PHYS212 : Lectures

Subsections

• 1.4.1 Beads on a String
• 1.4.2 Compound Pendulum or Hanging Chain

1.4 Many Coupled Oscillators

1.4.1 Beads on a String

Figure 1.4: Transverse displacements of $n$ beads on a massless string.

• Equations of Motion
  • As with the two-bead case, we will neglect gravity and consider small angles for which the tension $T_o$ in the strings is approximately constant leading to restoring forces of the form $-T_o \sin \theta$.

  • Consider bead 3 in Fig. 1.4. It is subject to only two forces - the tensions in the strings connecting it to its nearest neighbors,
    

    $$m \ddot{y_3} = F_{3 net y} = -T_o \sin \theta_3 + T_o \sin \theta_4$$

    $$= -T_o \left( \frac{y_3 - y_2}{\ell} + \frac{y_3 - y_4}{\ell} \right)$$ (1.34)

    
    
    leading to the equation of motion
    

    $$\ddot{y_3} = -\frac{T_o}{m \ell}(2 y_3 - y_2 - y_4)$$ (1.35)

    
    

  • We obtain an equation of motion with the same form as Eq. 1.35 for the $b$th bead,
    

    $$\ddot{y_b} = -\frac{T_o}{m \ell}(2 y_b - y_{b-1} - y_{b+1})$$ (1.36)

    
    
    provided $b \neq 1$ and $b \neq n$.

  • Eq. 1.36 can also be applied to the first and $n$th beads once we realize that fixed ends of the string can be described by $y_0 = y_{n+1} = 0$.

  • Exercise : Convince yourself that Eq. 1.36 gives Eq.'s 1.27 for $n=2$.

• Normal Modes

  Figure 1.5: Amplitudes of the normal modes of 5 beads on a massless string are represented by filled circles. The smooth curves are plots of Eq. 1.40. (The straight segments of string connecting the beads are not shown.) The amplitudes of neighboring beads are related by Eq. 1.39. The scales are arbitrary. However, the small angle approximation is used in these calculations, so the $x$ scale is greatly exaggerated relative to the $y$ scale.

  
  • Rather than solve $n$ simultaneous equations, we will make a good guess based on intuition and observation, and check it with a numerical solution (for homework).

  • We expect $n$ normal modes with
    

    $$y_b = C_b \cos(\omega t + \phi)$$ (1.37)

    
    
    for which Eq. 1.36 becomes
    

    $$-\omega^2 y_b = -\frac{T_o}{m \ell}(2 y_b - y_{b-1} - y_{b+1})$$ (1.38)

    
    
    which we can re-index taking $b \rightarrow b-1$ and express the position of one bead in terms of the positions of the two beads to its left,
    

    $$y_b = \left( 2-\frac{\omega^2 m \ell}{T_o} \right) y_{b-1} - y_{b-2}$$ (1.39)

    
    

  • Eq. 1.39 is a recursion formula that can be used to find the amplitudes of the oscillations of the beads. (You will try this for homework.)
    • The process can be started by using $y_0 = 0$ and setting $y_1$ to any non-zero value.
    • Then the normal modes can be found by adjusting $\omega$ to find values giving $y_{n+1} = 0$.
    • The value chosen for $y_1$ is independent of the normal mode frequencies. It sets the overall amplitude of the system.

  • Our good guess : The amplitudes lie along a sinusoidal shape with nodes at the endpoints of the system as shown in Fig. 1.5. The system has total length $L = (n+1) \ell$, and the $b$th bead is a distance $b \ell$ from the left wall. In the $M$th mode, $M$ half-periods of the sine function fit into the total length giving
    

    $$C_b = C \sin \left( \frac{M b \pi}{n+1} \right),$$ (1.40)

    
    
    where $C$ is approximately the overall amplitude of the system (approximately, because beads do not always fall at the maxima of the sine function of Eq. 1.40).

  • Fortunately, we can use Eq.'s 1.39 and 1.40 find the normal mode frequencies of the system without solving $n$ simultaneous equations. We take advantage of the simplified form of Eq. 1.39 for the first pair of beads
    

    $$y_2 = \left( 2-\frac{\omega^2 m \ell}{T_o} \right) y_1$$ (1.41)

    
    
    since $y_0 = 0$.

