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Chapter 17: Problem 18

Two pianos each sound the same note simultaneously, but they are both out of tune. On a day when the speed of sound is \(343 \mathrm{~m} / \mathrm{s}\), piano A produces a wavelength of \(0.769 \mathrm{~m}\), while piano B produces a wavelength of \(0.776 \mathrm{~m}\). How much time separates successive beats?

Short Answer

Expert verified
The time between successive beats is approximately 0.308 seconds.

Step by step solution

01

Calculate Frequency of Piano A

The frequency of a sound wave is calculated using the formula \( f = \frac{v}{\lambda} \), where \( v \) is the speed of sound and \( \lambda \) is the wavelength. For Piano A, \( v = 343 \text{ m/s} \) and \( \lambda = 0.769 \text{ m} \). Substitute these values into the formula:\[ f_A = \frac{343}{0.769} \approx 446.03 \text{ Hz} \]
02

Calculate Frequency of Piano B

Use the same formula for frequency for Piano B, where \( \lambda = 0.776 \text{ m} \):\[ f_B = \frac{343}{0.776} \approx 442.78 \text{ Hz} \]
03

Calculate Beat Frequency

The beat frequency is the absolute difference between the two frequencies. Using the frequencies calculated for Piano A and B:\[ f_{beat} = |f_A - f_B| = |446.03 - 442.78| = 3.25 \text{ Hz} \]
04

Determine Time Interval Between Successive Beats

The time interval between successive beats is the reciprocal of the beat frequency. Use the formula:\[ T = \frac{1}{f_{beat}} = \frac{1}{3.25} \approx 0.308 \text{ s} \]

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Key Concepts

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

Sound Waves
Sound waves are vibrations that travel through the air or any other medium. They are created by vibrating objects, such as the strings of a piano. As these vibrations move through the air, they carry energy with them. This is why we can hear sounds.
  • Sound waves are longitudinal, meaning the particles in the medium move parallel to the direction of wave travel.
  • They transfer energy from one place to another without transferring matter.
  • Sound waves can vary in frequency and wavelength, affecting how we perceive sounds, such as pitch.
The speed of sound depends on the medium and its temperature. For example, on the day of the exercise, the speed of sound in the air is given as \(343 \, \mathrm{m/s}\). This is a typical speed for sound in air at room temperature.
Frequency Calculation
Frequency refers to the number of wave cycles that pass a point per second. It is measured in hertz (Hz). In the exercise, we calculate the frequency for each piano using the formula \( f = \frac{v}{\lambda} \). Here鈥檚 how it works:
  • \(v\) is the speed of sound, which is \(343 \, \mathrm{m/s}\) in this case.
  • \(\lambda\) is the wavelength, or the distance between consecutive crests or troughs of a wave.
  • The result is a frequency that tells us how many complete cycles of the wave occur in one second.
For Piano A with a wavelength of \(0.769 \, \mathrm{m}\), the frequency is calculated as \( f_A \approx 446.03 \, \mathrm{Hz}\).
For Piano B, using a wavelength of \(0.776 \, \mathrm{m}\), the frequency is \( f_B \approx 442.78 \, \mathrm{Hz}\).
Wavelength
Wavelength is the distance between similar points in consecutive cycles of a wave, such as crest to crest or trough to trough.
This metric is crucial in determining the wave's frequency and therefore the pitch of a sound.
  • A shorter wavelength produces a higher frequency, leading to a higher-pitched sound.
  • A longer wavelength results in a lower frequency and thus a lower-pitched sound.
In the exercise, Piano A and B have different wavelengths, which leads to slightly different frequencies.
This difference in frequencies is what creates the phenomenon of beats, where two slightly different frequencies interfere with each other.
Speed of Sound
The speed at which sound waves move through a medium like air is called the speed of sound.
It can vary based on factors such as the medium itself and its temperature. For this exercise, we consider the speed of sound to be \(343 \, \mathrm{m/s}\).
  • Warmer air allows sound to travel faster. This is due to increased energy in air molecules, which helps transmit the vibrations more quickly.
  • The speed of sound is also affected by humidity and atmospheric pressure.
Accurately knowing the speed of sound is vital for calculating the frequency of the sounds produced by the pianos.
In our calculations for frequency, maintaining the correct value for the speed of sound ensures we get precise results.

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Most popular questions from this chapter

Sound enters the ear, travels through the auditory canal, and reaches the eardrum. The auditory canal is approximately a tube open at only one end. The other end is closed by the eardrum. A typical length for the auditory canal in an adult is about \(2.9 \mathrm{~cm} .\) The speed of sound is \(343 \mathrm{~m} / \mathrm{s}\). What is the fundamental frequency of the canal? (Interestingly, the fundamental frequency is in the frequency range where human hearing is most sensitive.)

A person hums into the top of a well and finds that standing waves are established at frequencies of \(42,70.0,\) and \(98 \mathrm{~Hz} .\) The frequency of \(42 \mathrm{~Hz}\) is not necessarily the fundamental frequency. The speed of sound is \(343 \mathrm{~m} / \mathrm{s} .\) How deep is the well?

Standing waves are set up on two strings fixed at each end, as shown in the drawing. The tensions in the strings are the same, and each string has the same mass per unit length. However, one string is longer than the other, (a) Do the waves on the longer string have a larger speed, a smaller speed, or the same speed as those on the shorter string? Justify your answer. (b) Will the longer string vibrate at a higher frequency, a lower frequency, or the same frequency as the shorter string? Provide a reason for your answer. (c) Will the beat frequency produced by the two standing waves increase or decrease if the longer string is increased in length? Why? The two strings in the drawing have the same tension and mass per unit length, but they differ in length by \(0.57 \mathrm{~cm} .\) The waves on the shorter string propagate with a speed of \(41.8 \mathrm{~m} / \mathrm{s},\) and the fundamental frequency of the shorter string is \(225 \mathrm{~Hz}\). Determine the beat frequency produced by the two standing waves.

A tube is open only at one end. A certain harmonic produced by the tube has a frequency of \(450 \mathrm{~Hz}\). The next higher harmonic has a frequency of \(750 \mathrm{~Hz}\). The speed of sound in air is \(343 \mathrm{~m} / \mathrm{s}\). (a) What is the integer \(n\) that describes the harmonic whose frequency is \(450 \mathrm{~Hz} ?\) (b) What is the length of the tube?

The fundamental frequency of a vibrating system is \(400 \mathrm{~Hz}\). For each of the following systems, give the three lowest frequencies (excluding the fundamental) at which standing waves can occur: (a) a string fixed at both ends, (b) a cylindrical pipe with both ends open, and (c) a cylindrical pipe with only one end open.
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