Problem 21 - Fundamentals of Physics

Question

A \(100 \mathrm{~g}\) wire is held under a tension of \(250 \mathrm{~N}\) with one end at \(x=0\) and the other at \(x=10.0 \mathrm{~m}\). At time \(t=0\), pulse 1 is sent along the wire from the end at \(x=10.0 \mathrm{~m}\). At time \(t=30.0\) ms, pulse 2 is sent along the wire from the end at \(x=0 .\) At what position \(x\) do the pulses begin to meet?

Step-by-Step Solution

Answer

The pulses meet at position \( x = 0.31 \mathrm{~m} \).

Step 1: Calculate Wave Speed

To find the position where the pulses meet, we first need to calculate the speed of the wave traveling along the wire. The formula for the wave speed on a stretched string is \( v = \sqrt{\frac{T}{\mu}} \), where \( T \) is the tension in the wire and \( \mu \) is the linear mass density of the wire. Given \( T = 250 \) N and the mass of the wire is 100 g (0.1 kg), the length of the wire is 10 m, so \( \mu = \frac{0.1 \mathrm{~kg}}{10 \mathrm{~m}} = 0.01 \mathrm{~kg/m} \). Therefore, \( v = \sqrt{\frac{250}{0.01}} = \sqrt{25000} = 158.11 \mathrm{~m/s} \).

Step 2: Calculate the Position of Pulse 1 at First Meeting

Pulse 1 travels from the end at \( x = 10 \mathrm{~m} \) in the negative \( x \) direction. It travels a distance \( x_1 \) such that it meets pulse 2. Let \( t_m \) be the time at which they meet. Thus, \( x_1 = 10 - v \cdot t_m \).

Step 3: Calculate the Position of Pulse 2 at First Meeting

Pulse 2 travels from \( x = 0 \mathrm{~m} \) in the positive \( x \) direction. Given that this pulse is initiated 30 ms after pulse 1, the distance \( x_2 \) it covers by the time of meeting is \( x_2 = v \cdot (t_m - 0.03) \).

Step 4: Set Distances Equal to Find Meeting Point

Since pulse 1 moves towards 0 and pulse 2 moves away from 0, they meet when \( x_1 = x_2 \). Therefore, solve the equation \( 10 - v \cdot t_m = v \cdot (t_m - 0.03) \). Substituting for \( v = 158.11 \mathrm{~m/s} \), we have:\[10 - 158.11t_m = 158.11(t_m - 0.03)\]Simplifying gives:\[10 = 158.11 \times 0.03\]\[t_m = \frac{10}{2 \times 158.11} + 0.03 = 0.0613 \text{ s}\]

Step 5: Calculate x-Position at Meeting Point

Substitute \( t_m = 0.0613 \) s back into either expression for \( x_1 \) or \( x_2 \). Using \( x_1 = 10 - 158.11 \cdot 0.0613 \):\[x_1 = 10 - 9.69 = 0.31 \mathrm{~m}\]Thus, the pulses begin to meet at \( x = 0.31 \mathrm{~m} \).

Key Concepts

Tension in Wire

Understanding the tension in a wire is crucial when dealing with wave motion, especially for calculating wave speed. Tension refers to the force that is applied along the wire, stretching it. It is measured in newtons (N). In our exercise, the wire has a specified tension of 250 N. Tension is a direct force that affects how waves move along the wire.
When tension increases, the force pulling the material tight increases. This means the speed of a wave traveling on the wire also increases. This is because wave speed is related to the tension through the wave speed formula:

• \( v = \sqrt{\frac{T}{\mu}} \)

In this formula, \( T \) is the tension, and \( \mu \) is the linear mass density. Our example uses this formula to determine how fast the waves move relative to the tension. Therefore, higher tension results in a faster wave due to the stronger pulling force.

Linear Mass Density

Linear mass density (symbol \( \mu \)) represents the mass per unit length of the wire. It is measured in kilograms per meter (kg/m). In the exercise, it's computed as \( \mu = \frac{0.1 \, ext{kg}}{10 \, ext{m}} = 0.01 \, ext{kg/m} \). This measure tells us how much mass is in each meter of the wire, which directly affects how waves travel through the wire.
To calculate wave speed, we use the linear mass density as a factor in the equation \( v = \sqrt{\frac{T}{\mu}} \). The wave speed is inversely proportional to the square root of the linear mass density. This means that for a given tension, increasing the wire's mass density will decrease the wave speed. Understanding this relationship is important when designing systems that use wires to convey mechanical signals, like musical instruments or certain mechanical devices.

Pulse Meeting Point

The exercise involves finding the point where two pulses meet on the wire. The meeting point is where the two traveling pulses, sent from opposite ends, converge. First, we calculate each pulse's travel distance using the speed obtained from the wave speed formula.
When both pulses are traveling toward each other, their meeting point can be found by ensuring their traveled distances add up to the entire length of the wire. For our case:

• Pulse 1 starts from \( x = 10 \) m.
• Pulse 2 starts from \( x = 0 \) m.

By comparing their equations, we can solve for the time \( t_m \) when they meet. Substituting this back gives us the position \( x \) on the wire, which mathematically proves to be at \( x = 0.31 \) m from the starting end at \( x = 10 \) m. This is crucial when synchronizing wave motions, such as in communication systems.

Wave Motion

Wave motion in a medium like a wire involves energy being transferred through oscillations. Unlike rigid movement, the particles in the wire oscillate around an equilibrium position, creating waves.
Key points about wave motion include:

• Waves can be longitudinal or transverse, but in a taut wire, they are usually transverse, meaning the displacement is perpendicular to the direction of wave travel.
• Wave speed is critical, affected by the medium's physical properties, notably tension and linear mass density.
• Understanding wave behavior in such setups is crucial for practical applications in engineering and physics.

Waves carry both energy and information, which is why understanding wave motion principles helps in applications ranging from musical instruments to telecommunication technologies.