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Two bodies with springs

Does the above system perform SHM when displaced slightly?

Mathematically I couldn't prove that it will, though generally such spring systems do perform SHM.

Is this just an exception?

P.S. You can ignore gravity.

P.S.2 This is not a homework question; I was just solving some problems on SHM and this popped into my head.

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  • $\begingroup$ Why do you think this might be an exception to the norm for spring systems? $\endgroup$ Feb 14, 2017 at 17:53
  • $\begingroup$ @sammygerbil Because I couldn't prove it mathematically. Suppose the central mass is displaced by x and the other by y. If I write the instantaneous acceleration of both particles, there is no indication that SHM will occur. $\endgroup$ Feb 14, 2017 at 17:58
  • $\begingroup$ There is a free PDF of Mary Boas's Math for the Physical Sciences, (sorry,my tablet won't copy URLs), that covers modes of oscillation of systems just like your drawing. $\endgroup$
    – user140606
    Feb 14, 2017 at 17:58
  • $\begingroup$ x and y are both dependent on each other due to d2x/dt2 and d2y/dt2. But for SHM we need a single variable x such that d2x/dt2 proportional to -x $\endgroup$ Feb 14, 2017 at 17:59
  • $\begingroup$ @Countto10 I went through the relevant chapters. I couldn't find too many problems on SHM and I certainly didn't find mine. $\endgroup$ Feb 14, 2017 at 18:10

2 Answers 2

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It would depend what do you mean by SHM. It seems from your comment that you associate a personal meaning to the term: "for SHM we need a single variable x such that d2x/dt2 proportional to -x".

The system shown has two degrees of freedom so it is described by two variables and two differential equations which in general are coupled (both variables show in each equation). However you can decouple the two equations by a change of variables (normal variables) so you will have two independent differential equations, describing the two "normal modes". So if you want to have only one variable to call it SHM then this system it is not one. And the general motion is not described by a simple sine or cosine. However, any motion of the system can be described as a superposition of normal modes and in each normal mode the motion of a specific object (ball here) looks pretty much like a SHM.

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This problem, with arbitrary masses, can indeed be reduced to a system of two coupled simple harmonic oscillators. Let's see how the equations of motion are solved, to give some insight to the solutions themselves:

Let the masses be$m_1$ for the upper ball and $m_2$ for the lower ball, let the heights of the balls be $y_1$ and $y_2$, and let the equilibrium positions of the two balls be at heights $Y_1$ and $Y_2$, respectively. It is easy to find the values of $Y_1$ and $Y_2$, and they will have little to due with the nature of the simple harmonic motion so I leave that to you. Let $x_1 = y_1 - Y_1$ and $x_2 = y_2 - Y_2$; these are the deviation from the equilibrium heights, and the restoring forces (effective spring tensions) from the combinations of springs and gravity are $$T_1 = -k(x_1-x_2)\\ T_2 = -kx_2 $$ The equations of motion are $$ \frac{d^2x_1}{dx_1^2} = -\frac{k}{m_1}(x_1-x_2) \\ \frac{d^2x_2}{dx_2^2} = \frac{1}{m_2}\left( k(x_1-x_2) -kx_2 \right) = -\frac{k}{m_2}(2x_2-x_1) $$ Now here is the trick for separating these equations into two simple harmonic oscillators: Make the substitution $$ u = ax_1+bx_2\\v = cx_1+dx_2 $$ choosing $a,b,c,d$ cleverly so that $\frac{d^2u}{dt^2}$ contains no mention of $v$ and $\frac{d^2v}{dt^2}$ contains no mention of $u$. $$ u=x_2-x_2\\v=\frac13 x_1+\frac23 x_2 \\ x_1 = v+\frac23 u\\x_2 = v-\frac13 u \\ \frac{d^2u}{dt^2} = -\frac{k}{m_1} u \\ \frac{d^2v}{dt^2} = -\frac{k}{m_2} v $$ So the solution is that $x_1$ and $x_2$ are each the sum of two harmonic oscillators: $$ x_1 = v+\frac23 u = A_2\cos\left( \sqrt{\frac{k}{m_2}} t+\delta_2\right) +\frac23 A_1\cos\left( \sqrt{\frac{k}{m_1}} t+\delta_1\right)\\ x_2 = v-\frac13 u = A_2\cos\left( \sqrt{\frac{k}{m_2}} t+\delta_2\right) -\frac13 A_1\cos\left( \sqrt{\frac{k}{m_1}} t+\delta_1\right) $$

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  • $\begingroup$ Thank you so much. This was just what I needed. Not something that immediately strikes, but maybe I could have thought of that. $\endgroup$ Feb 14, 2017 at 21:21
  • $\begingroup$ Ok wait. I retried solving the question myself. Is $u=x_2-x_1$ or $u=x_1-x_2$? In either case, how is $\frac{d^2u}{dt^2}=-\frac km u$? $\endgroup$ Feb 15, 2017 at 6:08

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