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The speed of sound in air at \(20^{\circ} \mathrm{C}\) is 344 \(\mathrm{m} / \mathrm{s} .\) (a) What is the wavelength of a sound wave with a frequency of 784 \(\mathrm{Hz}\) , corresponding to the note \(\mathrm{G}_{5}\) on a piano, and how many milliseconds does each vibration take? (b) What is the wavelength of a sound wave one octave higher than the note in part (a)?

Short Answer

Expert verified
(a) Wavelength is 0.439 m, each vibration takes 1.28 ms. (b) Wavelength is 0.219 m.

Step by step solution

01

Understand the Relationship Between Frequency, Wavelength, and Speed

The speed of a wave (\(v\)), frequency (\(f\)), and wavelength (\(\lambda\)) are related by the formula:\[v = f \times \lambda\]In this problem, the speed of sound (\(v\)) is given as 344 m/s, and we need to find the wavelength for \(f = 784 \text{ Hz}\).
02

Calculate the Wavelength for Frequency 784 Hz

Rearrange the formula to solve for wavelength:\[\lambda = \frac{v}{f}\]Substitute the given values:\[\lambda = \frac{344 \, \text{m/s}}{784 \, \text{Hz}} = 0.4388 \, \text{m}\]So the wavelength of the sound wave is approximately 0.439 meters.
03

Calculate the Period of the Wave

The period (\(T\)) of a wave is the reciprocal of the frequency:\[T = \frac{1}{f}\]Using the frequency 784 Hz:\[T = \frac{1}{784} \approx 0.00128 \, \text{s} = 1.28 \, \text{ms}\]Each vibration takes approximately 1.28 milliseconds.
04

Calculate the Wavelength for a Note One Octave Higher

An octave higher frequency is double the original frequency. Therefore, the frequency for one octave above 784 Hz is 1568 Hz:\[f_{\text{octave}} = 2 \times 784 \text{ Hz} = 1568 \text{ Hz}\]Using the formula for wavelength:\[\lambda_{\text{octave}} = \frac{v}{f_{\text{octave}}} = \frac{344 \, \text{m/s}}{1568 \, \text{Hz}} \approx 0.2194 \, \text{m}\]The wavelength of the sound wave one octave higher is approximately 0.219 meters.

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Key Concepts

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

Speed of Sound
In wave physics, the speed of sound is a key concept. It refers to how fast sound waves travel through a medium, such as air. At 20°C, the speed of sound in air is typically 344 meters per second (m/s). This speed can vary slightly due to changes in factors like temperature and pressure. Warm air allows sound to move faster because the molecules are more energetic and can transmit sound more briskly. By knowing the speed of sound, we can calculate various properties of sound waves, such as wavelength and frequency. Indeed, the formula \(v = f \times \lambda\) links the speed of sound \(v\) with the frequency \(f\) and the wavelength \(\lambda\). Hence, if we know the frequency of a sound and the medium it travels through, we can find out the wavelength.
Frequency and Wavelength Relationship
The relationship between frequency and wavelength is one of the foundational principles in wave physics. Frequency refers to how many wave cycles pass a fixed point in a second. It is measured in hertz (Hz). Wavelength is the distance between successive crests or troughs of a wave.
This relationship can be described by the equation \(v = f \times \lambda\), where \(v\) is the speed of sound, \(f\) is the frequency, and \(\lambda\) is the wavelength.
  • To find the wavelength, rearrange the formula: \(\lambda = \frac{v}{f}\).
  • An increase in frequency results in a shorter wavelength, assuming the wave speed in the medium remains constant.
  • This formula is useful for identifying the size of the sound wave based on its frequency, providing insights into the nature of the sound itself.
Therefore, for a note like \(\text{G}_5\) on a piano with a frequency of 784 Hz, knowing the speed of sound allows us to calculate its wavelength as approximately 0.439 meters.
Harmonics and Octaves
Understanding harmonics and octaves helps us grasp the concept of musical sound waves and their behaviors. An octave occurs when the frequency doubles. For instance, if a sound wave has a frequency of 784 Hz, an octave higher would be 1568 Hz. Doubling the frequency compresses the wavelength into half. As a result, the sound wave becomes shorter.
Harmonics are integral multiples of the fundamental frequency and enrich the timbre or quality of sound. They help extend and refine musical notes.
  • Musically, harmonics above a fundamental frequency contribute to the sound's distinctiveness.
  • An octave is a special harmonic ratio where the frequency is doubled.
When we calculate wavelengths for these octaves using \(\lambda = \frac{v}{f}\), the wavelength for the octave higher \(\text{G}_5\) is approximately 0.219 meters. Thus, knowing octaves and harmonics not only aids in the understanding of musical concepts but also in comprehending how sound waves function.

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Most popular questions from this chapter

A transverse wave on a rope is given by $$y(x, t)=(0.750 \mathrm{cm}) \cos \pi\left[\left(0.400 \mathrm{cm}^{-1}\right) x+\left(250 \mathrm{s}^{-1}\right) t\right]$$ (a) Find the amplitude, period, frequency, wavelength, and speed of propagation. (b) Sketch the shape of the rope at these values of \(t : 0.0005\) s, 0.0010 s. (c) Is the wave wave traveling in the or -direction? (d) The mass per unit length of the rope is 0.0500 \(\mathrm{kg} / \mathrm{m} .\) Find the tension. (e) Find the average power of this wave.

A 1.50 -m string of weight 0.0125 \(\mathrm{N}\) is tied to the ceiling at its upper end, and the lower end supports a weight \(W\) . Neglect the very small variation in tension along the length of the string that is produced by the weight of the string. When you pluck the string slightly, the waves traveling up the string obey the equation $$y(x, t)=(8.50 \mathrm{mm}) \cos \left(172 \mathrm{m}^{-1} x-4830 \mathrm{s}^{-1} t\right)$$ Assume that the tension of the string is constant and equal to \(W\) . (a) How much time does it take a pulse to travel the full length of the string? (b) What is the weight \(W ?\) (c) How many wavelengths are on the string at any instant of time? (d) What is the equation for wayes traveling down the string?

A piano wire with mass 3.00 \(\mathrm{g}\) and length 80.0 \(\mathrm{cm}\) is stretched with a tension of 25.0 \(\mathrm{N}\) . A wave with frequency 120.0 \(\mathrm{Hz}\) and amplitude 1.6 \(\mathrm{mm}\) travels along the wire. (a) Calculate the average power carried by the wave. (b) What happens to the average power if the wave amplitude is halved?

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 \(10,000 \mathrm{Hz} ?\)

Weightless Ant. Ant. An ant with mass \(m\) is standing peacefully on top of a horizontal, stretched rope. The rope has mass per unit length \(\mu\) and is under tension \(F .\) Without warning, Cousin Throckmorton starts a sinusoidal transverse wave of wave-length \(\lambda\) propagating along the rope. The motion of the rope is in a vertical plane. What minimum wave amplitude will make the ant become momentarily weightless? Assume that \(m\) is so small that the presence of the ant has no effect on the propagation of the wave.

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