- #1

YellowBiro

- 8

- 1

Thread moved from the technical forums, so no Homework Template is shown

$$m_1 \ddot{x} - m_1 g + \frac{k(d-l)}{d}x=0$$

$$m_2 \ddot{y} - m_2 \omega^2 y + \frac{k(d-l)}{d}y=0$$

It is two masses connected by a spring. ##d=\sqrt{x^2 + y^2}## and ##l## is the length of the relaxed spring (a constant).

What is the strategy to solve such a system? I tried substituting one in the other and got

$$y m_1\ddot{x} -m_1gy=xm_2\ddot{y}-m_2\omega^2 yx$$

I don't know how to continue from here. Can you maybe keep one coordinate constant and solve for the other? Doesn't seem to make much sense.

Also the question in the exam says, "Determine the two equilibrium solutions" but I presume, you first have to find the solutions first, right?

$$m_2 \ddot{y} - m_2 \omega^2 y + \frac{k(d-l)}{d}y=0$$

It is two masses connected by a spring. ##d=\sqrt{x^2 + y^2}## and ##l## is the length of the relaxed spring (a constant).

What is the strategy to solve such a system? I tried substituting one in the other and got

$$y m_1\ddot{x} -m_1gy=xm_2\ddot{y}-m_2\omega^2 yx$$

I don't know how to continue from here. Can you maybe keep one coordinate constant and solve for the other? Doesn't seem to make much sense.

Also the question in the exam says, "Determine the two equilibrium solutions" but I presume, you first have to find the solutions first, right?