# Does this system perform simple harmonic motion? [closed]

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.

• Why do you think this might be an exception to the norm for spring systems? Feb 14, 2017 at 17:53
• @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. Feb 14, 2017 at 17:58
• 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.
– user140606
Feb 14, 2017 at 17:58
• 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 Feb 14, 2017 at 17:59
• @Countto10 I went through the relevant chapters. I couldn't find too many problems on SHM and I certainly didn't find mine. Feb 14, 2017 at 18:10

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: $$x_1 = v+\frac23 u = A_2\cos\left( \sqrt{\frac{k}{m_2}} t+\delta_2\right) +\frac23 A_1\cos\left( \sqrt{\frac{k}{m_1}} t+\delta_1\right)\\ x_2 = v-\frac13 u = A_2\cos\left( \sqrt{\frac{k}{m_2}} t+\delta_2\right) -\frac13 A_1\cos\left( \sqrt{\frac{k}{m_1}} t+\delta_1\right)$$
• 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 $\frac{d^2u}{dt^2}=-\frac km u$? Feb 15, 2017 at 6:08