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Saitama
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Homework Statement
Homework Equations
The Attempt at a Solution
(see attachment 3)
If the middle charge is moved a y distance, then the other two move a distance y/2 in opposite direction. Similarly, the velocity in y direction of other two can be also calculated. As the rods are rigid, the component of velocities along the rod should be equal i.e
[tex]v_x\cos \theta-\frac{v}{2}\sin\theta=v\sin\theta \Rightarrow v_x=\frac{3v}{2}\tan\theta[/tex]
Calculating the energy of system and differentiating it w.r.t time and setting the derivative equal to zero gives the time period.
[tex]E=\frac{kq^2}{l}+\frac{kq^2}{l}+\frac{kq^2}{2l\cos\theta}+\frac{1}{2}mv^2+2 \times \frac{1}{2}m(v_x^2+\frac{v^2}{4})[/tex]
[tex]\Rightarrow E=\frac{kq^2}{l}+\frac{kq^2}{l}+\frac{kq^2}{2l\cos\theta}+\frac{3}{4}mv^2+\frac{9mv^2}{4}\tan^2\theta[/tex]
Since y<<<l,
[tex]\frac{1}{\cos\theta}=1/(\sqrt{1-\frac{9y^2}{4l^2}})=1+\frac{9y^2}{8l^2}[/tex]
I don't understand how should I write ##\tan^2\theta##
[tex]\tan\theta=\cfrac{\cfrac{3y}{2}}{\sqrt{1-\frac{9y^2}{4l^2}}}[/tex]
[tex]\tan^2\theta=\frac{9x^2}{4}\left(1-\frac{9y^2}{4l^2}\right)^{-1}[/tex]
[tex]\Rightarrow \tan^2\theta = \frac{9x^2}{4}\left(1+\frac{9y^2}{4l^2}\right)[/tex]
But I don't think substituting this in the energy expression is a good idea because I will end up with terms consisting ##v^2x^2## and ##v^2x^4##.
Any help is appreciated. Thanks!
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