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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

A sound level of 50 decibels is a) 2.5 times as intense as a sound of 20 decibels. b) 6.25 times as intense as a sound of 20 decibels. c) 10 times as intense as a sound of 20 decibels. d) 100 times as intense as a sound of 20 decibels. e) 1000 times as intense as a sound of 20 decibels.

Two 100.0-W speakers, A and B, are separated by a distance \(D=3.6 \mathrm{~m} .\) The speakers emit in-phase sound waves at a frequency \(f=10,000.0 \mathrm{~Hz}\). Point \(P_{1}\) is located at \(x_{1}=4.50 \mathrm{~m}\) and \(y_{1}=0 \mathrm{~m} ;\) point \(P_{2}\) is located at \(x_{2}=4.50 \mathrm{~m}\) and \(y_{2}=-\Delta y .\) Neglecting speaker \(\mathrm{B}\), what is the intensity, \(I_{\mathrm{A} 1}\) (in \(\mathrm{W} / \mathrm{m}^{2}\) ), of the sound at point \(P_{1}\) due to speaker \(\mathrm{A}\) ? Assume that the sound from the speaker is emitted uniformly in all directions. What is this intensity in terms of decibels (sound level, \(\beta_{\mathrm{A} 1}\) )? When both speakers are turned on, there is a maximum in their combined intensities at \(P_{1} .\) As one moves toward \(P_{2},\) this intensity reaches a single minimum and then becomes maximized again at \(P_{2}\). How far is \(P_{2}\) from \(P_{1},\) that is, what is \(\Delta y ?\) You may assume that \(L \gg \Delta y\) and that \(D \gg \Delta y\), which will allow you to simplify the algebra by using \(\sqrt{a \pm b} \approx a^{1 / 2} \pm \frac{b}{2 a^{1 / 2}}\) when \(a \gg b\).

Two vehicles carrying speakers that produce a tone of frequency \(1000.0 \mathrm{~Hz}\) are moving directly toward each other. Vehicle \(\mathrm{A}\) is moving at \(10.00 \mathrm{~m} / \mathrm{s}\) and vehicle \(\mathrm{B}\) is moving at \(20.00 \mathrm{~m} / \mathrm{s}\). Assume the speed of sound in air is \(343.0 \mathrm{~m} / \mathrm{s}\), and find the frequencies that the driver of each vehicle hears.

A sound meter placed \(3 \mathrm{~m}\) from a speaker registers a sound level of \(80 \mathrm{~dB}\). If the volume on the speaker is then turned down so that the power is reduced by a factor of 25 , what will the sound meter read? a) \(3.2 \mathrm{~dB}\) c) \(32 \mathrm{~dB}\) e) \(66 \mathrm{~dB}\) b) \(11 \mathrm{~dB}\) d) \(55 \mathrm{~dB}\)

Electromagnetic radiation (light) consists of waves. More than a century ago, scientists thought that light, like other waves, required a medium (called the ether) to support its transmission. Glass, having a typical mass density of \(\rho=2500 \mathrm{~kg} / \mathrm{m}^{3},\) also supports the transmission of light. What would the elastic modulus of glass have to be to support the transmission of light waves at a speed of \(v=2.0 \cdot 10^{8} \mathrm{~m} / \mathrm{s} ?\) Compare this to the actual elastic modulus of window glass, which is \(5 \cdot 10^{10} \mathrm{~N} / \mathrm{m}^{2}\).
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