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A violin string \(15.0 \mathrm{~cm}\) long and fixed at both ends oscillates in its \(n=1\) mode. The speed of waves on the string is \(250 \mathrm{~m} / \mathrm{s}\), and the speed of sound in air is \(348 \mathrm{~m} / \mathrm{s}\). What are the (a) frequency and (b) wavelength of the emitted sound wave?

Short Answer

Expert verified
The frequency is 833.33 Hz, and the wavelength is 0.418 m.

Step by step solution

01

Understanding the Fundamental Frequency

The fundamental frequency (or first harmonic) occurs when the string vibrates in its simplest mode, or with one antinode in the middle. The wavelength of the wave on the string, in this case, is twice the length of the string. Therefore, for the fundamental frequency, the wavelength of the wave on the string is calculated as: \[ \lambda = 2L \] where \( L = 15.0 \text{ cm} = 0.150 \text{ m} \). Thus, \( \lambda = 2 \times 0.150 \text{ m} = 0.300 \text{ m} \).
02

Calculating the Frequency on the String

The frequency of the wave on the string can be calculated using the wave speed formula: \[ v = f \lambda \] where \( v = 250 \text{ m/s} \), \( \lambda = 0.300 \text{ m} \), and \( f \) is the frequency. Rearrange the formula to solve for \( f \): \[ f = \frac{v}{\lambda} = \frac{250 \text{ m/s}}{0.300 \text{ m}} = \frac{250}{0.300} \]. Calculate to find \( f \).
03

Calculating the Frequency of the Emitted Sound

The frequency of the emitted sound wave in air is the same as the frequency of the wave on the string since the string and air vibrations are synchronized at the same frequency. Thus, the frequency of the sound wave is the same as calculated in Step 2 which is \( 833.33 \text{ Hz} \).
04

Calculating the Wavelength of the Emitted Sound

To find the wavelength of the sound wave in air, use the wave speed formula again with the speed of sound in air: \[ v = f \lambda_{sound} \] where \( v = 348 \text{ m/s} \) and \( f = 833.33 \text{ Hz} \). Rearrange the formula to solve for \( \lambda_{sound} \): \[ \lambda_{sound} = \frac{v}{f} = \frac{348 \text{ m/s}}{833.33 \text{ Hz}} \]. Calculate to find the wavelength in air.

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Key Concepts

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

Harmonics
When it comes to string instruments like violins or guitars, harmonics play a significant role in the sound that is produced. Harmonics refer to the different modes of vibration that a string can have. Each mode, known as a harmonic, corresponds to a standing wave pattern. The fundamental frequency is the lowest frequency at which the string vibrates, creating the first mode of vibration, or the first harmonic. Higher harmonics represent modes where the string vibrates with more complicated patterns, showing additional nodes or points of no displacement between the ends.
  • The first harmonic or fundamental contains a single antinode, implying one complete wave fits along the string length twice.
  • Subsequent harmonics, such as the second and third, can be imagined as squeezing more waves into the same length.
Understanding harmonics is crucial, as these vibrations form the foundation for calculating frequencies and wavelengths necessary for music and acoustics.
Fundamental Frequency
The fundamental frequency is vital in determining the sound's pitch produced by a vibrating string. It is the lowest frequency mode of vibration, occurring when there is one antinode along the length of the string. This frequency can be calculated by understanding the relationship between wave speed, wavelength, and frequency itself.
To find the fundamental frequency of a string that's fixed at both ends, use the formula:
  • For a string oscillating in its fundamental mode, the wavelength is twice the length of the string, as given by the equation: \[\lambda = 2L\] where \( L \) is the string length.
  • The fundamental frequency \( f \) can then be calculated using the wave speed \( v \) equation: \[ v = f \lambda \] which rearranges to: \[ f = \frac{v}{\lambda} \]
Calculating the fundamental frequency helps in setting the correct pitch for the string instrument.
Speed of Sound
The speed of sound is a fundamental concept in physics, representing how quickly sound waves travel through a medium such as air. It is influenced by factors like the medium's composition, temperature, and pressure. For air at room temperature, this speed is typically around 343 m/s, however in this exercise, it has been given as 348 m/s.
  • This value indicates how fast the sound waves generated by the vibrating violin string will travel through the air.
  • It serves as a constant in calculations involving sound wave frequency and wavelength in air.
  • The equation we're interested in is: \[ v = f \lambda_{sound} \]
Understanding the speed of sound is crucial for fields such as acoustics, meteorology, and engineering, as it affects how sound is perceived across different environments.
Wavelength Calculation
The concept of wavelength is paramount when discussing wave mechanics, as it defines the physical length of one complete wave cycle. In the context of sound waves in the air, wavelength calculation is essential for understanding how those waves propagate.
To calculate the wavelength of a sound, the formula used is:
  • Use the formula \[ \lambda = \frac{v}{f} \] where \( v \) is the speed of sound, and \( f \) is the frequency. This formula rearranges the wave speed equation to solve for wavelength.
  • In the given exercise, the speed of sound in air is 348 m/s and the frequency of the emitted sound wave is 833.33 Hz. Use these values to find the wavelength in air.
By calculating the wavelength, one can determine the distance over which the wave's shape repeats. This information is crucial for various practical applications, involving sound design, speaker setup, and even musical instrument construction.

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Most popular questions from this chapter

Hot chocolate effect. Tap a metal spoon inside a mug of water and note the frequency \(f_{i}\) you hear. Then add a spoonful of powder (say, chocolate mix or instant coffee) and tap again as you stir the powder. The frequency you hear has a lower value \(f_{s}\) because the tiny air bubbles released by the powder change the water's bulk modulus. As the bubbles reach the water surface and disappear, the frequency gradually shifts back to its initial value. During the effect, the bubbles don't appreciably change the water's density or volume or the sound's wavelength.

An ambulance with a siren emitting a whine at \(1600 \mathrm{~Hz}\) overtakes and passes a cyclist pedaling a bike at \(2.44 \mathrm{~m} / \mathrm{s}\). After being passed, the cyclist hears a frequency of \(1590 \mathrm{~Hz}\). How fast is the ambulance moving?

A tube \(1.20 \mathrm{~m}\) long is closed at one end. A stretched wire is placed near the open end. The wire is \(0.330 \mathrm{~m}\) long and has a mass of \(9.60 \mathrm{~g}\). It is fixed at both ends and oscillates in its fundamental mode. By resonance, it sets the air column in the tube into oscillation at that column's fundamental frequency. Find (a) that frequency and (b) the tension in the wire.

A siren emitting a sound of frequency \(1000 \mathrm{~Hz}\) moves away from you toward the face of a cliff at a speed of \(10 \mathrm{~m} / \mathrm{s}\). Take the speed of sound in air as \(330 \mathrm{~m} / \mathrm{s}\). (a) What is the frequency of the sound you hear coming directly from the siren? (b) What is the frequency of the sound you hear reflected off the cliff? (c) What is the beat frequency between the two sounds? Is it perceptible (less than \(20 \mathrm{~Hz}\) )?

Two trains are traveling toward each other at \(30.5 \mathrm{~m} / \mathrm{s}\) relative to the ground. One train is blowing a whistle at \(500 \mathrm{~Hz}\). (a) What frequency is heard on the other train in still air? (b) What frequency is heard on the other train if the wind is blowing at \(30.5 \mathrm{~m} / \mathrm{s}\) toward the whistle and away from the listener? (c) What frequency is heard if the wind direction is reversed?

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