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One end of a horizontal rope is attached to a prong of an electrically driven tuning fork that vibrates the rope transversely at 120 \(\mathrm{Hz}\) . The other end passes over a pulley and supports a \(1.50-\mathrm{kg}\) mass. The linear mass density of the rope is 0.0550 \(\mathrm{kg} / \mathrm{m}\) . (a) What is the speed of a transverse wave on the rope? (b) What is the wavelength? (c) How would your answers to parts (a) and (b) change if the mass were increased to 3.00 \(\mathrm{kg}\) ?

Short Answer

Expert verified
(a) 16.4 m/s, (b) 0.137 m, (a) 23.5 m/s, (b) 0.196 m with 3.00 kg.

Step by step solution

01

Understanding the Problem

We are tasked to find the speed of a transverse wave on a rope and its wavelength. We know the frequency of vibration, the mass at the end of the rope, and the linear mass density of the rope. Additionally, we need to determine how these values change if the mass were increased.
02

Calculating the Tension in the Rope

The tension in the rope is caused by the weight of the mass hanging from it. Tension can be calculated using the formula: \( T = mg \), where \( m \) is the mass and \( g \) is the acceleration due to gravity (approximated as \( 9.8 \ \mathrm{m/s^2} \)). For the initial mass of 1.50 kg: \( T = 1.50 \times 9.8 = 14.7 \ \mathrm{N} \).
03

Finding the Speed of the Wave

The speed \( v \) of a wave on a rope is given by the formula: \[ v = \sqrt{\frac{T}{\mu}} \] where \( T \) is the tension and \( \mu \) is the linear mass density. With \( T = 14.7 \ \mathrm{N} \) and \( \mu = 0.0550 \ \mathrm{kg/m} \), the speed is: \[ v = \sqrt{\frac{14.7}{0.0550}} \approx 16.4 \ \mathrm{m/s} \].
04

Calculating the Wavelength

The wavelength \( \lambda \) of the wave is related to the speed and frequency \( f \) through the formula: \[ \lambda = \frac{v}{f} \] Given \( v = 16.4 \ \mathrm{m/s} \) and \( f = 120 \ \mathrm{Hz} \), the wavelength is: \[ \lambda = \frac{16.4}{120} \approx 0.137 \ \mathrm{m} \].
05

Updating the Tension with Increased Mass

With a mass of 3.00 kg, the new tension \( T \) will be: \( T = 3.00 \times 9.8 = 29.4 \ \mathrm{N} \).
06

Recalculate Speed with New Tension

Using the updated tension \( T = 29.4 \ \mathrm{N} \), find the new speed: \[ v' = \sqrt{\frac{29.4}{0.0550}} \approx 23.5 \ \mathrm{m/s} \].
07

Recalculate Wavelength with New Speed

With the new speed \( v' = 23.5 \ \mathrm{m/s} \), find the new wavelength: \[ \lambda' = \frac{23.5}{120} \approx 0.196 \ \mathrm{m} \].