  • The normal mode solutions of Eq. 1.37 in Eq. 1.41 give
    

    $$\frac{C_2}{C_1} = \left( 2-\frac{\omega^2 m \ell}{T_o} \right)$$ (1.42)

    
    
    which we can relate to the periodic amplitude condition using Eq. 1.40 to yield
    

    $$\left( 2-\frac{\omega^2 m \ell}{T_o} \right) = \frac{\sin \... ... M \pi}{n+1} \right)} {\sin \left( \frac{M \pi}{n+1} \right)}$$ (1.43)

    
    

  • The trigonometric identities
    

    $$\sin (2 \theta) = 2 \sin \theta \cos \theta$$

    $$\sin^2 (\theta / 2) = \frac{1}{2} (1 - \cos \theta)$$ (1.44)

    
    
    with $\theta = \frac{M \pi}{n+1}$ can be used to show that Eq. 1.43 gives normal mode frequencies
    

    $$\omega_M = 2 \sqrt{\frac{T_o}{m \ell}} \sin \left( \frac{M \pi}{2(n+1)} \right)$$ (1.45)

    
    

  • Exercise : Convince yourself that Eq. 1.45 with $n=2$ gives the normal mode frequencies of of the two-bead system (Eq. 1.31) and that Eq. 1.42 gives the corresponding amplitude ratios (Eq. 1.32).

1.4.2 Compound Pendulum or Hanging Chain

Figure 1.6: A compound pendulum consisting of $n$ simple pendulums in series.

• Equations of Motion
  • The calculation is cleaner if we index the masses from the bottom up as shown in Fig. 1.6.

  • There are $b$ beads hanging from the $b$th string, so its tension, for small angles is given by
    

    $$T_b \approx bmg$$ (1.46)

    
    

  • The net horizontal force on bead $b$ is given by the horizontal components of the two strings attached to it,
    

    $$F_{b net x} \approx -bmg \theta_b + (b-1) mg \theta_{b-1}$$ (1.47)

    
    

  • We will describe the motion in terms of horizontal displacements instead of angles,
    

    $$m \ddot{x}_b = -bmg \frac{(x_b - x_{b+1})}{\ell} + (b-1) mg \frac{(x_{b-1}-x_b)}{\ell}$$ (1.48)

    
    
    yielding equations of motion
    

    $$\ddot{x}_b = -\frac{g}{\ell} \left[ (2b-1) x_b - b x_{b+1} - (b-1) x_{b-1} \right]$$ (1.49)

    
    

  • The fixed end of the chain is described by $x_{b+1} = 0$.

  • Exercise : Convince yourself that Eq. 1.49 with $n=2$ is equivalent to Eq.'s 1.18.

• Normal Modes

  Figure 1.7: The first three modes of a compound pendulum (hanging chain) of 20 masses (links). The scales are arbitrary. However, the small angle approximation is used in these calculations, so the $x$ scale is greatly exaggerated relative to the $y$ scale.

  
  • Intuition and observation are less helpful here. The symmetry of the horizontal beaded string is broken by gravity and by allowing one end to swing freely. As a result, the amplitude ratios are not periodic, and the nodes are not evenly spaced.

  • We will take a numerical approach. That is, we will use a computer to find the normal mode frequencies and amplitudes of the system via a recursion formula, similar to Eq. 1.39, derived from Eq. 1.49.

  • Noting that in a normal mode, $\ddot{x}_b = -\omega^2 x_b$, solving Eq. 1.49 for $x_{b+1}$, and re-indexing ( $b \rightarrow b-1$), we find
    

    $$x_b = \frac{1}{b-1} \left[ \left( 2b-3 - \frac{\omega^2 \ell}{g} \right) x_{b-1} - (b-2) x_{b-2} \right]$$ (1.50)

    
    

  • As with the horizontal beaded string, Eq. 1.50 is a recursion formula that can be used to find the amplitudes of the oscillations of the beads.
    • Start the process by setting $x_1$ to any non-zero value.
    • Then, we can calculate a value for $x_2$, because the factor of $b-2$ in the second term of Eq. 1.50 saves us from needing a value for $x_0$.
    • The normal mode frequencies and relative amplitudes can be found by adjusting $\omega$ to find values giving $y_{n+1} = 0$.
    • The value chosen for $y_1$ is independent of the normal mode frequencies. It sets the overall amplitude of the system.
    • Fig. 1.7 shows the amplitudes of oscillation of the first three normal modes of a system with 20 masses generated using this procedure. (You will try this as part of a lab exercise.)

    Figure 1.8: The amplitudes of oscillation of the masses of a compound pendulum with 10, 20, and 100 masses (links) in the second normal mode. The scales are arbitrary. However, the small angle approximation is used in these calculations, so the $x$ scale is greatly exaggerated relative to the $y$ scale.

    

  • Taking the limit as $n \rightarrow \infty$ while holding the length of the system constant yields a continuous cable.
    • Fig. 1.8 shows the amplitudes of oscillation of the masses of a compound pendulum with 10, 20, and 100 masses (links) in the second normal mode.
    • Note that lower density of masses corresponds to larger relative amplitudes toward the middle of the system. (The relative amplitudes of the horizontal beaded string is independent of mass density.)

Copyright © 2003-2009, Lewis A. Riley

Updated Wed Jan 18 09:51:28 2006