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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 (twice the frequency) than the note in part (a)?

Short Answer

Expert verified
The wavelength of the sound wave for the note G5 on a piano is \(0.44 m\). Each vibration takes approximately 1.28 milliseconds. The wavelength of a sound wave one octave higher than the note in part (a) is \(0.22 m\).

Step by step solution

01

Calculate the wavelength of the sound wave

Use the formula \(Wavelength = Speed of Sound / Frequency\). Substitute the given values: Wavelength = \(344 m/s / 784 Hz\).
02

Calculate the time duration of each vibration

Use the formula \(Time = 1 / Frequency\). Substitute the given values: Time = \(1 / 784 s^{-1}\).
03

Calculate the wavelength of a sound wave one octave higher

The doubled frequency is \(2 * 784 Hz\). Use the formula for wavelength again. Now the Frequency is \(2 * 784 Hz\). So, Wavelength = \(344 m/s / (2 * 784 Hz)\).

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Key Concepts

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

Speed of Sound
The speed of sound refers to how fast sound waves travel through a medium. At a comfortable room temperature of around 20°C (68°F), the speed of sound in air is approximately 344 meters per second (m/s). This speed can vary with temperature, humidity, and air pressure. In general, sound travels faster in warmer air because the molecules move more quickly, allowing sound waves to transmit energy more efficiently.

Knowing the speed of sound is essential for understanding other properties of sound waves, such as their wavelength and frequency. In this exercise, you need the speed of sound to calculate both the wavelength of sound waves at different frequencies, and to understand how changes in frequency impact the wavelength.
Wavelength Calculation
To find the wavelength of a sound wave, you need to know the speed of sound and the frequency of the wave. The wavelength formula is depicted as:
  • Wavelength = Speed of Sound / Frequency
Substituting the given values from the exercise, for a frequency of 784 Hz, the wavelength calculation will be:
  • Wavelength = 344 m/s / 784 Hz = 0.438 meters
The wavelength is the distance between two consecutive points of a wave (like crest to crest), and it signifies how "stretched" or "compressed" the waves are in the medium. The same formula applies when calculating the wavelength for frequencies that are an octave higher.
Frequency and Pitch
Frequency refers to the number of vibrations or waves that occur per second, measured in hertz (Hz). The pitch of a sound is how high or low it sounds to our ears, and it is directly related to frequency:
  • Higher frequency means a higher pitch.
  • Lower frequency means a lower pitch.
In the exercise, a frequency of 784 Hz corresponds to the note Gâ‚… on the piano, which is relatively higher on the scale. By computing the time each vibration takes using the formula:
  • Time = 1 / Frequency
we get the duration of each cycle of vibration for 784 Hz as approximately 0.00128 seconds or 1.28 milliseconds. This indicates how rapidly the sound wave oscillates, contributing to its brisk pitch perceived as Gâ‚….
Octaves in Music
An octave is a unique musical interval where the frequency of the sound doubles. This creates a sense of repetitive similarity while the pitch increases. When you take the original frequency and double it, the sound climbs one octave higher. From the exercise, if the original note is at 784 Hz (Gâ‚…), then one octave higher would be at 1568 Hz.
  • Frequency for an octave higher = 2 * Original Frequency = 2 * 784 Hz = 1568 Hz
For this new frequency, you recalculate the wavelength:
  • Wavelength = 344 m/s / 1568 Hz = 0.219 meters
This halved wavelength compared to the original shows that sound waves compress as frequencies double. This forms the basis of octave intervals, used widely across musical compositions to create harmonic structures and elaborate musical textures.

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Most popular questions from this chapter

Tsunami! On December \(26,2004,\) a great earthquake occurred off the coast of Sumatra and triggered immense waves (tsunami) that killed more than 200,000 people. Satellites observing these waves from space measured \(800 \mathrm{~km}\) from one wave crest to the next and a period between waves of 1.0 hour. What was the speed of these waves in \(\mathrm{m} / \mathrm{s}\) and in \(\mathrm{km} / \mathrm{h}\) ? Does your answer help you understand why the waves caused such devastation?

Adjacent antinodes of a standing wave on a string are \(15.0 \mathrm{~cm}\) apart. A particle at an antinode oscillates in simple harmonic motion with amplitude \(0.850 \mathrm{~cm}\) and period \(0.0750 \mathrm{~s}\). The string lies along the \(+x\) -axis and is fixed at \(x=0 .\) (a) How far apart are the adjacent nodes? (b) What are the wavelength, amplitude, and speed of the two traveling waves that form this pattern? (c) Find the maximum and minimum transverse speeds of a point at an antinode. (d) What is the shortest distance along the string between a node and an antinode?

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 \(0.365 \mathrm{~W} ?\)

A thin, taut string tied at both ends and oscillating in its third harmonic has its shape described by the equation \(y(x, t)=(5.60 \mathrm{~cm}) \sin [(0.0340 \mathrm{rad} / \mathrm{cm}) x] \sin [(50.0 \mathrm{rad} / \mathrm{s}) t],\) where the origin is at the left end of the string, the \(x\) -axis is along the string, and the \(y\) -axis is perpendicular to the string. (a) Draw a sketch that shows the standing-wave pattern. (b) Find the amplitude of the two traveling waves that make up this standing wave. (c) What is the length of the string? (d) Find the wavelength, frequency, period, and speed of the traveling waves. (e) Find the maximum transverse speed of a point on the string. (f) What would be the equation \(y(x, t)\) for this string if it were vibrating in its eighth harmonic?

For a string stretched between two supports, two successive standing-wave frequencies are \(525 \mathrm{~Hz}\) and \(630 \mathrm{~Hz}\). There are other standing-wave frequencies lower than \(525 \mathrm{~Hz}\) and higher than \(630 \mathrm{~Hz}\). If the speed of transverse waves on the string is \(384 \mathrm{~m} / \mathrm{s},\) what is the length of the string? Assume that the mass of the wire is small enough for its effect on the tension in the wire to be ignored.

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