Oscillations and Waves > The superposition of oscillations > Problem 3.4.12

Statement

$3.4.12.$ [The ends of a spring of stiffness $k$ are moved in the longitudinal direction according to the harmonic law: $x_1 = A_1 \cos(\omega t + \phi_1)$, $x_2 = A_2 \cos(\omega t + \phi_2)$; the average force of the spring tension over the period is zero. How does this force change with time? Determine the maximum and average energy of the spring over a long period. At which phase difference $\phi_2 - \phi_1$ is the average energy of the spring the greatest?]

Solution

1. Force as a Function of Time

The displacements of the two ends of the spring are given by:
$$x_1(t) = A_1 \cos(\omega t + \phi_1)$$
$$x_2(t) = A_2 \cos(\omega t + \phi_2)$$

According to Hooke's Law, the instantaneous tension force $F(t)$ is proportional to the elongation $\Delta x(t) = x_2(t) - x_1(t)$:
$$F(t) = k [A_2 \cos(\omega t + \phi_2) - A_1 \cos(\omega t + \phi_1)]$$

Using the trigonometric identity $\cos(\alpha + \beta) = \cos\alpha\cos\beta - \sin\alpha\sin\beta$, we expand and regroup the terms with respect to $\cos\omega t$ and $\sin\omega t$:
$$F(t) = k [(A_2 \cos\phi_2 - A_1 \cos\phi_1)\cos\omega t - (A_2 \sin\phi_2 - A_1 \sin\phi_1)\sin\omega t]$$

This harmonic expression can be simplified into $F(t) = F_0 \cos(\omega t + \Phi)$, where the peak force amplitude $F_0$ is derived using the identity $\sin^2\theta + \cos^2\theta = 1$ and the cosine subtraction formula:
$$F_0 = k \sqrt{A_1^2 + A_2^2 - 2A_1A_2\cos(\phi_2 - \phi_1)}$$

Thus, the force changes harmonically over time as:
$$F(t) = k \sqrt{A_1^2 + A_2^2 - 2A_1A_2\cos(\phi_2 - \phi_1)} \cos(\omega t + \Phi)$$

2. Maximum and Average Energy

The potential energy stored in the spring at any instant is given by $E(t) = \frac{F(t)^2}{2k}$.

1. Maximum Energy ($E_{\max}$):**
   The maximum energy occurs when the squared cosine function equals 1:
   $$E_{\max} = \frac{F_0^2}{2k} = \frac{1}{2}k [A_1^2 + A_2^2 - 2A_1A_2\cos(\phi_2 - \phi_1)]$$

2. **Average Energy ($\langle E \rangle$):**
   Over a long period, the time-average of a squared harmonic function is exactly $\frac{1}{2}$ ($\langle \cos^2(\omega t + \Phi) \rangle = \frac{1}{2}$). Therefore:
   $$\langle E \rangle = \frac{F_0^2}{4k} = \frac{1}{4}k [A_1^2 + A_2^2 - 2A_1A_2\cos(\phi_2 - \phi_1)]$$

3. Condition for Maximum Average Energy

To maximize the average energy $\langle E \rangle$, the subtracted term inside the bracket must be minimized. Since the minimum value of the cosine function is $-1$:
$$\cos(\phi_2 - \phi_1) = -1 \implies \phi_2 - \phi_1 = \pi$$

Conclusion: The average energy of the spring is greatest when the phase difference is $\phi_2 - \phi_1 = \pi$ (or $180^\circ$), which yields:

$$\langle E \rangle_{\max} = \frac{1}{4}k(A_1 + A_2)^2$$

Answer

$$\boxed{F(t) = k \sqrt{A_1^2 + A_2^2 - 2A_1A_2\cos(\phi_2 - \phi_1)} \cos(\omega t + \Phi)}$$

$$\boxed{E_{\max} = \frac{1}{2}k [A_1^2 + A_2^2 - 2A_1A_2\cos(\phi_2 - \phi_1)]}$$

$$\boxed{\langle E \rangle = \frac{1}{4}k [A_1^2 + A_2^2 - 2A_1A_2\cos(\phi_2 - \phi_1)]}$$

$$\boxed{\phi_2 - \phi_1 = \pi}$$