Solution

Statement

$3.9.8^\ast.$ a. An external longitudinal force $F = F_0 \cos \omega t$ acts on an infinite rod in a certain cross-section. What velocity and deformation waves arise in the rod? The cross-section of the rod is $S$, the density of its material is $\rho$, and the wave speed in the rod is $c$.
b. An external longitudinal force $F = F_0 \cos \omega t$ acts in two cross-sections of an infinite rod located at a distance $l$ from each other. What waves arise in the rod? At what values of $l$ is the wave power in the rod greatest? least? Why is the energy of the resulting wave in the rod not equal to the sum of the energies of the waves emitted by each source separately?

Solution

Part A. Single Source

1. Velocity waves
Let's set the origin $x=0$ at the cross-section where the force is applied. The external force $F(t)$ generates two elastic waves traveling in opposite directions.
The wave (acoustic) impedance of one semi-infinite rod is $Z = \rho c S$. Since the source must drive two semi-infinite rods simultaneously (to the right and left), the total resistance force of the medium is $2Zv(0, t)$.
By Newton's third law, the external force is balanced by this resistance:
$$F_0 \cos\omega t = 2 \rho c S v(0, t)$$
From this, we find the velocity of the cross-section $x=0$:
$$v(0, t) = \frac{F_0}{2 \rho c S} \cos\omega t$$
Let's denote the velocity amplitude as $v_m = \frac{F_0}{2 \rho c S}$ and the wave number as $k = \frac{\omega}{c}$. Taking into account the phase delay $\pm kx$, the equations for the traveling velocity waves are:
For $x > 0$:
$$v(x,t) = v_m \cos(\omega t - kx)$$
For $x < 0$:
$$v(x,t) = v_m \cos(\omega t + kx)$$

2. Deformation waves
The relative deformation $\varepsilon = \frac{\partial u}{\partial x}$ is related to the particle velocity $v = \frac{\partial u}{\partial t}$.
For a plane wave traveling right ($x>0$), the relationship is $\varepsilon = -v/c$. For a wave traveling left ($x<0$), it is $\varepsilon = v/c$.
The deformation waves are described by the equations:
For $x > 0$:
$$\varepsilon(x,t) = -\frac{F_0}{2 \rho c^2 S} \cos(\omega t - kx)$$
For $x < 0$:
$$\varepsilon(x,t) = \frac{F_0}{2 \rho c^2 S} \cos(\omega t + kx)$$

Part B. Two Sources

Let the sources be located at coordinates $x_1 = -l/2$ and $x_2 = l/2$. By the superposition principle, the resulting velocity is the sum of the waves from each source.

1. Outside the sources (traveling waves)
Consider the region to the right of both sources ($x > l/2$). Summing the two waves traveling to the right:
$$v_{\text{res}} = v_m \cos\left(\omega t - k\left(x + \frac{l}{2}\right)\right) + v_m \cos\left(\omega t - k\left(x - \frac{l}{2}\right)\right)$$
Applying the sum-to-product trigonometric formula, we obtain the equation of the two traveling waves outside the sources:
$$v_{\text{res}} = 2 v_m \cos\left(\frac{kl}{2}\right) \cos(\omega t - kx) = \frac{F_0}{\rho c S} \cos\left(\frac{\omega l}{2c}\right) \cos\left(\omega \left(t - \frac{x}{c}\right)\right)$$

2. Between the sources (standing wave)
In the region $-l/2 < x < l/2$, the wave from the left source travels right, and from the right source travels left:
$$v_{\text{int}} = v_m \cos\left(\omega t - k\left(x + \frac{l}{2}\right)\right) + v_m \cos\left(\omega t - k\left(\frac{l}{2} - x\right)\right)$$
$$v_{\text{int}} = 2 v_m \cos\left(\omega t - \frac{kl}{2}\right) \cos(kx) = \frac{F_0}{\rho c S} \cos\left(\omega \left(t - \frac{l}{2c}\right)\right) \cos\left(\frac{\omega x}{c}\right)$$
A classical standing wave is formed here.

3. Power analysis
The power emitted into the outer region is proportional to the square of the amplitude of the traveling wave $v_{\text{res}}$.

Maximum power is achieved when $\cos\left(\frac{kl}{2}\right) = \pm 1$, which corresponds to $\frac{\omega l}{2c} = \pi n$.
Hence $l = n \frac{2\pi c}{\omega} = n\lambda$. An even number of half-waves fits in the distance $l$. The waves reinforce each other.

Minimum power (equal to zero) is achieved when $\cos\left(\frac{kl}{2}\right) = 0$, which corresponds to $\frac{\omega l}{2c} = \frac{\pi}{2} + \pi n$.
Hence $l = \left(n + \frac{1}{2}\right)\lambda$. An odd number of half-waves fits in the distance $l$. The waves completely cancel each other out in the outer region.

4. Why is the energy of the resulting wave not equal to the sum of the energies of the sources?
The power that a source transfers to the rod is equal to the dot product of the force and the velocity of its point of application: $P = F \cdot v$.
The peculiarity of the system is that the velocity $v$ at the point of application of the first force is the sum of the velocity initiated by that force itself and the velocity of the wave arriving from the second source.

The sources interact with each other through the elastic medium of the rod. If $l = (n+0.5)\lambda$, the wave from the second source arrives at the first one in anti-phase. As a result, the velocity of the metal at the point of force application becomes zero. The source acts with a force $F_0 \cos\omega t$, but the point of application does not move; therefore, no work is done ($P = 0$). The sources "block" each other's work, which is why the total emitted energy is zero.