# Oscillation on Angled Rails (Diff Equation)

This problem was taken from David Morin's Introduction to Classical Mechanics

My attempt at solving the problem:

First, I labeled all the relevant forces acting only on one of the particles of mass $$m$$, which were gravity and the force of the spring acting on said mass.

The forces contributing to the movement of the object along the rails were:

$$F_{\text g}=mg\cos(\theta) \\ F_{\text{spring}}=-k(l-l_i)$$ $$l$$ notates the length of the spring at any given moment while $$l_i$$ is a constant that represents the initial length of the spring in its equilibrium. $$x = 0$$ at the point where the two rails meet and $$x$$ notates the distance along the rail to the particle $$m$$. Now I shall proceed to solve the differential equation for this motion. First, I would like to invoke the law of sines to relate the length of the spring and the distance $$x$$. Since the triangle bounded by the spring is isosceles, the two identical angles would measure $$\frac{\pi}{2}-\theta$$

$$\frac{l}{\sin(2\theta)}=\frac{x}{\sin(\frac{\pi}{2}-\theta)} \\ l=\frac{2x\sin(\theta)\cos(\theta)}{\cos(\theta)} \\ l=2x\sin(\theta)$$ Now, we will move onto the differential equation. We must take the force of the spring in the direction of the rail, so we have to multiply it by cosine. $$x$$ is the current distance along the rail while $$x_i$$ is a constant that represents the initial distance of the masses from the bottom: $$\sum F=m\ddot{x}=-mg\cos\theta - 2k\sin(\theta)(x-x_i)\cos(\frac{\pi}{2}-\theta) \\ m\ddot{x} + 2kx\sin^2(\theta) = 2kx_i\sin^2(\theta) - mg\cos(\theta)$$

Now, I don't know whether I should continue to solve it like an in-homogeneous differential equation because I feel like I'm over-complicating this just to solve for the frequency. Also, the only "variables" here are $$\ddot{x}$$ and $$x$$. Everything else are constants, including the trig functions. Any help on how to move forward on this problem or another way of solving this would be high appreciated. Thank you

• Isn't the last equation is the equation for simple harmonic motion? May 4, 2020 at 18:46
• @sslucifer, I forgot to add the cosine component of the force of the spring and I edited it now. But, yes, it should be an SMH differential equation. The thing is, I don't know if I am approaching this problem in the right way
– LVST
May 4, 2020 at 18:56
• Seems like its the correct approach (well you can also use Lagrangian approach, but I think that will eventually leads up to $F=ma$), use $x(t)=Asin(\omega t)+Bcos(\omega t)$ for further solution. May 4, 2020 at 19:00
• Ok, but since the right side of the equation contains a $sin^2\theta$, does your solution $x(t)=Asin(\omega t)+Bcos(\omega t)$ still apply?
– LVST
May 4, 2020 at 19:09
• Alright, thanks for the clarification. I will try that. If you want to write down what you just said as an answer, I could check it and you could get some reputation points or whatever :) @sslucifer
– LVST
May 4, 2020 at 19:22

The last equation is just inhomogeneous differential equation for the simple harmonic motion. So use, $$x(t)=x_{inhm}(t)+Asin(\omega t)+Bcos(\omega t)$$