{30\times {10}^{-3}}dx=3600dx$

Since, direction of acceleration is opposite to that of displacement $dx$ and its magnitude is directly proportional to displacement, therefore, the particle performs SHM. Comparing equation (4) with acceleration $=$$\omega$². displacement,

Angular frequency,$\omega =\sqrt{3600}=60rad/sec$

Frequency, f =$\frac{30}{\mathrm{\pi }} Hz$