1
$\begingroup$

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} $.

enter image description here

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

$\endgroup$
3
  • $\begingroup$ What do you mean by "I diferentiated this potential"? How? What equations did you get? $\endgroup$
    – nasu
    Aug 11, 2021 at 14:16
  • $\begingroup$ 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 $2l$ $\endgroup$
    – CedricL
    Aug 11, 2021 at 14:27
  • $\begingroup$ There are two partial derivatives and you get two coupled equations. It looks quite messy to solve them. You really did this? $\endgroup$
    – nasu
    Aug 11, 2021 at 14:55

1 Answer 1

4
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.