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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.
In the exercise, the second overtone corresponds to the third harmonic of the string. This means there are three antinodes along the length of the wire. The length of the wire contains one and a half wavelengths in this configuration.
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:
  • \(v\) 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 \(L\).
For example, in the exercise provided, the mass of the wire is 5.625 grams and its length is 75 centimeters. To find the linear mass density, we convert the mass to kilograms and length to meters. This results in a linear mass density of 0.0075 kg/m. Using all given values, the wave speed is approximately 68.06 m/s.
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.
  • In the third harmonic, the wire supports three antinodes.
  • The length of the wire is divided into sections that correspond to three half-wavelengths.
This third harmonic setup signifies that the entire length of the wire is filled with 1.5 wavelengths of the wave.
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.

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Most popular questions from this chapter

A car alarm is emitting sound waves of frequency 520 Hz. You are on a motorcycle, traveling directly away from the parked car. How fast must you be traveling if you detect a frequency of 490 Hz?

How fast (as a percentage of light speed) would a star have to be moving so that the frequency of the light we receive from it is 10.0% higher than the frequency of the light it is emitting? Would it be moving away from us or toward us? (Assume it is moving either directly away from us or directly toward us.)

Horseshoe bats (genus \(Rhinolophus\)) emit sounds from their nostrils and then listen to the frequency of the sound reflected from their prey to determine the prey's speed. (The "horseshoe" that gives the bat its name is a depression around the nostrils that acts like a focusing mirror, so that the bat emits sound in a narrow beam like a flashlight.) A \(Rhinolophus\) flying at speed \(v_{bat}\) emits sound of frequency \(f_{bat}\); the sound it hears reflected from an insect flying toward it has a higher frequency \(f_{refl}\). (a) Show that the speed of the insect is $$vinsect = v\Bigg[\frac{f_{refl}(v - v_{bat}) - f_{bat}(v + v_{bat})}{f_{refl}(v - v_{bat}) + f_{bat}(v + v_{bat})}\Bigg] $$ where \(v\) is the speed of sound. (b) If \(f_{bat} =\) 80.7 kHz, \(f_{refl} =\) 83.5 kHz, and \(v_{bat} =\) 3.9 m/s, calculate the speed of the insect.

Singing in the Shower. A pipe closed at both ends can have standing waves inside of it, but you normally don't hear them because little of the sound can get out. But you \(can\) hear them if you are \(inside\) the pipe, such as someone singing in the shower. (a) Show that the wavelengths of standing waves in a pipe of length \(L\) that is closed at both ends are \(\lambda$$_n =\) \(2L /n\) and the frequencies are given by \({f_n}\) \(= n$$v/2L = nf$$_1\), where \(n =\) 1, 2, 3, . . . . (b) Modeling it as a pipe, find the frequency of the fundamental and the first two overtones for a shower 2.50 m tall. Are these frequencies audible?

You live on a busy street, but as a music lover, you want to reduce the traffic noise. (a) If you install special soundreflecting windows that reduce the sound intensity level (in dB) by 30 dB, by what fraction have you lowered the sound intensity (in W\(/\)m\(^2\))? (b) If, instead, you reduce the intensity by half, what change (in dB) do you make in the sound intensity level?
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