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Chapter 15: Problem 19

A thin, 75.0 -cm wire has a mass of 16.5 \(\mathrm{g} .\) One end is tied to a nail, and the other end is attached to a screw that can be adjusted to vary the tension in the wire. (a) To what tension (in newtons) must you adjust the screw so that a transverse wave of wavelength 3.33 \(\mathrm{cm}\) makes 875 vibrations per second? (b) How fast would this wave travel?

Short Answer

Expert verified
(a) Adjust tension to 18.67 N. (b) Wave speed is 29.14 m/s.

Step by step solution

01

Convert Units

Firstly, convert the units to the standard metric system. The length of the wire, \(75.0\, \text{cm}\), is converted to meters: \(0.75\, \text{m}\). The mass of the wire, \(16.5\, \text{g}\), is converted to kilograms: \(0.0165\, \text{kg}\). The wavelength \(3.33\, \text{cm}\) is converted to meters: \(0.0333\, \text{m}\).
02

Determine the Wave Velocity

Use the formula for the velocity of a wave, \( v = f \lambda \), where \( f \) is the frequency and \( \lambda \) is the wavelength. The frequency is \(875\, \text{Hz}\) and the wavelength is \(0.0333\, \text{m}\), so the velocity \( v \) is \( v = 875 \times 0.0333 \). Calculating this gives \( v = 29.1375\, \text{m/s}\).
03

Calculate Linear Mass Density

The linear mass density \( \mu \) of the wire is \( \mu = \frac{m}{L} \), where \( m \) is the mass and \( L \) is the length of the wire. Substituting the known values, \( \mu = \frac{0.0165}{0.75} = 0.022\, \text{kg/m}\).
04

Calculate Tension in the Wire

Using the wave velocity and linear mass density, calculate the tension \( T \) in the wire using the formula \( v = \sqrt{\frac{T}{\mu}} \). Rearrange to solve for \( T \): \( T = v^2 \mu \). Substituting the values, \( T = (29.1375)^2 \times 0.022\). Calculating gives \( T \approx 18.67\, \text{N}\).
05

Verify Wave Speed Calculation

Use the derived tension to verify the speed of the wave using \( v = \sqrt{\frac{T}{\mu}} \). Substituting \( T = 18.67 \) and \( \mu = 0.022 \), verify that \( v \approx 29.1375\, \text{m/s}\), consistent with the previous calculation.

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Key Concepts

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

Wave Velocity
Wave velocity tells us how fast a wave travels through a medium. In the context of a wire, the velocity of a transverse wave can be calculated using the formula:\[ v = f \lambda \]where:
  • \( v \) is the wave velocity in meters per second (m/s)
  • \( f \) is the frequency of the wave in hertz (Hz)
  • \( \lambda \) is the wavelength in meters (m)
Understanding wave velocity is crucial in applications like musical instruments, where the speed of waves on strings determines the pitch we hear. With a frequency of \( 875 \, \text{Hz} \) and a wavelength of \( 0.0333 \, \text{m} \), the wave velocity can be clearly computed as \( v = 29.1375 \, \text{m/s} \). This tells us how quickly disturbances propagate along the wire.
Tension in a Wire
Tension is a force stretching the wire linearly that's applied to either end. The tension in a wire affects the wave velocity, and can be calculated if you know the wave velocity and the linear mass density:\[ T = v^2 \mu \]Here:
  • \( T \) is the tension in newtons (N)
  • \( v \) is the wave velocity
  • \( \mu \) is the linear mass density
In this exercise, with a wave velocity of \( 29.1375 \, \text{m/s} \) and linear mass density \( 0.022 \, \text{kg/m} \), the tension comes out to be around \( 18.67 \, \text{N} \). Ensuring correct tension is important for performance in structures and instruments made with wires or strings.
Linear Mass Density
Linear mass density \( \mu \) describes how much mass is distributed along a unit length of the wire:\[ \mu = \frac{m}{L} \]Where:
  • \( m \) is the mass of the wire in kilograms (kg)
  • \( L \) is the length of the wire in meters (m)
For the given wire with a mass of \( 0.0165 \, \text{kg} \) and a length of \( 0.75 \, \text{m} \), the linear mass density is computed as \( 0.022 \, \text{kg/m} \). Linear mass density plays a vital role in understanding how a wave travels through the wire 鈥� higher density wires will generally result in slower wave propagation.
Transverse Wave
Transverse waves are characterized by particle displacement perpendicular to the direction of wave travel. They're common in strings, where the vibration creates peaks and troughs as the wave propagates. Transverse waves are responsible for the musical notes produced by instruments like guitars and violins. In this exercise, we're looking at a transverse wave on a wire, with properties shaped by both tension and mass distribution:
  • Unlike longitudinal waves, transverse waves can travel in media like strings and surfaces but not through fluids.
  • The frequency and wavelength determine both the pitch and tone quality of notes played.
      • Understanding transverse waves helps us appreciate how vibrations translate into sound and how engineers can manipulate such properties to alter wave behaviors.

