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Chapter 16: Problem 49

A soprano sings the note \(C 6(1046 \mathrm{~Hz})\) across the mouth of a soda bottle. To excite a fundamental frequency in the soda bottle equal to this note, describe how far the top of the liquid must be below the top of the bottle.

Short Answer

Expert verified
Answer: The liquid's surface should be placed 8.20 cm below the top of the bottle to excite a fundamental frequency equal to the sung note C6 (1046 Hz).

Step by step solution

01

Calculate the wavelength of the sound wave

To find the wavelength of the sound wave, we need to know the speed of sound in air. The speed of sound (\(v\)) is approximately \(343 \frac{m}{s}\) at room temperature. Given the frequency (\(f\)) of the sound wave, we can use the formula \(v = f\lambda\) to calculate the wavelength (\(\lambda\)) of the sound wave. Rearranging the formula for wavelength, \[\lambda = \frac{v}{f}\] We are given the frequency \(f = 1046 \mathrm{~Hz}\), and the speed of sound \(v = 343 \frac{m}{s}\). So, \[\lambda = \frac{343 \frac{\mathrm{m}}{\mathrm{s}}}{1046 \mathrm{~Hz}}\]
02

Calculate the wavelength of the sound wave

Now we substitute the given values of \(f\) and \(v\) in the formula, \[\lambda = \frac{343 \frac{\mathrm{m}}{\mathrm{s}}}{1046 \mathrm{~Hz}} \approx 0.3279 \mathrm{~m}\] This is the wavelength of the sound wave produced by the singing soprano.
03

Calculate the length of the air column in the bottle

To find the length of the air column inside the bottle that would resonate at this given frequency, we need to consider the fact that, for a closed-end air column, the length corresponds to one-fourth the wavelength. For an air column with one closed end (such as a bottle with liquid inside), the length (\(L\)) is given by: \[L = \frac{1}{4} \lambda\] Here, the length \(L\) is the distance from the top of the liquid to the top of the bottle. Now substitute the calculated wavelength value: \[L = \frac{1}{4} \cdot 0.3279 \mathrm{~m} \approx 0.0820 \mathrm{~m}\]
04

Convert the length to centimeters

Finally, we convert the calculated length of the air column in meters to centimeters for a more practical representation of the distance: \[L = 0.0820 \mathrm{~m} \cdot \frac{100 \mathrm{~cm}}{1 \mathrm{~m}} = 8.20 \mathrm{~cm}\] Thus, the liquid's surface should be \(8.20 \mathrm{~cm}\) below the top of the bottle in order to excite a fundamental frequency equal to the sung note \(C 6(1046 \mathrm{~Hz})\).

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Most popular questions from this chapter

Standing on the sidewalk, you listen to the horn of a passing car. As the car passes, the frequency of the sound changes from high to low in a continuous manner; that is, there is no abrupt change in the perceived frequency. This occurs because a) the pitch of the sound of the horn changes continuously. b) the intensity of the observed sound changes continuously. c) you are not standing directly in the path of the moving car. d) of all of the above reasons.

A half-open pipe is constructed to produce a fundamental frequency of \(262 \mathrm{~Hz}\) when the air temperature is \(22^{\circ} \mathrm{C} .\) It is used in an overheated building when the temperature is \(35^{\circ} \mathrm{C} .\) Neglecting thermal expansion in the pipe, what frequency will be heard?

In a suspense-thriller movie, two submarines, \(X\) and Y, approach each other, traveling at \(10.0 \mathrm{~m} / \mathrm{s}\) and \(15.0 \mathrm{~m} / \mathrm{s}\), respectively. Submarine X "pings" submarine Y by sending a sonar wave of frequency \(2000.0 \mathrm{~Hz}\). Assume that the sound travels at \(1500.0 \mathrm{~m} / \mathrm{s}\) in the water. a) Determine the frequency of the sonar wave detected by submarine Y. b) What is the frequency detected by submarine \(X\) for the sonar wave reflected off submarine Y? c) Suppose the submarines barely miss each other and begin to move away from each other. What frequency does submarine Y detect from the pings sent by X? How much is the Doppler shift?

A source traveling to the right at a speed of \(10.00 \mathrm{~m} / \mathrm{s}\) emits a sound wave at a frequency of \(100.0 \mathrm{~Hz}\). The sound wave bounces off of a reflector, which is traveling to the left at a speed of \(5.00 \mathrm{~m} / \mathrm{s}\). What is the frequency of the reflected sound wave detected by a listener back at the source?

Two sources, \(A\) and \(B\), emit a sound of a certain wavelength. The sound emitted from both sources is detected at a point away from the sources. The sound from source \(\mathrm{A}\) is a distance \(d\) from the observation point, whereas the sound from source \(\mathrm{B}\) has to travel a distance of \(3 \lambda .\) What is the largest value of the wavelength, in terms of \(d\), for the maximum sound intensity to be detected at the observation point? If \(d=10.0 \mathrm{~m}\) and the speed of sound is \(340 \mathrm{~m} / \mathrm{s}\), what is the frequency of the emitted sound?
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