Oscillations and Waves > The superposition of oscillations > Problem 3.4.18

Statement

$3.4.18.$ Small oscillations of pendulums connected by a spring occur according to the law: $x_1 = B \cos (\omega_0 t + \phi) + A \cos \omega t$, $x_2 = B \cos (\omega_0 t + \phi) - A \cos \omega t$. Determine the stiffness of the spring linking the pendulums. In the equilibrium position the pendulums are vertical, the mass of each ball is $m$.

Solution

1. Identification of Normal Modes

The given equations describing the positions of the two coupled pendulums are:
$$x_1(t) = B \cos (\omega_0 t + \phi) + A \cos \omega t$$
$$x_2(t) = B \cos (\omega_0 t + \phi) - A \cos \omega t$$

These equations show that the motion is a linear combination (superposition) of two distinct normal modes:

1. **Symmetric Mode ($\omega_0$):** When $A = 0$, both pendulums move in perfect phase with identical displacements ($x_1 = x_2$). In this mode, the connecting spring undergoes no stretching or compression, meaning it exerts zero force. The frequency is purely dictated by gravity and the length $l$ of the pendulums:
   $$\omega_0^2 = \frac{g}{l}$$

2. **Antisymmetric Mode ($\omega$):** When $B = 0$, the pendulums move completely out of phase with opposite displacements ($x_1 = -x_2$). Here, the spring is actively stretched and compressed, providing an additional restoring force that increases the oscillation frequency ($\omega > \omega_0$).

2. Equations of Motion for Small Oscillations

Let $m$ be the mass of each pendulum bob, $l$ be the length of the suspension strings, and $k$ be the stiffness of the coupling spring. For small angular deflections, the restoring force due to gravity is approximately $-mg\frac{x}{l} = -m\omega_0^2 x$. The force exerted by the spring on the first mass is $-k(x_1 - x_2)$ and on the second mass is $-k(x_2 - x_1)$.

Setting up the equations of motion for both masses:
$$m \ddot{x}_1 = -m\omega_0^2 x_1 - k(x_1 - x_2)$$
$$m \ddot{x}_2 = -m\omega_0^2 x_2 - k(x_2 - x_1)$$

3. Deriving Stiffness from the Antisymmetric Channel

To isolate the antisymmetric parameters, we subtract the second equation of motion from the first:
$$m(\ddot{x}_1 - \ddot{x}_2) = -m\omega_0^2(x_1 - x_2) - 2k(x_1 - x_2)$$

Let us define the coordinate for relative displacement as $u = x_1 - x_2$. Dividing the entire relation by $m$ gives:
$$\ddot{u} = -\left(\omega_0^2 + \frac{2k}{m}\right)u \implies \ddot{u} + \left(\omega_0^2 + \frac{2k}{m}\right)u = 0$$

This represents a simple harmonic equation governing the antisymmetric mode. The square of its natural angular frequency must match the coefficient of $u$:
$$\omega^2 = \omega_0^2 + \frac{2k}{m}$$

4. Solving for Spring Stiffness $k$

Isolating the term containing the unknown stiffness $k$ from the frequency equation:
$$\omega^2 - \omega_0^2 = \frac{2k}{m}$$

Multiplying both sides by $m$ and dividing by $2$ directly yields the stiffness of the connecting spring:
$$k = \frac{m(\omega^2 - \omega_0^2)}{2}$$

Answer

$$\boxed{k = \frac{m(\omega^2 - \omega_0^2)}{2}}$$