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Most popular questions from this chapter

A sinusoidal transverse wave travels on a string. The string has length 8.00 \(\mathrm{m}\) and mass 6.00 \(\mathrm{g} .\) The wave speed is \(30.0 \mathrm{m} / \mathrm{s},\) and the wavelength is 0.200 \(\mathrm{m} .\) (a) If the wave is to have an average power of \(50.0 \mathrm{W},\) what must be the amplitude of the wave? (b) For this same string, if the amplitude and wave-length are the same as in part (a), what is the average power for the wave if the tension is increased such the wave speed is doubled?

Threshold of Pain. You are investigating the report of a UFO landing in an isolated portion of New Mexico, and you encounter a strange object that is radiating sound waves uniformly in all directions. Assume that the sound comes from a point source and that you can ignore reflections. You are slowly walking toward the source. When you are 7.5 \(\mathrm{m}\) from it, you measure its intensity to be 0.11 \(\mathrm{W} / \mathrm{m}^{2}\) . An intensity of 1.0 \(\mathrm{W} / \mathrm{m}^{2}\) is often used as the "threshold of pain." How much closer to the source can you move before the sound intensity reaches this threshold?

Audible Sound. Provided the amplitude is sufficiently great, the human ear can respond to longitudinal waves over a range of frequencies from about 20.0 \(\mathrm{Hz}\) to about 20.0 \(\mathrm{kHz}\) . (a) If you were to mark the beginning of each complete wave pattern with a red dot for the long-wavelength sound and a blue dot for the short-wavelength sound, how far apart would the red dots be, and how far apart would the blue dots be? (b) In reality would adjacent dots in each set be far enough apart for you to easily measure their separation with a meter stick? (c) Suppose you repeated part (a) in water, where sound travels at 1480 \(\mathrm{m} / \mathrm{s}\) . How far apart would the dots be in each set? Could you readily measure their separation with a meter stick?

A piano wire with mass 3.00 \(\mathrm{g}\) and length 80.0 \(\mathrm{cm}\) is stretched with a tension of 25.0 \(\mathrm{N}\) . A wave with frequency 120.0 \(\mathrm{Hz}\) and amplitude 1.6 \(\mathrm{mm}\) travels along the wire. (a) Calculate the average power carried by the wave. (b) What happens to the average power if the wave amplitude is halved?

A light wire is tightly stretched with tension \(F .\) Trans- verse traveling waves of amplitude \(A\) and wavelength \(\lambda_{1}\) carry average power \(P_{\mathrm{av}, 1}=0.400 \mathrm{W}\) . If the wavelength of the waves is doubled, so \(\lambda_{2}=2 \lambda_{1},\) while the tension \(F\) and amplitude \(A\) are not altered, what then is the average power \(P_{\mathrm{av}, 2}\) carried by the waves?
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