1. Conceptual Analysis

When a force is applied to one end of a long wire, the disturbance does not travel instantaneously to the other end. It propagates as a longitudinal elastic wave with a finite speed $v$.

First, we must determine if the disturbance reaches the other end of the wire within the given time interval $\Delta t$.

• If $\Delta t < \frac{l}{v}$, the wave has not reached the far end. The wire effectively behaves as a semi-infinite rod, and the far end remains stationary.
• If $\Delta t > \frac{l}{v}$, the wave has reflected, and the entire wire is in motion.

2. Calculating Wave Speed and Transit Time

The speed of a longitudinal elastic wave in a solid is given by:

$$v = \sqrt{\frac{Y}{\rho}}$$

Substituting the given values:

$$v = \sqrt{\frac{2.0 \times 10^{11}}{8.0 \times 10^3}} = \sqrt{0.25 \times 10^8} = \sqrt{25 \times 10^6} = 5000 \text{ m/s}$$

The time required for the wave to travel the length of the wire ($l = 800$ m) is:

$$t_{transit} = \frac{l}{v} = \frac{800}{5000} = 0.16 \text{ s}$$

Since the given time interval $\Delta t = 0.1 \text{ s}$ is less than $t_{transit}$ ($0.1 < 0.16$), the wave has not reached the other end. The rear part of the wire remains stationary, and the wire behaves as if it were infinitely long.

3. Determining Displacement

As the wave propagates, the section of the wire behind the wavefront is in a state of uniform stress $\sigma$ and moves with a constant particle velocity $v_p$.

From the impulse-momentum relationship for a wave propagating into a stationary medium:

$$F = Z v_p$$

Where $Z$ is the mechanical impedance of the wire, $Z = S\sqrt{\rho Y}$. Alternatively, we can derive this from strain:

The strain in the moving region is $\varepsilon = \frac{\sigma}{Y} = \frac{F}{SY}$.
In a traveling wave, particle velocity is related to strain by $v_p = v \varepsilon$.

Combining these:

$$v_p = v \left( \frac{F}{SY} \right) = \sqrt{\frac{Y}{\rho}} \cdot \frac{F}{SY} = \frac{F}{S\sqrt{\rho Y}}$$

Since the force $F$ is constant, the particle velocity $v_p$ of the end is constant. The total displacement $\Delta x$ is simply velocity multiplied by time.

$$\Delta x = v_p \Delta t = \frac{F \Delta t}{S\sqrt{\rho Y}}$$

4. Numerical Calculation

Let’s substitute the values:

• $F = 1.0 \text{ N}$
• $\Delta t = 0.1 \text{ s}$
• $S = 1.0 \text{ mm}^2 = 1.0 \times 10^{-6} \text{ m}^2$
• $\rho = 8.0 \times 10^3 \text{ kg/m}^3$
• $Y = 2.0 \times 10^{11} \text{ N/m}^2$

First, calculate the impedance term $S\sqrt{\rho Y}$:

$$S\sqrt{\rho Y} = (1.0 \times 10^{-6}) \sqrt{(8.0 \times 10^3)(2.0 \times 10^{11})}$$ $$= 10^{-6} \sqrt{16 \times 10^{14}}$$ $$= 10^{-6} \times (4 \times 10^7)$$ $$= 40 \text{ kg/s}$$

Now, calculate the displacement:

$$\Delta x = \frac{1.0 \times 0.1}{40} = \frac{0.1}{40} = \frac{1}{400} \text{ m}$$ $$\Delta x = 0.0025 \text{ m} = 2.5 \text{ mm}$$