Chapter 16: Problem 33
A 75.0-cm-long wire of mass 5.625 g is tied at both ends and adjusted to a tension of 35.0 N. When it is vibrating in its second overtone, find (a) the frequency and wavelength at which it is vibrating and (b) the frequency and wavelength of the sound waves it is producing.
Short Answer
Expert verified
(a) Frequency: 136.12 Hz; Wavelength on wire: 0.50 m. (b) Frequency: 136.12 Hz; Wavelength in air: 2.52 m.
Step by step solution
01
Identify the Problem
We have a wire vibrating under tension, and we need to find the frequency and wavelength when it's at its second overtone. We also need to determine the frequency and wavelength of the sound waves it produces.
02
Understand Second Overtone
The second overtone of a vibrating string is its third harmonic. For a string fixed at both ends, the number of antinodes equals the harmonic number. Therefore, in the second overtone (third harmonic), there will be three antinodes.
03
Calculate Wavelength on the Wire
For the third harmonic, the wire length contains 1.5 wavelengths. If the wire is 75.0 cm long, the length of one wavelength is given by: \[\text{Length of one wavelength} = \frac{2}{3} \times 75.0 \, \text{cm} = 50.0 \, \text{cm} = 0.50 \, \text{m}\]
04
Calculate the Wave Speed in the Wire
The wave speed \( v \) in a wire under tension \( T \) is given by \[v = \sqrt{\frac{T}{\mu}} \]where \( \mu = \frac{m}{L} \) is the linear mass density of the wire. Given:- \( m = 5.625 \, \text{g} = 0.005625 \, \text{kg} \)- \( L = 0.75 \, \text{m} \)- \( T = 35.0 \, \text{N} \)The linear mass density \( \mu \) is:\[\mu = \frac{0.005625}{0.75} = 0.0075 \, \text{kg/m}\]Thus, the wave speed is:\[v = \sqrt{\frac{35.0}{0.0075}} = 68.06 \, \text{m/s}\]
05
Calculate Frequency on the Wire
Using the wave speed and wavelength, we calculate the frequency \( f \) using:\[f = \frac{v}{\lambda} = \frac{68.06}{0.50} = 136.12 \, \text{Hz}\]
06
Relate to Sound Waves
The frequency of the sound waves produced is the same as the frequency of the vibrating wire as sound waves are produced by the vibrating wire.
07
Calculate Wavelength of Sound Waves
The speed of sound in air at room temperature is approximately 343 m/s. Using this speed and the previously calculated frequency, the wavelength \( \lambda_{sound} \) in the air is:\[\lambda_{sound} = \frac{343}{136.12} = 2.52 \, \text{m}\]
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Vibrating String
A vibrating string is a key concept in physics, especially when understanding wave phenomena. When a string is fixed at both ends, it can produce standing waves. These are waves that appear to be stationary, but in reality, they are a result of two waves traveling in opposite directions. The fixed points at the ends are called nodes, where there is no movement, and the points of maximum movement are antinodes.
- For any given harmonic, the number of antinodes is the same as the harmonic number.
- The first harmonic is the simplest form, with one antinode in the middle.
- Each subsequent harmonic adds an antinode and creates a more complex wave pattern.
Wave Speed Calculation
Calculating the wave speed on a string under tension is fundamental to solving problems involving vibrations. The wave speed on a string depends on its tension and linear mass density. The formula to find the speed of a wave traveling along a string is:
\[v = \sqrt{\frac{T}{\mu}} \] Where:
This speed of the wave on the wire will help determine the frequency with which it is vibrating in a specific harmonic.
\[v = \sqrt{\frac{T}{\mu}} \] Where:
- \(v\) is the wave speed
- \(T\) is the tension applied to the string
- \(\mu\) (mu) is the linear mass density, calculated as mass \(m\) divided by length \(L\).
This speed of the wave on the wire will help determine the frequency with which it is vibrating in a specific harmonic.
Second Overtone
The term "second overtone" can be a bit tricky because it's often mistaken for the second harmonic, but it is actually the third harmonic in the context of a vibrating string fixed at both ends.
To find the wavelength in this setting, if the total length of the wire is 0.75 meters, each wavelength occupies two-thirds of the string's total length, as calculated \[ \lambda = \frac{2}{3} \times \text{Total Length} = 0.5 \, \text{m}\] With this wavelength and the calculated wave speed of 68.06 m/s, we derive the frequency as \[f = \frac{v}{\lambda} = 136.12 \, \text{Hz}\].
This frequency is also the frequency of the sound waves produced by the string, as the vibrations from the wire produce sound waves in the surrounding air. The sound will have a different wavelength due to the different speed of sound in air, calculated as 2.52 m.
- In the third harmonic, the wire supports three antinodes.
- The length of the wire is divided into sections that correspond to three half-wavelengths.
To find the wavelength in this setting, if the total length of the wire is 0.75 meters, each wavelength occupies two-thirds of the string's total length, as calculated \[ \lambda = \frac{2}{3} \times \text{Total Length} = 0.5 \, \text{m}\] With this wavelength and the calculated wave speed of 68.06 m/s, we derive the frequency as \[f = \frac{v}{\lambda} = 136.12 \, \text{Hz}\].
This frequency is also the frequency of the sound waves produced by the string, as the vibrations from the wire produce sound waves in the surrounding air. The sound will have a different wavelength due to the different speed of sound in air, calculated as 2.52 m.