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Chapter 15: Problem 42

One string of a certain musical instrument is \(75.0 \mathrm{~cm}\) long and has a mass of \(8.75 \mathrm{~g}\). It is being played in a room where the speed of sound is \(344 \mathrm{~m} / \mathrm{s}\). (a) To what tension must you adjust the string so that, when vibrating in its second overtone, it produces sound of wavelength \(0.765 \mathrm{~m} ?\) (Assume that the breaking stress of the wire is very large and isn't exceeded.) (b) What frequency sound does this string produce in its fundamental mode of vibration?

Short Answer

Expert verified
a) The tension required to produce sound of wavelength \( 0.765 \: m \) when vibrating in its second overtone is approximately \( 591 \: N \). b) The frequency of the sound produced in the fundamental mode of vibration is approximately \( 144 \: Hz \). Note: The exact values can somewhat differ due to the rounding in the calculations.

Step by step solution

01

Calculate the tension for the second overtone

To find the needed tension, first recognize that the second overtone corresponds to the third harmonic, which means the string is divided into three segments. Each segment represents a wave with a wavelength equal to \(2/3\) of the length of the string. So, one can write \( 位_{segment}=\frac{2L}{3} \) where \( 位_{segment} \) is the wavelength of each segment and \( L \) is the length of the string. Given that the speed of sound \( v \) is equal to the product of the wavelength of the sound \( 位_{sound} \) and frequency \( f \), write \( 位_{sound}=v/f=2v/3f_{1} \) where \( f_{1} \) is the fundamental frequency and the factor 2/3 comes from the division into three segments for the third harmonic. Solve this equation for \( f_{1} \), substitute it into the formula \( f_{1}=\frac{1}{2L}\sqrt{\frac{T}{\mu}} \) where \( T \) is the tension and \( \mu \) is the linear mass density, and solve for \( T \).
02

Calculate the fundamental frequency

To find the fundamental frequency \( f_{1} \), use the formula \( f_{1}=\frac{1}{2L}\sqrt{\frac{T}{\mu}} \), substituting the calculated tension \( T \) and the provided length \( L \) and mass \( m \). To calculate the linear mass density \( \mu \), use the formula \( \mu = m/L \), substituting the given mass and length.

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Key Concepts

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

Harmonic Frequency
Harmonic frequency refers to the frequencies at which standing waves are established in a medium like a string. In this context, the **second overtone** corresponds to the third harmonic. For a string fixed at both ends, the harmonics are integer multiples of the fundamental frequency.
  • The fundamental frequency (\(f_1\)) is the lowest frequency at which the string vibrates.
  • The nth harmonic (\(f_n\)) is \(n\) times the fundamental frequency.
For example, if the fundamental frequency is 100 Hz, then:
  • The second harmonic is 200 Hz.
  • The third harmonic (or second overtone) is 300 Hz.
Recognizing harmonics is essential in understanding how musical instruments produce different pitches. It dictates the vibration modes and helps in calculating the wavelength for various notes.In our exercise, the third harmonic is critical to understand how the string can appropriately be tuned for precise sound quality. It's essential to use the wave equations accurately to solve for tension and frequency.
String Tension
String tension is an essential factor when determining the pitch and quality of the sound produced by a string instrument. String tension (\(T\)) defines how tight the string is, impacting the speed at which waves travel along it. The formula used to investigate tension is derived from the fundamental frequency:\[f_1 = \frac{1}{2L}\sqrt{\frac{T}{\mu}}\]Here:
  • \(f_1\) is the fundamental frequency.
  • \(L\) is the length of the string.
  • \(T\) is the tension in the string.
  • \(\mu\) is the linear mass density, calculated as the mass (\(m\)) divided by length (\(L\)) of the string: \(\mu = \frac{m}{L}\).
The correct string tension ensures that the instrumental cord vibrates at its intended frequency. By maintaining appropriate tension, musicians can control pitch perfectly, achieving the desired sound, especially in string instruments like guitars or violins.In this problem, students learn how to adjust string tension to attain a specific sound, aiding in effective tuning of musical instruments.
Speed of Sound
Speed of sound in a medium is the rate at which sound waves travel through it. In our exercise, the given speed of sound is 344 m/s. This value is typical at room temperature when sound travels through air. Sound鈥檚 speed varies with temperature, medium, and other factors.Understanding the relationship between wave speed, frequency, and wavelength is critical:\[v = f \cdot \lambda\]Where:
  • \(v\) is the speed of sound.
  • \(f\) is the frequency of the sound wave.
  • \(\lambda\) is the wavelength of the sound.
Given this relationship, altering one aspect of the equation requires a change in the others to maintain balance. If frequency increases, wavelength must decrease to keep \(v\) constant.For the exercise, this understanding helps in determining the frequency produced by a vibrating string as well as explaining why particular sound frequencies occur at specific tension levels.

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Most popular questions from this chapter

A piano tuner stretches a steel piano wire with a tension of \(800 \mathrm{~N}\). The steel wire is \(0.400 \mathrm{~m}\) long and has a mass of \(3.00 \mathrm{~g}\). (a) What is the frequency of its fundamental mode of vibration? (b) What is the number of the highest harmonic that could be heard by a person who is capable of hearing frequencies up to \(10,000 \mathrm{~Hz} ?\)

A heavy rope \(6.00 \mathrm{~m}\) long and weighing \(29.4 \mathrm{~N}\) is attached at one end to a ceiling and hangs vertically. A \(0.500 \mathrm{~kg}\) mass is suspended from the lower end of the rope. What is the speed of transverse waves on the rope at the (a) bottom of the rope, (b) middle of the rope, and (c) top of the rope? (d) Is the tension in the middle of the rope the average of the tensions at the top and bottom of the rope? Is the wave speed at the middle of the rope the average of the wave speeds at the top and bottom? Explain.

A horizontal wire is stretched with a tension of \(94.0 \mathrm{~N},\) and the speed of transverse waves for the wire is \(406 \mathrm{~m} / \mathrm{s}\). What must the amplitude of a traveling wave of frequency \(69.0 \mathrm{~Hz}\) be for the average power carried by the wave to be \(0.365 \mathrm{~W} ?\)

You are exploring a newly discovered planet. The radius of the planet is \(7.20 \times 10^{7} \mathrm{~m}\). You suspend a lead weight from the lower end of a light string that is \(4.00 \mathrm{~m}\) long and has mass \(0.0280 \mathrm{~kg}\). You measure that it takes \(0.0685 \mathrm{~s}\) for a transverse pulse to travel from the lower end to the upper end of the string. On the earth, for the same string and lead weight, it takes \(0.0390 \mathrm{~s}\) for a transverse pulse to travel the length of the string. The weight of the string is small enough that you ignore its effect on the tension in the string. Assuming that the mass of the planet is distributed with spherical symmetry, what is its mass?

You must determine the length of a long, thin wire that is suspended from the ceiling in the atrium of a tall building. A \(2.00-\mathrm{cm}\) -long piece of the wire is left over from its installation. Using an analytical balance, you determine that the mass of the spare piece is \(14.5 \mu \mathrm{g}\). You then hang a \(0.400 \mathrm{~kg}\) mass from the lower end of the long, suspended wire. When a small-amplitude transverse wave pulse is sent up that wire, sensors at both ends measure that it takes the wave pulse \(26.7 \mathrm{~ms}\) to travel the length of the wire. (a) Use these measurements to calculate the length of the wire. Assume that the weight of the wire has a negligible effect on the speed of the transverse waves. (b) Discuss the accuracy of the approximation made in part (a).
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