/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 28 The water level in a vertical gl... [FREE SOLUTION] | 91Ó°ÊÓ

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The water level in a vertical glass tube \(1.00 \mathrm{~m}\) long can be adjusted to any position in the tube. A tuning fork vibrating at \(686 \mathrm{~Hz}\) is held just over the open top end of the tube, to set up a standing wave of sound in the air-filled top portion of the tube. (That air- filled top portion acts as a tube with one end closed and the other end open.) At what positions of the water level is there resonance?

Short Answer

Expert verified
The resonance positions are at \( 0.125 \mathrm{~m}, 0.375 \mathrm{~m}, \) and \( 0.625 \mathrm{~m} \) from the top.

Step by step solution

01

- Understand the resonance condition

In a tube with one end closed and the other end open, resonance occurs when the length of the air column equals an odd multiple of one-quarter of the wavelength of the sound (i.e., \( L = (2n + 1)\frac{\lambda}{4} \), where n = 0, 1, 2, ...).
02

- Find the speed of sound in air

Assume the speed of sound in air is approximately \( 343 \mathrm{~m/s} \).
03

- Calculate the wavelength

Use the formula \( \lambda = \frac{v}{f} \) where \( v \) is the speed of sound (\( 343 \mathrm{~m/s} \)) and \( f \) is the frequency (\( 686 \mathrm{~Hz} \)).
04

- Compute the wavelength value

Substitute the values into the formula to get \( \lambda = \frac{343 \mathrm{~m/s}}{686 \mathrm{~Hz}} = 0.5 \mathrm{~m} \).
05

- Find the positions of resonance

Using the resonance condition \( L = (2n + 1)\frac{0.5 \mathrm{~m}}{4} \), rewrite it as \( L = (2n + 1)\frac{0.125 \mathrm{~m}}{2} \). For \( n = 0, L = 0.125 \mathrm{~m} \), for \( n = 1, L = 0.375 \mathrm{~m} \), and for \( n = 2, L = 0.625 \mathrm{~m} \).
06

- Conclusion

The water level positions for resonance are at \( 0.125 \mathrm{~m}, 0.375 \mathrm{~m}, \) and \( 0.625 \mathrm{~m} \) from the open top end of the tube.

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Key Concepts

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

Standing Waves
When sound waves reflect off a surface, they can interfere with incoming waves, creating standing waves. These waves oscillate in place and form patterns of nodes (points that stay still) and antinodes (points that vibrate the most). In a tube with one closed end and one open end, resonance occurs when a standing wave fits perfectly within the tube, causing the air column at specific lengths to vibrate strongly. This phenomenon can be easily observed by adjusting the water level in the tube and noting the resonance positions.
Speed of Sound
The speed of sound in air is essential for understanding resonance. It is influenced by various factors like temperature and pressure but is commonly approximated as \texttt{343 m/s}. This value helps in calculating the wavelength of the sound waves produced by the tuning fork. The speed of sound can be determined using the formula: \(v = \texttt{343 m/s}\), where \texttt{v} is the speed of sound.
Wavelength Calculation
To calculate the wavelength (\texttt{ λ }) of a sound wave, we use the formula: \(\texttt{λ = v / f}\), where \texttt{ v } is the speed of sound and \texttt{ f } is the frequency of the tuning fork. In our problem, \texttt{ v = 343 m/s } and \texttt{ f = 686 Hz }. So, the wavelength: \(\texttt{λ = 343 m/s / 686 Hz = 0.5 m}\). Knowing the wavelength helps determine the exact positions of resonance in the tube.
Resonance Conditions
For resonance to occur in a tube with one closed end and one open end, the length (\texttt{ L }) of the air column must be an odd multiple of one-quarter of the wavelength. This condition can be written as: \(\texttt{L = (2n + 1) λ / 4}\), where \texttt{ n } is an integer (0, 1, 2, ...). Substituting \( λ = 0.5 m \) into the formula, we get: \(L = (2n + 1) 0.5 m / 4 = (2n + 1) 0.125 m \). For \(n = 0\), \texttt{ L = 0.125 m }; for \( n = 1 \), \texttt{ L = 0.375 m }; and for \(n = 2\), \texttt{ L = 0.625 m }. These are the positions of the water level where resonance occurs.

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Most popular questions from this chapter

(a) Find the speed of waves on a violin string of mass \(800 \mathrm{mg}\) and length \(22.0 \mathrm{~cm}\) if the fundamental frequency is \(920 \mathrm{~Hz}\). (b) What is the tension in the string? For the fundamental, what is the wavelength of (c) the waves on the string and (d) the sound waves emitted by the string?

Trooper \(B\) is chasing speeder \(A\) along a straight stretch of road. Both are moving at a speed of \(160 \mathrm{~km} / \mathrm{h}\). Trooper \(B\), failing to catch up, sounds his siren again. Take the speed of sound in air to be \(343 \mathrm{~m} / \mathrm{s}\) and the frequency of the source to be \(500 \mathrm{~Hz}\). What is the Doppler shift in the frequency heard by speeder \(A\) ?

Organ pipe \(A\), with both ends open, had a fundamental frequency of \(300 \mathrm{~Hz}\). The third harmonic of organ pipe \(B\), with one end open, has the same frequency as the second harmonic of pipe \(A\). How long are (a) pipe \(A\) and (b) pipe \(B\) ?

A sound source \(A\) and a reflecting surface \(B\) move directly toward each other. Relative to the air, the speed of source \(A\) is \(29.9 \mathrm{~m} / \mathrm{s}\), the speed of surface \(B\) is \(65.8 \mathrm{~m} / \mathrm{s}\), and the speed of sound is \(329 \mathrm{~m} / \mathrm{s}\). The source emits waves at frequency \(1200 \mathrm{~Hz}\) as measured in the source frame. In the reflector frame, what are (a) the frequency and (b) the wavelength of the arriving sound waves? In the source frame, what are (c) the frequency and (d) the wavelength of the sound waves reflected back to the source?

Pipe \(A\), which is \(1.2 \mathrm{~m}\) long and open at both ends, oscillates at its third lowest harmonic frequency. It is filled with air for which the speed of sound is \(343 \mathrm{~m} / \mathrm{s}\). Pipe \(B\), which is closed at one end, oscillates at its second lowest harmonic frequency. These frequencies of pipes \(A\) and \(B\) happen to match. (a) If an \(x\) axis extends along the interior of pipe \(A\), with \(x=0\) at one end, where along the axis are the displacement nodes? (b) How long is pipe \(B ?\) (c) What is the lowest harmonic frequency of pipe \(A\) ?

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