Skip to main content
Ch 17: Superposition
Knight Calc - Physics for Scientists and Engineers 5th Edition
Knight Calc5th EditionPhysics for Scientists and EngineersISBN: 9780137344796Not the one you use?Change textbook
Chapter 17, Problem 58

An old mining tunnel disappears into a hillside. You would like to know how long the tunnel is, but it's too dangerous to go inside. Recalling your recent physics class, you decide to try setting up standing-wave resonances inside the tunnel. Using your subsonic amplifier and loudspeaker, you find resonances at 4.5 Hz and 6.3 Hz, and at no frequencies between these. It's rather chilly inside the tunnel, so you estimate the sound speed to be 335 m/s . Based on your measurements, how far is it to the end of the tunnel?

Verified step by step guidance
1
Step 1: Understand the concept of standing-wave resonances. Standing waves occur when sound waves reflect back and forth in a confined space, creating nodes and antinodes. The resonant frequencies correspond to the natural frequencies of the tunnel, which depend on its length and the speed of sound.
Step 2: Recognize that the resonances at 4.5 Hz and 6.3 Hz represent consecutive harmonics. The difference between these frequencies gives the fundamental frequency of the tunnel. Calculate the fundamental frequency using the formula: \( f_1 = f_2 - f_1 \), where \( f_2 \) and \( f_1 \) are consecutive resonant frequencies.
Step 3: Use the relationship between the fundamental frequency, the speed of sound, and the length of the tunnel. The formula is \( f_1 = \frac{v}{2L} \), where \( v \) is the speed of sound and \( L \) is the length of the tunnel. Rearrange the formula to solve for \( L \): \( L = \frac{v}{2f_1} \).
Step 4: Substitute the given values into the formula. The speed of sound \( v \) is 335 m/s, and the fundamental frequency \( f_1 \) was calculated in Step 2. Perform the substitution to find \( L \).
Step 5: Interpret the result. The calculated \( L \) represents the distance to the end of the tunnel. Ensure the units are consistent and verify the calculation steps for accuracy.

Verified video answer for a similar problem:

This video solution was recommended by our tutors as helpful for the problem above.
Video duration:
4m
Was this helpful?

Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Standing Waves

Standing waves occur when two waves of the same frequency and amplitude traveling in opposite directions interfere with each other, creating nodes and antinodes. In a confined space like a tunnel, standing waves can form at specific frequencies, known as resonant frequencies, which depend on the length of the tunnel and the speed of sound in the medium.
Recommended video:
Guided course
07:58
Intro to Transverse Standing Waves

Resonant Frequencies

Resonant frequencies are specific frequencies at which a system naturally oscillates with greater amplitude. In the context of the tunnel, the observed frequencies (4.5 Hz and 6.3 Hz) indicate the fundamental and harmonic modes of vibration, which can be used to determine the length of the tunnel based on the relationship between frequency, wavelength, and the speed of sound.
Recommended video:
Guided course
05:23
Resonance in Series LRC Circuits

Speed of Sound

The speed of sound is the rate at which sound waves propagate through a medium, influenced by factors such as temperature and density. In this scenario, the estimated speed of sound is 335 m/s, which is crucial for calculating the wavelength associated with the resonant frequencies and ultimately determining the length of the tunnel using the formula: length = (n * wavelength) / 2, where n is the mode number.
Recommended video:
Guided course
05:17
Standing Sound Waves
Related Practice
Textbook Question

BIO Deep-sea divers often breathe a mixture of helium and oxygen to avoid getting the 'bends' from breathing high-pressure nitrogen. The helium has the side effect of making the divers' voices sound odd. Although your vocal tract can be roughly described as an open-closed tube, the way you hold your mouth and position your lips greatly affects the standing-wave frequencies of the vocal tract. This is what allows different vowels to sound different. The 'ee' sound is made by shaping your vocal tract to have standing-wave frequencies at, normally, 270 Hz and 2300 Hz. What will these frequencies be for a helium-oxygen mixture in which the speed of sound at body temperature is 750m/s ? The speed of sound in air at body temperature is 350m/s .

1977
views
Textbook Question

A 280 Hz sound wave is directed into one end of the trombone slide seen in FIGURE P17.55. A microphone is placed at the other end to record the intensity of sound waves that are transmitted through the tube. The straight sides of the slide are 80 cm in length and 10 cm apart with a semicircular bend at the end. For what slide extensions s will the microphone detect a maximum of sound intensity?

2051
views
Textbook Question

Two loudspeakers emit sound waves of the same frequency along the x-axis. The amplitude of each wave is a. The sound intensity is minimum when speaker 2 is 10 cm behind speaker 1. The intensity increases as speaker 2 is moved forward and first reaches maximum, with amplitude 2a, when it is 30 cm in front of speaker 1. What is The amplitude of the sound (as a multiple of a) if the speakers are placed side by side?

324
views
Textbook Question

Microwaves with a frequency of 10.5 GHz are aimed downward into a flat-bottomed beaker that contains sunflower oil. A microwave detector above the beaker finds that there are strong reflections when the oil depth is 2.76 cm and 3.68 cm but at no depths in between. What is the index of refraction of sunflower oil at microwave frequencies?

1643
views
Textbook Question

A 1.0-m-tall vertical tube is filled with 20°C water. A tuning fork vibrating at 580 Hz is held just over the top of the tube as the water is slowly drained from the bottom. At what water heights, measured from the bottom of the tube, will there be a standing wave in the tube above the water?

1988
views
Textbook Question

Engineers are testing a new thin-film coating whose index of refraction is less than that of glass. They deposit a 560-nm-thick layer on glass, then shine lasers on it. A red laser with a wavelength of 640 nm has no reflection at all, but a violet laser with a wavelength of 400 nm has a maximum reflection. How the coating behaves at other wavelengths is unknown. What is the coating’s index of refraction?

1516
views