IIT JEE , AIEEE Forum - Physics , Electricity

punnima

Let the spring compress a distance x due to the given force F. So By Work-energy theorem we have,
1\2 Mass*Velocity^2 =Fx-1\2 Kx^2+(1/2C2V^2-1\2 C1V^2) ............(1)
Where C1 =4ea^2/d and C2=xae/d +4e(a-x)a/d
We know that dielectric experiences a force inwards given by 1\2V^2ea(k-1)/d.(k=dielectric constant) ( This can be easily got by -dU=Force*dx) when F tries to displace it .
So in the equilibrium of forces, we have displacement s
F=Ks+ 1\2V^2 (3ea/d) (V=Voltage).... ..............................(2)
So we get s i.e. equilibrium position where no net force act on it but from (1) it possesses some K.E., i.e. velocity i.e. momentum. But as the other two forces remain constant in magnitude the unbalanced spring force tries to bring the dielectric in the equilibrium position Such that F(net )= -Kx where x is the displacement from the Mean Position . Hence it performs SHM.
The Amplitude can be got from (1)
0=FA-1/2 KA^2 +1/2 V^2*(-3eaA)/d
So on Solving we get
A=2F/K-(3eaV^2)/Kd
Again From (2) we have s= F/K - 3eaAV^2/2Kd
So the amplitude is A-s i.e.
A/2 == F/K - 3eaAV^2/2Kd
And Time Period is given by 2P(M/K)^1/2

punnima

do u find this answer satiable ????
for further doubts plizz nudge me

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