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Chapter 15: Problem 41

A wire with mass 40.0 \(\mathrm{g}\) is stretched so that its ends are tied down at points 80.0 \(\mathrm{cm}\) apart. The wire vibrates in its fundamental mode with frequency 60.0 \(\mathrm{Hz}\) and with an amplitude at the antinodes of 0.300 \(\mathrm{cm} .\) (a) What is the speed of propagation oftransverse waves in the wire? (b) Compute the tension in the wire. (c) Find the maximum transverse velocity and acceleration of particles in the wire.

Short Answer

Expert verified
Wave speed is 96.0 m/s, tension is 460.8 N, max velocity is 1.131 m/s, and max acceleration is 426.0 m/s虏.

Step by step solution

01

Understanding the Fundamental Mode

The fundamental mode of vibration has a single antinode in the middle and nodes at both ends of the wire. This means the wavelength \( \lambda \) is twice the length of the wire. Given the length of the wire is 80.0 cm, or 0.8 m, the wavelength is \( \lambda = 2 \times 0.8 = 1.6 \text{ m} \).
02

Finding the Wave Speed

Using the relationship between frequency \( f \), wavelength \( \lambda \), and wave speed \( v \), defined by \( v = f \lambda \), we can substitute the given values: \( f = 60.0 \text{ Hz} \) and \( \lambda = 1.6 \text{ m} \). So \( v = 60.0 \times 1.6 = 96.0 \text{ m/s} \).
03

Calculating Tension in the Wire

The speed of a wave in a string is given by \( v = \sqrt{\frac{T}{\mu}} \), where \( T \) is the tension in the wire and \( \mu \) is the linear mass density, \( \mu = \frac{m}{L} \). The mass of the wire is 40.0 g (0.040 kg) and length is 0.8 m, so \( \mu = \frac{0.040}{0.8} = 0.050 \text{ kg/m} \). Solving for \( T \), we have \( T = v^2 \cdot \mu = 96.0^2 \times 0.050 = 460.8 \text{ N} \).
04

Determining Maximum Transverse Velocity

The maximum transverse velocity \( v_{max} \) in a vibrating string is \( v_{max} = 2\pi f A \), where \( A \) is the amplitude. Given \( A = 0.300 \text{ cm} = 0.003 \text{ m} \), \( v_{max} = 2\pi \times 60.0 \times 0.003 = 1.131 \text{ m/s} \).
05

Calculating Maximum Transverse Acceleration

The maximum transverse acceleration \( a_{max} \) is \( a_{max} = (2\pi f)^2 A \). Using the values \( A = 0.003 \text{ m} \) and \( f = 60.0 \), we calculate \( a_{max} = (2\pi \times 60.0)^2 \times 0.003 = 426.0 \text{ m/s}^2 \).

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Key Concepts

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

Fundamental Mode
In wave mechanics, the fundamental mode of a string or wire refers to the simplest form of vibration. In this mode, there is only one antinode, which is a point of maximum displacement in the middle of the wire and nodes, points of no displacement, at both ends. To visualize, think of a jump rope moving in one large hump when it's swung up and down. For the given problem, the wire's length is 0.8 meters, and in its fundamental mode, the wavelength is twice this length. Hence, the wavelength \( \lambda \) is 1.6 meters. This understanding is crucial for calculating wave properties such as speed and tension.
Transverse Waves
Transverse waves are waves where the particle displacement is perpendicular to the direction of wave propagation. Imagine flicking a slinky side-to-side; the waves move along the slinky, but each coil moves up and down. In the context of a vibrating wire, these are the waves traveling along the wire, causing it to move up and down while the wave travels horizontally. Understanding transverse waves helps in analyzing how vibrations are sustained and propagated in strings and wires.
Wave Speed
Wave speed in a string relates highly to both the frequency and the wavelength of the wave. In equation form, wave speed \( v \) is represented as \( v = f \lambda \), where \( f \) is the frequency and \( \lambda \) is the wavelength. Here, the given frequency is 60 Hz, and the calculated wavelength is 1.6 meters, resulting in a wave speed of 96 m/s. Wave speed is essential in determining how quickly the wave travels from one point to another along the string.
Tension in Strings
The tension in a string plays a pivotal role in how waves propagate through it. In simple terms, tension is the force that is pulling the string tight. Practically, it's like how tightly a guitar string is wound. The formula for wave speed, \( v = \sqrt{\frac{T}{\mu}} \), where \( T \) is the tension and \( \mu \) is the linear mass density, shows the relationship. In this exercise, with a wave speed of 96 m/s and a linear mass density \( \mu = 0.050 \text{ kg/m} \), the tension \( T \) computes to 460.8 N. The higher the tension, the faster the wave travels on the string.
Transverse Velocity and Acceleration
In vibrating strings, particles move up and down as the wave passes. This motion is characterized by transverse velocity and acceleration. The maximum transverse velocity \( v_{max} \) is calculated by the equation \( v_{max} = 2\pi f A \). In the case of this wire, with an amplitude \( A \) of 0.003 meters, it results in a velocity of 1.131 m/s. Transverse acceleration, on the other hand, is the rate of change of this velocity with time, given by \( a_{max} = (2\pi f)^2 A \), yielding an acceleration of 426.0 m/s虏. These values describe how fast and intensely the wire moves as it vibrates.

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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?

A \(5.00-\mathrm{m}, 0.732-\mathrm{kg}\) wire is used to support two uni-? form \(235-\mathrm{N}\) posts of equal length (Fig. P15.62). Assume that the wire is essentially horizontal and that the speed of sound is 344 \(\mathrm{m} / \mathrm{s} . \mathrm{A}\) strong wind is blowing, causing the wire to vibrate in its 5 th overtone. What are the frequency and wavelength of the sound this wire produces?

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 m from the source? (b) How much sound energy does the source emit in one hour if its power output remains constant?

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?

A simple harmonic oscillator at the point \(x=0\) generates a wave on a rope. The oscillator operates at a frequency of 40.0 \(\mathrm{Hz}\) and with an amplitude of 3.00 \(\mathrm{cm} .\) The rope has a linear mass density of 50.0 \(\mathrm{g} / \mathrm{m}\) and is stretched with a tension of 5.00 \(\mathrm{N}\) .(a) Determine the speed of the wave. (b) Find the wavelength. (c) Write the wave function \(y(x, t)\) for the wave, Assume that the oscillator has its maximum upward displacement at time \(t=0\) . (d) Find the maximum transverse acceleration of points on the rope. (e) In the discussion of transverse waves in this chapter, the force of gravity was ignored. Is that a reasonable approximation for this wave? Explain.
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