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Chapter 12: Problem 50

\(\cdot\) Two guitarists attempt to play the same note of wavelength 6.50 \(\mathrm{cm}\) at the same time, but one of the instruments is slightly out of tune and plays a note of wavelength 6.52 \(\mathrm{cm}\) instead. What is the frequency of the beat these musicians hear when they play together?

Short Answer

Expert verified
The beat frequency is approximately 14.58 Hz.

Step by step solution

01

Calculate Frequency of First Note

First, find the frequency of the note with wavelength 6.50 cm using the formula for wave speed, which is \( v = f \cdot \lambda \), where \( v \) is the speed of sound in air (approximately 343 m/s) at room temperature.Convert wavelength from cm to m: 6.50 cm = 0.0650 m.Rearrange the formula to solve for frequency \( f_1 \):\[ f_1 = \frac{v}{\lambda_1} = \frac{343}{0.0650} \approx 5276.92 \text{ Hz} \]
02

Calculate Frequency of Second Note

Using the same method as in Step 1, calculate the frequency of the second note with wavelength 6.52 cm.Convert wavelength from cm to m: 6.52 cm = 0.0652 m.\[ f_2 = \frac{v}{\lambda_2} = \frac{343}{0.0652} \approx 5262.34 \text{ Hz} \]
03

Calculate Beat Frequency

The beat frequency is the absolute difference between the two frequencies:\[ f_{\text{beat}} = |f_1 - f_2| = |5276.92 - 5262.34| \approx 14.58 \text{ Hz} \]

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Key Concepts

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

Wave Speed
Understanding wave speed is essential to solving problems like the one where two guitarists play slightly different notes. Wave speed, denoted as \( v \), refers to the distance a wave travels in a given amount of time. It depends on both the frequency of the wave and its wavelength. The formula to calculate wave speed is \( v = f \cdot \lambda \), where:
  • \( v \) is the wave speed
  • \( f \) is the frequency
  • \( \lambda \) (lambda) is the wavelength
To find either the frequency or the wavelength, knowing the wave speed allows for the rearrangement of this equation, making it a powerful tool in acoustic problems.
Frequency Calculation
Frequency, designated by \( f \), is how often the particles of the medium vibrate when a sound wave passes through. It's measured in Hertz (Hz), which represents cycles per second. To calculate the frequency of a sound wave, you can rearrange the wave speed formula to \( f = \frac{v}{\lambda} \). For example, in the case of two slightly different guitar string notes, you'd convert the given wavelengths from centimeters to meters. Then, use the speed of sound in air (343 m/s) to calculate their respective frequencies:
  • For 6.50 cm (or 0.0650 m): \( f_1 \approx 5276.92 \text{ Hz} \)
  • For 6.52 cm (or 0.0652 m): \( f_2 \approx 5262.34 \text{ Hz} \)
These calculations show how precise the vibration rate in the air molecules is for different wavelengths.
Sound Wavelength
Wavelength is the distance between consecutive peaks of a wave. In sound waves, it helps define the pitch we hear. When two sound waves have slightly different wavelengths, as with the two guitarists, we can hear beats. To use wavelength in equations, it is crucial to convert from units like centimeters to meters (1 cm = 0.01 m). This ensures compatibility with the standard speed of sound. Also, wavelength is inversely proportional to frequency; longer wavelengths mean lower frequencies and vice versa.
Speed of Sound
The speed of sound is an essential factor in calculating the properties of sound waves. Typically, the speed of sound in air is around 343 m/s at room temperature. This speed can vary with temperature, altitude, and the composition of the air. However, for basic calculations like finding beat frequency from musical notes, using the average speed simplifies the process. By knowing the wave speed, frequencies can be determined for given wavelengths, as seen in the notes played by the guitarists. A solid grasp of the speed of sound helps in predicting how sound travels through different mediums and over various distances.

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Most popular questions from this chapter

\(\bullet\) Voiceprints. In this chapter, we have been concentrating on sinusoidal waves. But most waves in the real world are far more complicated. However, many complicated waves can be created by adding together sine waves of varying amplitude and frequency. When a singer, for example, sings a note, the pitch we hear is the fundamental frequency at which his or her larynx is vibrating. But the larynx also vibrates in other frequencies (the overtones) at the same time. So the sound we hear is a superposition of the fundamental frequency plus all the overtones. This set of all the frequencies (with their respective amplitudes) is called the person's voice print. (a) To see how this works, carefully graph a sine wave of frequency 440 Hz (concert \(A ),\) with time on the horizontal axis and displacement on the vertical axis. Let the amplitude be 1 unit. On the same set of axes, graph the first overtone of 880 \(\mathrm{Hz}\) , but with an amplitude of \(\frac{1}{2}\) unit. (b) Now add the two waves to find their superposition. Notice that the shape is no longer a Isine wave.

Standing sound waves are produced in a pipe that is 1.20 \(\mathrm{m}\) long. For the fundamental frequency and the first two over-tones, determine the locations along the pipe (measured from the left end) of the displacement nodes if (a) the pipe is open atboth ends; (b) the pipe is closed at the left end and open at the right end.

\(\bullet\) A person is playing a small flute 10.75 \(\mathrm{cm}\) long, open at one end and closed at the other, near a taut string having a fundamental frequency of 600.0 \(\mathrm{Hz}\) . If the speed of sound is \(344.0 \mathrm{m} / \mathrm{s},\) for which harmonics of the flute will the string res- onate? In each case, which harmonic of the string is in resonance?

\(\cdot\) Find the intensity \(\left(\) in \(W / m^{2}\right)\) of (a) a 55.0 dB sound, (b) a 92.0 dB sound, (c) a \(-2.0\) dB sound.

\(\bullet\) (a) By what factor must the sound intensity be increased to raise the sound intensity level by 13.0 dB? (b) Explain why you don't need to know the original sound intensity.
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