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Key Concepts

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

Transverse Waves
Transverse waves are a type of wave where the oscillations move perpendicular to the direction of the wave's advance. In simpler terms, the movement of the particles in the medium is at right angles to the direction of wave travel. A common analogy is waves on a guitar string or ripples on a pond; the particles move up and down while the wave moves horizontally. When a tuning fork vibrates a rope, it creates these transverse waves, making them visible as the rope moves up and down.
Understanding the nature of transverse waves is crucial when analyzing wave mechanics because they provide an intuitive grasp of how energy can propagate through a medium without the medium itself traveling with the wave. In this example, the vibration generated by the tuning fork causes the rope to oscillate, allowing us to observe and measure the properties of the transverse wave.
Wave Speed Calculation
Wave speed calculation is a fundamental aspect of understanding wave mechanics. The speed of a wave on a string or rope is determined by two main factors - the tension in the rope and the linear mass density (mass per unit length) of the rope. This relationship is described by the formula:
\[v = \sqrt{\frac{T}{\mu}}\]
where:
  • \(v\) is the wave speed,
  • \(T\) is the tension in the rope, usually calculated as the weight of the mass causing the tension, and
  • \(\mu\) is the linear mass density of the rope.
Calculating wave speed helps in predicting how quickly waves will move along the rope. By using the tension derived from the rope's weight and the linear mass density, you can determine how fast these transverse waves will travel under particular conditions.
Tension and Wave Speed
The relationship between tension and wave speed is quite direct - as tension increases, so does the wave speed. This is because tension essentially acts as a driving force affecting the energy and speed of the wave propagation along the rope.
In our example, when a 1.50 kg mass is used, the tension, calculated with gravity taken into account, is 14.7 N. Plugging this value into the wave speed equation with a given linear mass density allows us to calculate a specific speed. When the weight is increased to 3.00 kg, the tension doubles to 29.4 N. This increased tension results in a higher wave speed, highlighting the proportional relationship between these two variables.
A practical understanding of this concept is valuable, especially in musical instruments and engineering, where controlling wave speed is vital.
Mass and Tension Relationship
The mass and tension relationship is a key concept in determining wave behavior on ropes and strings. The weight of the mass hanging from the rope directly influences the tension force experienced by the rope.
To calculate tension, the formula used is:
\[T = m \cdot g\]
where:
  • \(T\) is the tension,
  • \(m\) is the mass suspended from the rope, and
  • \(g\) is the acceleration due to gravity (approximately 9.8 m/s²).
The mass suspended plays a crucial role because it determines how much force is pulling on the rope. For larger masses, the tension is greater, resulting in higher wave speeds as demonstrated in the exercise with different masses (1.50 kg and 3.00 kg). Understanding this relationship allows you to control both tension and wave properties depending on the application, whether in scientific experiments, manufacturing, or engineering scenarios.

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Most popular questions from this chapter

A fellow student with a mathematical bent tells you that the wave function of a traveling wave on a thin rope is \(y(x, t)=\) 2.30 \(\mathrm{mm} \cos [(6.98 \mathrm{rad} / \mathrm{m}) x+(742 \mathrm{rad} / \mathrm{s}) t] .\) Being more practical, you measure the rope to have a length of 1.35 \(\mathrm{m}\) and a mass of 0.00338 \(\mathrm{kg}\) . You are then asked to determine the following: (a) amphtude; (b) frequency; (c) wavelength; (d) wave speed; (e) direction the wave is traveling; (f) tension in the rope; (g) aver- age power transmitted by the wave.

The wave function of a standing wave is \(y(x, t)=\) 4.44 \(\mathrm{mm} \sin [(32.5 \mathrm{rad} / \mathrm{m}) x] \sin [(754 \mathrm{rad} / \mathrm{s}) t] .\) For the two traveling waves that make up this standing wave, find the (a) amplitude; (b) wavelength; (c) frequency; (d) wave speed; (e) wave functions. (f) From the information given, can you determine which harmonic this is? Explain.

Ultrasound Imaging. Sound having frequencies above the range of human hearing (about \(20,000 \mathrm{Hz}\) ) is called ultrasound. Waves above this frequency can be used to penetrate the body and to produce images by reflecting from surfaces. In a typical ultrasound scan, the waves travel through body tissue with a speed of 1500 \(\mathrm{m} / \mathrm{s}\) . For a good, detailed image, the wavelength should be no more than 1.0 \(\mathrm{mm}\) . What frequency sound is required for a good scan?

Energy Output. By measurement you determine that sound waves are spreading out equally in all directions from a point source and that the intensity is 0.026 \(\mathrm{W} / \mathrm{m}^{2}\) at a distance of 4.3 \(\mathrm{m}\) from the source. (a) What is the intensity at a distance of 3.1 \(\mathrm{m}\) from the source? (b) How much sound energy does the source emit in one hour if its power output remains constant?

A string with both ends held fixed is vibrating in its third harmonic. The waves have a speed of 192 \(\mathrm{m} / \mathrm{s}\) and a frequency of 240 \(\mathrm{Hz}\) . The amplitude of the standing wave at an antinode is 0.400 \(\mathrm{cm}\) . (a) Calculate the amplitude at points on the string a distance of (i) \(40.0 \mathrm{cm} ;\) (ii) \(20.0 \mathrm{cm} ;\) and (iii) 10.0 \(\mathrm{cm}\) from the left end of the string. (b) At each point in part (a), how much time does it take the string to go from its largest upward displacement to its largest downward displacement? (c) Calculate the maximum transverse velocity and the maximum transverse acceleration of the string at each of the points in part (a).

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