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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 of \(0.300 \mathrm{~cm}\) at the antinodes. (a) What is the speed of propagation of transverse waves in the wire? (b) Compute the tension in the wire.

Short Answer

Expert verified
(a) The speed of the wave is 96.0 m/s. (b) The tension in the wire is 460.8 N.

Step by step solution

01

Understanding the Fundamental Mode

In the fundamental mode of vibration for a string fixed at both ends, the length of the wire corresponds to half of the wavelength. So if the length of the wire is L, then the wavelength \( \lambda \) is given by \( \lambda = 2L \).
02

Calculate Wavelength

Given that the length \( L = 80.0 \ \mathrm{cm} = 0.8 \ \mathrm{m} \), the wavelength \( \lambda \) can be calculated using the formula from Step 1: \( \lambda = 2L = 2 \times 0.8 = 1.6 \ \mathrm{m} \).
03

Use the Wave Speed Formula

The speed of a wave is given by the formula \( v = f \lambda \), where \( f \) is the frequency. Here, \( f = 60.0 \ \mathrm{Hz} \). Substituting in the values, we get: \( v = 60.0 \times 1.6 = 96.0 \ \mathrm{m/s} \).
04

Calculate Linear Mass Density

Linear mass density \( \mu \) is mass per unit length. Given that the mass of the wire is \( 40.0 \mathrm{~g} = 0.040 \mathrm{~kg} \) and the length is \( 0.8 \mathrm{~m} \), \( \mu = \frac{0.040}{0.8} = 0.050 \mathrm{~kg/m} \).
05

Calculate the Tension in the Wire

The wave speed on the wire can also be expressed in terms of the tension \( T \) and linear mass density \( \mu \): \( v = \sqrt{\frac{T}{\mu}} \). Rearranging and solving for \( T \), we have \( T = \mu v^2 = 0.050 \times (96.0)^2 \). Calculating gives \( T = 0.050 \times 9216 = 460.8 \mathrm{~N} \).

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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 fascinating phenomena where particles of the medium move perpendicular to the direction of the wave's propagation. Think about how waves on a string or water surface propagate. While the wave travels horizontally, individual particles move up and down or side to side, creating the characteristic peaks and troughs of a wave.

In a stretched wire like the one in our exercise, transverse waves carry energy without transporting matter over long distances. The particles simply oscillate around fixed positions due to the tension in the wire. This tension, along with the properties of the wire, determines the speed at which the wave travels. Understanding transverse waves is crucial since they reveal much about the medium they're traveling through.
Wave Tension
Wave tension is a critical factor that affects the speed of a wave traveling through a string or wire. Tension can be thought of as the force that keeps the wire taut, and it greatly influences how quickly the wave can move from one end to another.

In this context, the wave speed equation summarizes the relationship: \[ v = \sqrt{\frac{T}{\mu}} \] where \( v \) is the wave speed, \( T \) is the tension, and \( \mu \) is the linear mass density.

From the exercise, we know that the wave speed is given, and by solving for tension using the calculated linear mass density, we find that tension effectively controls the wave propagation speed. Knowing the tension can help us analyze the force distribution within the wire, which is essential for designing instruments or structures involving tensioned wires.
Fundamental Mode
The fundamental mode represents the simplest vibration pattern of a string fixed at both ends, commonly referred to as the first harmonic. This basic mode has only one antinode, located at the center, and nodes at each end where the string is fixed.

In this simplest vibrational state, the wavelength \( \lambda \) is twice the length of the string: \[ \lambda = 2L \] This relationship helps us calculate the wavelength from the known length of the wire in the exercise.

Understanding the fundamental mode is important because it determines the natural frequency, or the pitch, of the wire's vibration. Musicians, for example, rely on this concept to tune their instruments.
Linear Mass Density
Linear mass density \( \mu \) is the measure of mass distributed per unit length of a material, like our wire in the exercise. It defines how dense the wire is linearly, which significantly affects how it behaves when a wave travels through it.

The equation for linear mass density is: \[ \mu = \frac{m}{L} \] where \( m \) is the mass of the wire and \( L \) is its length. This value directly influences the wave speed due to its appearance in the wave speed formula involving tension: \[ v = \sqrt{\frac{T}{\mu}} \]

A higher linear mass density usually results in a slower wave speed, given a constant tension, because more mass per length means more inertia resisting the wave's propagation. This is crucial for understanding how varying material properties affect wave behavior on different scales.

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Most popular questions from this chapter

A \(1.50-\mathrm{m}\) -long rope is stretched between two supports with a tension that makes the speed of transverse waves \(48.0 \mathrm{~m} / \mathrm{s}\). What are the wavelength and frequency of (a) the fundamental tone? (b) the second overtone? (c) the fourth harmonic?

On the planet Arrakis, a male ornithoid is flying toward his stationary mate at \(25.0 \mathrm{~m} / \mathrm{s}\) while singing at a frequency of \(1200 \mathrm{~Hz}\). If the female hears a tone of \(1240 \mathrm{~Hz}\), what is the speed of sound in the atmosphere of Arrakis?

With what tension must a rope with length \(2.50 \mathrm{~m}\) and mass \(0.120 \mathrm{~kg}\) be stretched for transverse waves of frequency \(40.0 \mathrm{~Hz}\) to have a wavelength of \(0.750 \mathrm{~m} ?\)

For cranial ultrasound, why is it advantageous to use frequencies in the kHZ range rather than in the MHz range? A. The antinodes of the standing waves will be closer together at the lower frequencies than at the higher frequencies. B. There will be no standing waves at the lower frequencies. C. Cranial bones will attenuate the ultrasound more at the lower frequencies than at the higher frequencies. D. Cranial bones will attenuate the ultrasound less at the lower frequencies than at the higher frequencies.

A baseball park is filled with 5000 home-team fans. In the bottom of the ninth inning, the home team's leading batter hits a game winning home run and all 5000 fans stand and cheer. Assume that each individual fan produces a sound intensity of \(10^{-5} \mathrm{~W} / \mathrm{m}^{2},\) as heard from the pitching mound. (a) What is the total sound intensity heard at the pitching mound? (b) What is the total sound intensity level at the pitching mound?

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