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Chapter 16: Problem 59

A rope of total mass \(m\) and length \(L\) is suspended vertically. Show that a transverse pulse travels the length of the rope in a time interval \(\Delta t=2 \sqrt{L} / g .\) (Suggestion: First find an expression for the wave speed at any point a distance \(x\) from the lower end by considering the tension in the rope as resulting from the weight of the segment below that point.)

Short Answer

Expert verified
The time interval needed for a transverse pulse to travel the length of a suspended rope is \(\Delta t = 2\sqrt{L/g}\).

Step by step solution

01

Expression for Tension

We start by considering an x-length segment of the rope hanging freely. The weight of this segment equals its mass times gravity, which provides the tensile force, or tension. The mass of the segment can be represented as \(mx/L\), where m is the total mass and L is the total length of the rope. Therefore, the tension \(T\) at this point is \(mgx/L\). Since the mass per unit length is m/L, the tension is thus equal to \(T = (m/L) \cdot g \cdot x\).
02

Calculate Wave Speed

Next, the speed of the wave is analyzed. The speed of a wave in a rope under tension is given by \(v = \sqrt{T/\mu}\), where T is the tension in the rope and \( \mu = m/L \) is the linear mass density of the rope. Substituting the expression for T from the previous step, we get \(v = \sqrt{(m/L) \cdot g \cdot x / (m/L)}\) = \(\sqrt{g \cdot x}\).
03

Evaluate Time Interval

To obtain the travel time of the pulse through the varying tensions along the length of the rope, the concept of infinitesimal elements is employed. It means the wave propagates across each small length dx of the rope in a small time interval dt given by \(dt = dx/v\). Substituting the expression for v, this leads to \(dt = dx/\sqrt{g \cdot x}\). To find the total time interval \( \Delta t\), integrate this expression from 0 to L, which will give us \(\Delta t = 2\sqrt{L/g}\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Tension in a Rope
When considering the tension in a rope, think of it as the force that keeps the rope stretched, providing a certain rigidity to support itself. The tension arises primarily from the weight of the rope if it is hanging vertically. Here鈥檚 how:

  • If you consider a small part of the rope at a distance \( x \) from the lower end, the tension at that point results from the weight of the rope below \( x \).
  • The weight of this segment is equivalent to its mass multiplied by gravity (\( g \)).
  • The segment鈥檚 mass can be calculated using the fraction of total mass: \( \frac{mx}{L} \), where \( m \) is the total mass and \( L \) is the total length of the rope.

This means the tension \( T \) at distance \( x \) is \( T = (\frac{m}{L}) \cdot g \cdot x \). This formula shows how the tension varies with the position along the rope due to the force of gravity. This relationship is crucial for understanding the wave dynamics in ropes.
Transverse Wave
A transverse wave is a type of wave where the motion of the medium is perpendicular to the direction of the wave itself. In the case of a rope, such a wave causes the rope to move up and down as the wave moves along it.

The speed of a transverse wave on a rope is determined by the tension within the rope and its linear mass density. Understanding the tension, as we discussed, helps us predict how quickly the wave will propagate. For simplification:

  • Wave speed \( v \) is described by the formula \( v = \sqrt{\frac{T}{\mu}} \), where \( \mu \) is the linear mass density.
  • In a suspended rope, the wave speed changes with position due to variations in tension.
  • This speed of the wave influences how quickly a signal or pulse can travel from one end of the rope to the other.

This concept highlights the interaction between mechanical properties of the rope and the nature of the wave.
Linear Mass Density
Linear mass density \( \mu \) refers to the distribution of mass along the length of the rope. It simplifies to mass per unit length, often expressed as \( \mu = \frac{m}{L} \). Here鈥檚 why it鈥檚 important:

  • It helps determine the wave speed as it acts as a denominator when calculating wave speed: \( v = \sqrt{\frac{T}{\mu}} \).
  • This density ensures that we understand how concentrated the mass is along the rope.
  • A greater linear mass density implies a slower wave speed given the same tension, due to the increased inertia.

The concept of linear mass density is particularly vital in interpreting how different ropes with varying material properties could affect wave propagation. This factor emphasizes the need to consider both material and geometric properties in wave dynamics.

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Most popular questions from this chapter

If a loop of chain is spun at high speed, it can roll along the ground like a circular hoop without slipping or collapsing. Consider a chain of uniform linear mass density \(\mu\) whose center of mass travels to the right at a high speed \(v_{0}\). (a) Determine the tension in the chain in terms of \(\mu\) and \(v_{0}\). (b) If the loop rolls over a bump, the resulting deformation of the chain causes two transverse pulses to propagate along the chain, one moving clockwise and one moving counterclockwise. What is the speed of the pulses traveling along the chain? (c) Through what angle does each pulse travel during the time it takes the loop to make one revolution?

It is found that a \(6.00-\mathrm{m}\) segment of a long string contains four complete waves and has a mass of \(180 \mathrm{g} .\) The string is vibrating sinusoidally with a frequency of \(50.0 \mathrm{Hz}\) and a peak-to-valley distance of \(15.0 \mathrm{cm} .\) (The "peak-to-valley" distance is the vertical distance from the farthest positive position to the farthest negative position.) (a) Write the function that describes this wave traveling in the positive \(x\) direction. (b) Determine the power being supplied to the string.

A string of length \(L\) consists of two sections. The left half has mass per unit length \(\mu=\mu_{0} / 2,\) while the right has a mass per unit length \(\mu^{\prime}=3 \mu=3 \mu_{0} / 2 .\) Tension in the string is \(T_{0} .\) Notice from the data given that this string has the same total mass as a uniform string of length \(L\) and mass per unit length \(\mu_{0}\). (a) Find the speeds \(v\) and \(v^{\prime}\) at which transverse pulses travel in the two sections. Express the speeds in terms of \(T_{0}\) and \(\mu_{0},\) and also as multiples of the speed \(v_{0}=\left(T_{0} / \mu_{0}\right)^{1 / 2} .\) (b) Find the time interval required for a pulse to travel from one end of the string to the other. Give your result as a multiple of \(\Delta t_{0}=L / v_{0}\).

Motion picture film is projected at 24.0 frames per second. Each frame is a photograph \(19.0 \mathrm{mm}\) high. At what constant speed does the film pass into the projector?

A student taking a quiz finds on a reference sheet the two equations $$f=1 / T \text { and } v=\sqrt{T / \mu}$$ She has forgotten what \(T\) represents in each equation. (a) Use dimensional analysis to determine the units required for \(T\) in each equation. (b) Identify the physical quantity each \(T\) represents.
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