Oscillation of system of three charges

[Bling Fizikst]

Homework Statement

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Relevant Equations

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I tried to take angles and proceed by energy conservation

But this doesn't seem to lead me anywhere .
Here , the length of threads is $l$ each and $2\theta$ is the central angle. $y_1$ is the displacement of the charges attached at the extreme ends of the threads respectively while $y$ is the displacement of the middle charge .

 

[TSny]

If possible, please type your equations using Latex rather than post images of your hand-written work. I'm having difficulty deciphering parts of your diagram and some of the terms in your equations.

Using energy is a nice way to approach the problem. Can you find a simple relationship between $y$ and $y_1$, where $y$ is the displacement from equilibrium of the central particle and $y_1$ is the displacement of the left and right particles? Hint: Think about the center of mass of the system.

The third term in your energy equation doesn't look right if it's meant to represent the potential energy between the two outer particles. Since the oscillations are small, I would express the horizontal distance $x$ between the end particles in terms of a small angle instead of the large angle $\theta$ in your diagram. You can then use small-angle approximations.

Do the end particles move horizontally as well as vertically? If so, you will either need to include this horizontal motion in the kinetic energy of the system or show that it can be neglected.

 

[Bling Fizikst]

Intuitively , the center of mass of the system should be the origin . Hence , $$x_{\text{cm}}=0$$ $$y_{\text{cm}}=\frac{my_1+m(-y)+my_1}{3m}=0\implies y_1=\frac{y}{2}$$ Writing the energy conservation equation : $$E=\frac{kq^2}{l}\cdot 2 + \frac{kq^2}{2l\sin\beta} + 2\cdot \frac{1}{2}m \dot{y_1}^2+\frac{1}{2}m\dot{y}^2$$ Differentiating and re-arranging gives : $$ \frac{3m}{2}\dot{y}\ddot{y}=\frac{kq^2}{2l}\cot\beta \csc \beta \dot{\beta}$$ Now using the relation : $$l\cos\beta = y_1+y = \frac{3y}{2}$$ Using this to eliminate $\dot{\beta}$ , we get a trigonometric mess . I did try to move ahead with it assuming $\beta$ to be small and $y<<l$ but i got something of the form : $$\ddot{y}=ay^2+b$$ which i do not know how to deal with .

 

[TSny]

Intuitively , the center of mass of the system should be the origin . Hence , $$x_{\text{cm}}=0$$ $$y_{\text{cm}}=\frac{my_1+m(-y)+my_1}{3m}=0\implies y_1=\frac{y}{2}$$

That looks good. Does the position of the center of mass move during the oscillations? Why or why not?

Writing the energy conservation equation : $$E=\frac{kq^2}{l}\cdot 2 + \frac{kq^2}{2l\sin\beta} + 2\cdot \frac{1}{2}m \dot{y_1}^2+\frac{1}{2}m\dot{y}^2$$

OK. But, since the oscillations are small, you can see that the angle $\beta$ will only deviate a small amount from 90^o. So, $\beta$ will never be small. If you let $\alpha$ be the complement of $\beta$, will $\alpha$ be small? Can you express $x$ in terms of $\alpha$ instead of $\beta$?

Alternatively, you could try to express $x$ directly in terms of $y$ using geometry. Then you could make an approximation for small $y$.

 

[Bling Fizikst]

That looks good. Does the position of the center of mass move during the oscillations? Why or why not?


OK. But, since the oscillations are small, you can see that the angle $\beta$ will only deviate a small amount from 90^o. So, $\beta$ will never be small. If you let $\alpha$ be the complement of $\beta$, will $\alpha$ be small? Can you express $x$ in terms of $\alpha$ instead of $\beta$?

Alternatively, you could try to express $x$ directly in terms of $y$ using geometry. Then you could make an approximation for small $y$.

We can observe that there are two right triangles formed in the diagram each having a hypotenuse of $l$ and legs of $\frac{x}{2} , y+y_1$ where $y_1=\frac{y}{2}$ . By pythagoras theorem , $$l^2=\frac{x^2}{4}+(y+y_1)^2\implies x=\sqrt{4l^2-9y^2}$$ We can assume $y<<l$ and approximate to get $$x=2l\left(1-\frac{9y^2}{8l^2}\right)\implies \frac{1}{x}=\frac{1}{2l}\left(1+\frac{9y^2}{8l^2}\right)$$ Writing the total energy equation : $$E=\frac{2kq^2}{l}+\frac{kq^2}{x}+ \frac{3}{4}m\dot{y}^2$$ Differentiating and re-arranging to get : $$ m\ddot{y}=-\frac{3q^2}{16\pi\epsilon_{\circ}l^3}$$ $$\implies \omega=\sqrt{\frac{3q^2}{16\pi\epsilon_{\circ} ml^3}} $$ $$\boxed{T=\frac{8\pi}{q}\sqrt{\frac{\pi\epsilon_{\circ} ml^3}{3}}}$$

 

[TSny]

We can observe that there are two right triangles formed in the diagram each having a hypotenuse of $l$ and legs of $\frac{x}{2} , y+y_1$ where $y_1=\frac{y}{2}$ . By pythagoras theorem , $$l^2=\frac{x^2}{4}+(y+y_1)^2\implies x=\sqrt{4l^2-9y^2}$$ We can assume $y<<l$ and approximate to get $$x=2l\left(1-\frac{9y^2}{8l^2}\right)\implies \frac{1}{x}=\frac{1}{2l}\left(1+\frac{9y^2}{8l^2}\right)$$ Writing the total energy equation : $$E=\frac{2kq^2}{l}+\frac{kq^2}{x}+ \frac{3}{4}m\dot{y}^2$$ Differentiating and re-arranging to get : $$ m\ddot{y}=-\frac{3q^2}{16\pi\epsilon_{\circ}l^3}$$ $$\implies \omega=\sqrt{\frac{3q^2}{16\pi\epsilon_{\circ} ml^3}} $$ $$\boxed{T=\frac{8\pi}{q}\sqrt{\frac{\pi\epsilon_{\circ} ml^3}{3}}}$$

That all looks good to me.

[Edit: There's a typographical error in the equation $m\ddot{y}=-\dfrac{3q^2}{16\pi\epsilon_{\circ}l^3}$. It is missing the factor of $y$ on the right-hand side.]

 

[TSny]

Since $x$ varies with time, you might want to show why it’s not necessary to include $\dot x$ in the kinetic energy of the system.

 

[Bling Fizikst]

Since $x$ varies with time, you might want to show why it’s not necessary to include $\dot x$ in the kinetic energy of the system.

I guess it is because variation of $x$ with time is quite small . Hence , $\dot{x}$ is small , which implies that the second ordered term $\dot{x}^2\approx 0$ meaning the kinetic energy should be $0?$

 

[TSny]

I guess it is because variation of $x$ with time is quite small . Hence , $\dot{x}$ is small , which implies that the second ordered term $\dot{x}^2\approx 0$ meaning the kinetic energy should be $0?$

Yes, that's the idea. To show it quantitatively, use your approximation $x = 2l\left(1-\dfrac{9y^2}{8ml^2}\right)$.

Show $\dot x$ is of second order in small quantities. Hence, $\dot x^2$ is of fourth order in small quantities. But, you only need to keep quantities to second order in the energy equation when deriving the first order equation of motion.