Problem 25 - Fundamentals of Physics

Question

A uniform rope of mass \(m\) and length \(L\) hangs from a ceiling. (a) Show that the speed of a transverse wave on the rope is a function of \(y\), the distance from the lower end, and is given by \(v=\sqrt{g y}\). (b) Show that the time a transverse wave takes to travel the length of the rope is given by \(t=2 \sqrt{L / g} .\)

Step-by-Step Solution

Answer

Wave speed is \(v = \sqrt{g y}\). Travel time is \(t = 2\sqrt{L/g}\).

Step 1: Understanding the Problem

We are asked to calculate the speed of a transverse wave on a hanging rope and the time it takes for such a wave to travel the length of the rope. The speed is dependent on the distance from the lower end, and we need to find an expression for the time across the entire length.

Step 1: Consider the Force Diagram

When a wave travels along the rope at a distance y below the ceiling, the force acting at that point is due to the weight pulling from below. At distance y from the bottom, the mass of the rope segment below y is \(\frac{m}{L}(L-y)\) resulting in a force of \(F = \frac{m}{L} (L-y) g\).

Step 2: Relate Force to Wave Speed

The tension in the rope segment at position y is \(T = \frac{m}{L} (L-y) g\). The speed of a wave on a string is given by \(v = \sqrt{\frac{T}{\mu}}\), where \(\mu = \frac{m}{L}\) is the linear mass density.

Step 3: Derive Wave Speed as a Function of y

Substitute the tension \(T\) into the wave speed equation: \( v = \sqrt{\frac{\frac{m}{L} (L-y) g}{\frac{m}{L}}} = \sqrt{g y}\).\Thus, we have shown that \(v = \sqrt{g y}\).

Step 4: Setup the Integral for Wave Travel Time

To find the total travel time \(t\), notice that the speed changes along the rope, so we need to integrate over the rope's length. We use \(t = \int \frac{dy}{v(y)}\), where \(v(y) = \sqrt{g y}\).

Step 5: Evaluate the Integral

Substitute \(v(y) = \sqrt{g y}\) into \(t = \int \frac{dy}{\sqrt{g y}}\). The integral becomes \(t = \frac{1}{\sqrt{g}} \int \frac{dy}{\sqrt{y}}\). This evaluates to \(\frac{2\sqrt{y}}{\sqrt{g}}\) evaluated from 0 to L, giving \(t = 2 \sqrt{\frac{L}{g}}\).

Step 7: Conclusion

Hence, both parts of the problem have been solved, confirming that the speed of the wave varies with the square root of the distance from the lower end, and the time taken for a wave to travel the entire rope is given by \(t = 2 \sqrt{L/g}\).

Key Concepts

Transverse waves

Transverse waves are a type of wave where particle displacement is perpendicular to the direction of wave propagation. Imagine a rope being snapped up and down; the moving ripples along the rope are transverse waves. These waves are prevalent in nature, visible in surface water waves, and electromagnetic waves such as light.

• In transverse waves, peaks and troughs can clearly indicate movement direction.
• The wave's ability to transfer energy perpendicular to its motion is key in numerous physical processes.
• Understanding transverse waves helps in visualizing how waves carry energy without moving the material itself along with the wave.

Wave mechanics involves the study of these waves, helping us interpret the complex motion of particles within various environments.

Wave speed

Wave speed in a medium is determined by several factors such as the properties of the medium and wave type. For our uniform rope problem, the speed of a transverse wave in a rope depends on how the tension in the rope varies with the distance.

• The wave speed, expressed as \(v = \sqrt{gy}\), indicates that speed increases with the depth below the hanging point.
• Wave speed relies on the medium's tension and density, ensuring that waves move faster in tighter, less dense materials.

The exploration of wave speed in different scenarios is vital for applications ranging from musical instruments to seismic wave analysis, showing its significance in both everyday and scientific contexts.

Tension in a rope

The tension in a rope is a result of forces acting along its length, crucial in determining wave characteristics. For a rope hanging vertically, like our example, tension changes as a function of distance from the hanging point.

• Tension at point \(y\) is given by \(T = \frac{m}{L} (L-y)g\), reflecting how only the portion below adds to gravitational force.
• Engineers use this principle to design systems ensuring that tension supports weight without excess strain.
• Dynamic tension considerations are crucial in diverse applications from climbing ropes to long-span bridges.

This understanding is essential in solving problems where variable forces impact wave propagation, demonstrating practical applications in various engineering solutions.

Integral calculus in physics

Integral calculus is a powerful mathematical tool in physics used to calculate quantities that depend on continuous variables. When dealing with variable wave speeds along a rope, integration provides the precise total travel time for waves.

• The equation \(t = \int \frac{dy}{v(y)}\) allows us to sum small time differences along the rope length to find total time.
• Evaluating integrals helps find solutions for problems involving continuous change, like temperature gradients or cumulative force impacts.
• Integral calculus simplifies complex real-world situations into solvable models, essential in fields like quantum mechanics and fluid dynamics.

Mastering integration skills opens up a broader understanding of dynamic systems, enabling physics students to solve diverse and challenging problems with confidence.