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A string has mass \(2.00 \mathrm{~g}\), wave speed \(120 \mathrm{~m} / \mathrm{s}\), and tension \(7.00\) N. (a) What is its length? (b) What is the lowest resonant frequency of this string?

Short Answer

Expert verified
(a) Length = 4.114 m; (b) Lowest resonant frequency = 14.58 Hz.

Step by step solution

01

Convert Mass Units

To solve the problem, convert the mass of the string from grams to kilograms. Since there are 1000 grams in a kilogram, we have: \[ 2.00 \text{ g} = 0.002 \text{ kg} \]
02

Calculate Linear Mass Density

The linear mass density \( \mu \) of the string is the mass per unit length, given by \( \mu = \frac{m}{L} \). We'll use this formula later to solve for the length. For now, rearrange it to express length \( L \) as \( L = \frac{m}{\mu} \).
03

Use Wave Equation

The wave speed \( v \) on the string is related to the tension and linear mass density by the formula \( v = \sqrt{\frac{T}{\mu}} \). Rearrange this equation to solve for \( \mu \): \[ \mu = \frac{T}{v^2} \]Substitute \( T = 7.00 \text{ N} \) and \( v = 120 \text{ m/s} \) to find \( \mu \): \[ \mu = \frac{7.00}{(120)^2} \approx 0.0004861 \text{ kg/m} \]
04

Calculate Length of the String

Now that we have \( \mu \), use the expression for length \( L \) found in Step 2: \[ L = \frac{m}{\mu} \] Substitute \( m = 0.002 \text{ kg} \) and \( \mu = 0.0004861 \text{ kg/m} \): \[ L = \frac{0.002}{0.0004861} \approx 4.114 \text{ m} \]
05

Determine Lowest Resonant Frequency

The lowest resonant frequency, or the fundamental frequency, for a string fixed at both ends is given by: \[ f_1 = \frac{v}{2L} \] Substituting \( v = 120 \text{ m/s} \) and \( L = 4.114 \text{ m} \): \[ f_1 = \frac{120}{2 \times 4.114} \approx 14.58 \text{ Hz} \]

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Key Concepts

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

Wave Speed
Wave speed is an important concept in understanding how waves travel along a medium like a string. Wave speed, noted as \( v \), tells us how fast a wave crest moves along the string. In a given experiment, when the wave speed is noted as \( 120 \text{ m/s} \), it means that any disturbance or wave takes one second to travel 120 meters along the string.
Wave speed can be calculated using the formula: \( v = \sqrt{\frac{T}{\mu}} \). Here, \( T \) is the string tension, and \( \mu \) is the linear mass density. Let's break this down in simpler terms:
  • The higher the tension in the string, the faster the wave travels, leading to a larger wave speed.
  • The greater the mass per length of the string (linear mass density), the slower the wave speed, as it takes more effort to move a heavier string.
Understanding wave speed is crucial since it helps determine other characteristics of the wave, such as frequency and wavelength.
Linear Mass Density
Linear mass density, symbolized as \( \mu \), refers to the mass of the string per unit length, and it is expressed in kg/m. It tells us how much mass there is in one meter of the string. You can think of it as how 'heavy' one unit of length of the string is. This concept is essential for understanding how waves move through the string.
To find the linear mass density, use the formula: \( \mu = \frac{T}{v^2} \). Here:
  • \( T \) represents the tension in the string, given in Newtons (N).
  • \( v \) is the wave speed in meters per second (m/s).
In our specific case, with a tension of 7.00 N and a wave speed of 120 m/s, the calculation becomes \( \mu \approx 0.0004861 \text{ kg/m} \).
This information helps us further determine the string's length and understand the behavior of waves moving along it.
String Tension
String tension is the force applied to stretch the string and is crucial in determining the wave speed within it. The tension is measured in Newtons (N) and influences how tight or loose the string is. A higher tension tightens the string, which can increase the wave speed because it allows the wave to travel more efficiently through the medium.
In our scenario, the tension is specified as 7.00 N. Here's why that's important:
  • High tension results in a smoother and faster wave propagation.
  • Lower tension can cause the wave to travel more slowly and may produce other effects like increased wave amplitude.
An understanding of string tension aids in calculating other wave properties, such as the linear mass density and fundamental frequency, tying into the broader physics of waves.
Fundamental Frequency
The fundamental frequency is the lowest frequency at which a string vibrates when disturbed. It's also known as the first harmonic. This frequency is crucial when determining the notes produced by musical instruments since it is the main note that you hear.
For a string fixed at both ends, the fundamental frequency is given by: \( f_1 = \frac{v}{2L} \). Here:
  • \( v \) is the wave speed along the string in meters per second (m/s).
  • \( L \) is the length of the string in meters (m).
Using the known wave speed of 120 m/s and the calculated length of 4.114 m, the fundamental frequency turns out to be approximately 14.58 Hz.
This frequency is significant because any wave on the string can be seen as a combination of this frequency and its harmonics, making it fundamental to the string's vibrational characteristics.

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Most popular questions from this chapter

A string under tension \(\tau_{i}\) oscillates in the third harmonic at frequency \(f_{3}\), and the waves on the string have wavelength \(\lambda_{3}\). If the tension is increased to \(\tau_{f}=8 \tau_{i}\) and the string is again made to oscillate in the third harmonic, what then are (a) the frequency of oscillation in terms of \(f_{3}\) and (b) the wavelength of the waves in terms of \(\lambda_{3}\) ?

The heaviest and lightest strings on a certain violin have linear densities of \(3.2\) and \(0.26 \mathrm{~g} / \mathrm{m}\). What is the ratio of the diameter of the heaviest string to that of the lightest string, assuming that the strings are of the same material?

The speed of a transverse wave on a string is \(115 \mathrm{~m} / \mathrm{s}\) when the string tension is \(200 \mathrm{~N}\). To what value must the tension be changed to raise the wave speed to \(223 \mathrm{~m} / \mathrm{s}\) ?

A human wave. During sporting events within large, densely packed stadiums, spectators will send a wave (or pulse) around the stadium (Fig. 16-29). As the wave reaches a group of spectators, they stand with a cheer and then sit. At any instant, the width \(w\) of the wave is the distance from the leading edge (people are just about to stand) to the trailing edge (people have just sat down). Suppose a human wave travels a distance of 853 seats around a stadium in \(51 \mathrm{~s}\), with spectators requiring about \(1.8 \mathrm{~s}\) to respond to the wave's passage by standing and then sitting. What are (a) the wave speed \(v\) (in seats per second) and (b) width \(w\) (in number of seats)?

The equation of a transverse wave traveling along a very long string is \(y=3.0 \sin (0.020 \pi x-4.0 \pi t)\), where \(x\) and \(y\) are expressed in centimeters and \(t\) is in seconds. Determine (a) the amplitude, (b) the wavelength, (c) the frequency, (d) the speed, (e) the direction of propagation of the wave, and (f) the maximum transverse speed of a particle in the string. (g) What is the transverse displacement at \(x=3.5 \mathrm{~cm}\) when \(t=0.26 \mathrm{~s}\) ?

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