# Forced harmonic oscillator with two springs

Consider a vertical system of two springs in series, with a mass(50 g) between them. From below the system is driven by a vibration generator. The setup is shown here, but the picture is taken while the mass was moving, and therefore the mass is not in the equilibrium position: I expected the system to act close to a system of springs coupled in series, but it does not. For instance, there is a big difference in resonance frequency between a setup with the spring with elastic constant $K_1=15\:\mathrm{N}/\mathrm{m}$(m=3.876 g) on top and the one with $K_2=5\:\mathrm{N}/\mathrm{m}$(m=3.070 g) below, and the setup where these two springs change place.

EDIT: The results are shown under: Why is this? In the equilibrium position the two springs are not streched equally. Is that the reason?

• Can you confirm that both springs are in tension at all times? – Emilio Pisanty May 12 '14 at 9:49
• They are. When I change the height of the system,I always make sure that both the springs are streched. – user46330 May 12 '14 at 11:02
• Have you checked if you are in the linear region of the springs? – Davidmh May 12 '14 at 22:02
• What are the masses of the two springs and the weight in the middle between the two springs? – Qmechanic May 12 '14 at 22:06
• Did you repeat the violet experimental curve, which doesn't show a sharp resonance peak? – Qmechanic May 14 '14 at 17:35

The Lagrangian of the system is, $$L = \frac{1}{2}m\dot{x_1}^2 - \frac{1}{2}k_1(l_1 - l_0+x_1)^2-\frac{1}{2}k_2(l_2-l_0+x_2-x_1)^2+mgx_1$$ Here, $x_1$ is the downward distance from the equilibrium position of the mass, $x_2$ is the downward distance from the midpoint of the driving oscillator, $k_1$ and $k_2$ are spring constants of the top and bottom springs respectively, $l_0$ is the unstretched length of both springs, $l_1$ and $l_2$ are stretched lengths at equilibrium of the top and bottom springs respectively. The condition for equilibrium, when the driving force is not applied, is $$k_1(l_1-l_0) = k_2(l_2-l_0)+mg$$ We can take $x_2=a\ cos(\omega t)$ for the driving oscillation and then the equation of motion for the mass from Euler-Lagrange equation would be, $$m\ddot{x_1}+(k_1+k_2)x_1-ak_2cos(\omega t) = 0$$ Solving this equation, we can get the amplitude of the driven oscillation to be, $$A=\frac{ak_2}{k_1+k_2-m\omega^2}$$ As a result, the resonant frequency should be the same for both setups with springs interchanged as $\omega_r=\sqrt{(k_1+k_2)/m}$. The proportion between the resonant frequencies in your setups should be $\omega_{r(5,5)}:\omega_{r(5,15)}:\omega_{r(15,15)}=1:\sqrt{2}:\sqrt{3}$ which is apparent in your results. However, based on theoretical calculations the resonant frequencies should be the same for setups with just the springs interchanged.

Now, in this calculation damping is ignored while in all real systems damping is inevitable. The presence of damping is also the reason you are not getting an infinite amplitude at resonance. For a damping factor $\gamma$, the resonant frequency will be $$\omega_r = \sqrt{\frac{k_1+k_2}{m} - \frac{\gamma^2}{2}}$$

If the damping is not same for both the springs, then the resonant frequencies might not match within a certain limit. I would suggest you to check for any possible source of damping in this setup that might significantly affect the resonant frequency.

• Thank you for your thorough explanation! The data from four series, each using a uniqe spring configuration are now included in my original question. The oscillating mass was 50 g for all the series. – user46330 May 12 '14 at 16:39
• In my explanation, I assumed the unstretched lengths of both the springs are same. Is it not the case for your setup? – AJS May 12 '14 at 17:52
• That assumtpion is right. All strings used have exactly the same length when unstreched. – user46330 May 12 '14 at 18:34

1. Technically, the two springs are connected in parallel rather than in series since the equivalent spring constant is $$k_{\rm eq}~=~k_1+k_2.$$ (If the two springs had been truly in series, then $k_{\rm eq}=\frac{k_1k_2}{k_1+k_2}.$)
2. For spring masses $m_1,m_2\ll m$ much smaller than the weight $m$, the effective mass is $$m_{\rm eff}~:=~\frac{k_{\rm eq}}{\omega^2}~\approx~ m+ \frac{m_1+m_2}{3},$$ where $\omega=2\pi f$ is the resonance frequency.
3. Amusingly, gravity $g$ does theoretically not enter. The resonance frequency $\omega$ should in principle be the same for a horizontal oscillation, although for practical reasons, it may be difficult to test experimentally a horizontal oscillation without introducing additional friction.
• Here is an outline of a possible explanation of why the violet curve shows no large peak when the vibration generator (VG) is next to the dense (D) spring and far away from the sparse (S) spring: The generator frequency is at least an order of magnitude smaller than the characteristic frequency of the springs. Similarly, the wavelength is at least an order of magnitude longer than the springs. There will be reflections at the middle weight, especially when $k_1$ and $k_2$ are different. But there is no $\pi$ phase shift when reflecting D$\to$S$\to$D, only when reflecting S$\to$D$\to$S. (cont.) – Qmechanic May 14 '14 at 21:52
• (cont.) When the S spring is next to the VG, there will be two $\pi$ phase shifts, one in each end of the S spring, so the S string will be in phase. When the D spring is next to the VG, there will be only one $\pi$ phase shift, so the D spring will be out of phase. Thus the coupling to the middle weight will be suppressed in the violet case. – Qmechanic May 14 '14 at 21:52