# Particles connected with spring, sliding over lines. Question about the potential energy

I'm currently studying small oscillations with the Lagrangian formalism. I stumbled upon an exercise that I can't seem to understand the method of solving it.

Two particles $$P_1$$ and $$P_2$$, with the same mass $$m$$, slide over two smooth, frictionless lines (see the figure). The two lines intersect each other at an angle of $$60$$°. The particles experience gravity and are connected by an ideal spring with natural length $$l$$ and a spring constant $$k = \frac{\sqrt{3} mg}{l}$$.

a) Prove that the equillibrium position of the system is given by $$|OP_1 | = |OP_2| = 2l$$

b) Show that this equillibrium is stable and show that the frequencies of the small oscillations along the equillibrium position are given by $$\omega_1 = \sqrt{\frac{k}{2m}}$$ and $$\omega_2 = \sqrt{\frac{3k}{4m}}$$

To find the equillibrium position I calculated the potential energy $$V$$, both gravity and the elastic forces acting on the particle due to the spring contribute to the potential. I started by choosing the movement along both lines as generalised coordinates. $$p_1$$ and $$p_2$$ as the movement of particle $$P_1$$ and $$P_2$$ respectively.

The potential energy due to gravity is given by

$$V_g(y_1, y_2) = -mg y_1 - mgy_2$$ $$V_g(p_1, p_2) = -mg p_1 \cos(30°) - mg (p_2 \cos(30°))$$ $$V_g(p_1, p_2) = -\frac{\sqrt{3}}{2} mg ( p_1 + p_2 )$$

The potential elastic energy is given by

$$V_e(p_1, p_2) = \frac{1}{2}k (L-l)^2$$

Where I denoted the total length with $$L$$ and the natural (rest) length as $$l$$. With the cosine law, the abive can be written as

$$L = \sqrt{p_1^2 + p_2^2 - 2p_1p_2 \cos(60°)}$$

$$L = \sqrt{p_1^2 + p_2^2 - p_1p_2 }$$

Such that the elastic potential energy can be written as:

$$V (p_1, p_2) = V_g (p_1, p_2) + V_e (p_1, p_2)$$ $$V(p_1, p_2) = -\frac{\sqrt{3}}{2} mg ( p_1 + p_2 ) + \frac{\sqrt{2}}{2l}mg \left( \sqrt{p_1^2 + p_2^2 - p_1p_2 } - l \right)^2$$

I then differentiated this potential, but I didn't find the correct answer. Is the potential correct like this? How can you define it in systems like this, where you have to take the natural length in account

• What do you mean by "I diferentiated this potential"? How? What equations did you get?
– nasu
Aug 11, 2021 at 14:16
• Well, to find if the equillibrium position I differentiated wrt the generalised coordinates and I tried to find the roots of the derivative. But they weren't equal to $2l$ Aug 11, 2021 at 14:27
• There are two partial derivatives and you get two coupled equations. It looks quite messy to solve them. You really did this?
– nasu
Aug 11, 2021 at 14:55

It looks like you wrote $$\sqrt{2}$$ instead of $$\sqrt{3}$$ (look back at the expression for the spring constant) in the coefficient of the second term of the potential. With this correction, we get $$\begin{split} \frac{\partial V}{\partial q_1} &= \frac{\partial}{\partial q_1}\left[\frac{\sqrt{3}}{2}mg\left(\frac{1}{l}\left(\sqrt{q_1^2 + q_2^2 - q_1q_2} - l\right)^2-q_1-q_2\right)\right] \\ &= \frac{\sqrt{3}}{2}mg\left[\frac{1}{l}\frac{\left(\sqrt{q _1^2 + q_2^2 - q_1q_2} - l\right)\left(2q_1-q_2\right)}{\sqrt{q_1^2 + q_2^2 - q_1q_2}} - 1\right]. \end{split}$$ (I've changed $$p_1$$ and $$p_2$$ to $$q_1$$ and $$q_2$$, respectively, because the $$p$$'s look like generalized momenta to me.) By symmetry, the equilibrium positions must be the same for both masses. So we can set $$q_1 = q_2 = r_\text{eq}$$. Setting the derivative of the potential to zero and simplifying, we find $$\frac{(r_\text{eq}-l)r_\text{eq}}{lr_\text{eq}} - 1 = 0 \Rightarrow \boxed{r_\text{eq} = 2l},$$ as desired.