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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.

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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

  • $\begingroup$ Isn't the last equation is the equation for simple harmonic motion? $\endgroup$
    – sslucifer
    May 4, 2020 at 18:46
  • $\begingroup$ @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 $\endgroup$
    – LVST
    May 4, 2020 at 18:56
  • $\begingroup$ 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. $\endgroup$
    – sslucifer
    May 4, 2020 at 19:00
  • $\begingroup$ 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? $\endgroup$
    – LVST
    May 4, 2020 at 19:09
  • 1
    $\begingroup$ 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 $\endgroup$
    – LVST
    May 4, 2020 at 19:22

1 Answer 1


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

for further calculations.


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