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Chapter 15: Problem 1

The speed of sound in air at 20\(^\circ\)C is 344 m/s. (a) What is the wavelength of a sound wave with a frequency of 784 Hz, corresponding to the note 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 (twice the frequency) than the note in part (a)?

Short Answer

Expert verified
(a) Wavelength: 0.439 m; Each vibration: 1.28 ms. (b) Wavelength: 0.219 m.

Step by step solution

01

Understand the Relationship

The relationship between speed, frequency, and wavelength is given by the formula: \[ v = f \times \lambda \] where \( v \) is the speed of sound, \( f \) is the frequency, and \( \lambda \) is the wavelength.
02

Calculate the Wavelength for Note G鈧�

The frequency of the note G鈧� is given as 784 Hz. Using the formula \( v = f \times \lambda \), we can rearrange to find the wavelength: \[ \lambda = \frac{v}{f} = \frac{344 \, \text{m/s}}{784 \, \text{Hz}} \] Calculating this gives: \[ \lambda \approx 0.439 \text{ m} \]
03

Determining the Time for One Vibration

The time for one vibration (or one period) is the reciprocal of the frequency: \[ T = \frac{1}{f} = \frac{1}{784 \, \text{Hz}} \] Calculating this gives: \[ T \approx 0.00128 \, \text{s} = 1.28 \, \text{ms} \]
04

Calculate Wavelength for One Octave Higher

One octave higher means the frequency is doubled: \( 2 \times 784 \text{ Hz} = 1568 \text{ Hz} \). Using the wavelength formula: \[ \lambda = \frac{v}{f} = \frac{344 \, \text{m/s}}{1568 \, \text{Hz}} \] Calculating this gives: \[ \lambda \approx 0.219 \text{ m} \]

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Key Concepts

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

Frequency
Frequency refers to the number of sound wave cycles that occur in one second. It is measured in Hertz (Hz). In our exercise, the note G 5 on a piano has a frequency of 784 Hz. This means that the sound wave oscillates 784 times per second. Understanding frequency is crucial because it directly affects how we perceive sound: higher frequencies produce higher-pitched sounds.
  • Key fact: Human ears can typically hear frequencies ranging from 20 Hz to 20 kHz.
  • Higher frequency means a shorter wavelength for sound waves, provided the speed of sound remains constant.
  • Doubling the frequency (from one octave to the next) halves the wavelength if the speed is unchanged.
Wavelength
Wavelength is the distance between two consecutive points in phase on a wave (e.g., from crest to crest). It is denoted by the Greek letter lambda (\( \lambda \)) and measured in meters. For a frequency of 784 Hz, the wavelength of the sound wave is approximately 0.439 meters, using the speed of sound in air at 20掳C, which is 344 m/s.
Understanding wavelength is important because it helps us determine how sounds propagate through different mediums.
  • Longer wavelengths can travel further without losing energy and are effective in carrying sound over long distances.
  • In the provided exercise, doubling the frequency to 1568 Hz resulted in a wavelength of approximately 0.219 meters.
Sound Wave
A sound wave is a type of mechanical wave that travels through a medium, like air, by vibrating the particles in the medium. These waves are longitudinal, meaning the displacement of the medium is parallel to the direction of the wave's travel.
  • Sound waves require a medium to travel; they cannot move through a vacuum.
  • The speed of sound in a medium is influenced by the medium's properties, such as temperature and density.
  • In air at 20掳C, sound travels at 344 m/s, which assists in calculating both frequency and wavelength.
Octave
An octave in music refers to the interval between one musical pitch and another with half or double its frequency. Essentially, moving up one octave doubles the frequency. In our example, increasing the frequency of G 5 from 784 Hz to 1568 Hz means the sound wave is now one octave higher, changing both its pitch and wavelength.
  • This concept is fundamental in music theory, allowing for better understanding of scales and harmonics.
  • Doubling the frequency of a sound wave affects not only its pitch but also results in a proportionally shorter wavelength.
Vibration
Vibration is the periodic motion of particles in a medium through which the sound wave travels. When discussing vibration in sound, it often refers to the oscillations of particles that create waves of pressure, traveling back and forth in a medium like air.
  • The time taken for one complete vibration or oscillation is known as the period, calculated as the inverse of frequency.
  • In our exercise, each vibration of the 784 Hz sound wave corresponds to a period of approximately 1.28 milliseconds.

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Most popular questions from this chapter

The upper end of a 3.80-m-long steel wire is fastened to the ceiling, and a 54.0-kg object is suspended from the lower end of the wire. You observe that it takes a transverse pulse 0.0492 s to travel from the bottom to the top of the wire. What is the mass of the wire?

A transverse wave on a rope is given by $$y(x, t) = (0.750 \, \mathrm{cm}) \mathrm{cos} \space \pi[(10.400 \, \mathrm{cm}^{-1})x + (250 \mathrm s^{-1})t]$$ (a) Find the amplitude, period, frequency, wavelength, and speed of propagation. (b) Sketch the shape of the rope at these values of \(t:\) 0, 0.0005 s, 0.0010 s. (c) Is the wave traveling in the \(+x-\) or \(-x\)-direction? (d) The mass per unit length of the rope is 0.0500 kg/m. Find the tension. (e) Find the average power of this wave.

A wave on a string is described by \(y(x, t) = A \mathrm{cos}(kx - \omega t)\). (a) Graph \(y, v_y\), and \(a_y\) as functions of \(x\) for time \(t = 0\). (b) Consider the following points on the string: (i) \(x =\) 0; (ii) \(x = \pi/4k\); (iii) \(x = \pi/2k\); (iv) \(x = 3\pi/4k\); (v) \(x = \pi k\); (vi) \(x = 5\pi/4k\); (vii) \(x = 3\pi/2k\); (viii) \(x = 7\pi/4k\). For a particle at each of these points at \(t = 0\), describe in words whether the particle is moving and in what direction, and whether the particle is speeding up, slowing down, or instantaneously not accelerating.

A light wire is tightly stretched with tension F. Transverse traveling waves of amplitude \(A\) and wavelength \(\lambda_1\) carry average power \(P_{av,1} = 0.400\) W. If the wavelength of the waves is doubled, so \(\lambda_2 = 2\lambda_1\), while the tension \(F\) and amplitude \(A\) are not altered, what then is the average power \(P_{av,2}\) carried by the waves?

A large rock that weighs 164.0 N is suspended from the lower end of a thin wire that is 3.00 m long. The density of the rock is 3200 kg/m\(^3\). The mass of the wire is small enough that its effect on the tension in the wire can be ignored. The upper end of the wire is held fixed. When the rock is in air, the fundamental frequency for transverse standing waves on the wire is 42.0 Hz. When the rock is totally submerged in a liquid, with the top of the rock just below the surface, the fundamental frequency for the wire is 28.0 Hz. What is the density of the liquid?
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