Two small identical balls lying on a horizontal plane are connected by a weightless spring. One ball (ball 2) is fixed at O, and the other (ball 1) is free. The balls are charged identically as a result of which the spring length increases `eta=2` times. Determine the change in frequency.

Explanation

When the balls are unchangeed
`v_(0)=(1)/(2 pi)sqrt((k)/(m))`
where k is the force constant of the spring and ma is the mass of the oscillating ball (ball 1). When the balls are charged, ball 1 will oscillate about the new equilibrium position. At the equilibrium position of ball 1.
`(1)/(4 pi epsilon_(0))(q^(2))/((eta l)^(2))=k(etal-l)=k(eta-1)`
or `l^(3)=(q^(2))/(4 pi epsilon_(0)eta^(2)(eta-1)k)` ...(i)
When ball 1 is displaced by a small distance from the equilibrium position to the right. the unbalanced froce to the reight is given by
`F=(1)/(4 pi epsilon_(0))(q^(2))/((eta l+x)^(2))-k(eta l+x-1)`
From Newton law, we have
`m(d^(2)x)/(dt^(2))=(1)/(4 pi epsilon_(0))(q^(2))/(eta^(2)l^(2))[1+(x)/(eta l)]^(2)-kl(eta-1)-kx`
`=(1)/(4 pi epsilon_(0))(q^(2))/(eta^(2)l^(2))[1-(2x)/(eta l)]-kl(eta-1)-kx`
`=-[(1)/(4 pi epsilon_(0))(2q^(2))/(eta^(3)l^(3))+k]x`
From Eqs (i) and (ii), `m(d^(2)x)/(dt^(2))=-[(2(eta-1))/(eta)k+k]x=-(3 eta-2)/(eta)kx`
or `(d^(2)x)/(dt^(2))=-(3 eta-2)/(eta)(k)/(m)` or `a =-[(3eta-2 k)/(meta)]x`
Comparing with `a=-omega^(2)x`,
(i) `omega^(2)=(3 eta-2)/(eta)(k)/(m)` or `v=(1)/(2pi)sqrt(((3 eta-2)/(eta))(k)/(m))`
(ii) or `(v)/(v_(0))=sqrt((3 eta-2)/(eta))`
(iii) Thus the frequency is increased `sqrt((3 eta-2)//eta)` times . Hence `eta=2` and so frequency increases `sqrt(2)` times .