Solution :
We assume that there is a ring having a charge +Q and radius r. Electric field due to the ring at a point P on the axis is given by :
If we put an electron on point P, then force on point e is :
![F= \frac{-eKQx}{(r^2+x^2)^{3/2}}= \frac{-eKQx}{r^3[1+\frac{x^2}{r^2}]^{3/2}}](https://tex.z-dn.net/?f=F%3D%20%5Cfrac%7B-eKQx%7D%7B%28r%5E2%2Bx%5E2%29%5E%7B3%2F2%7D%7D%3D%20%5Cfrac%7B-eKQx%7D%7Br%5E3%5B1%2B%5Cfrac%7Bx%5E2%7D%7Br%5E2%7D%5D%5E%7B3%2F2%7D%7D)
If r >> x , then 
Then, 
Compare, a = -ω²x
We get,
<h2>Answer:</h2>
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