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Chapter 16: Problem 20

If you blow air across the mouth of an empty soda bottle, you hear a tone. Why is it that if you put some water in the bottle, the pitch of the tone increases?

Short Answer

Expert verified
Answer: The pitch of the tone increases when water is added to a soda bottle and air is blown across its mouth because the effective length of the air column inside the bottle decreases, which in turn increases the resonance frequency of the bottle. Higher resonance frequencies lead to higher pitch tones, thus explaining the observed change in pitch.

Step by step solution

01

Understand resonance frequency and standing waves

When air is blown across the mouth of the bottle, it creates a disturbance in the air column inside the bottle. This disturbance travels down the bottle and reflects back up, causing a series of waves to form within the bottle. The constructive interference of these waves results in a standing wave with a specific frequency known as the resonance frequency. The resonance frequency determines the pitch of the tone heard.
02

Examine how the effective length of the bottle changes with water

When water is poured into the bottle, the air column inside it gets shorter. This reduction in the effective length of the air column leads to a change in the resonance frequency. The resonance frequency can be related to the length of the bottle by the formula f = v / (4 * L), where f is the resonance frequency, v is the speed of sound in the air, and L is the effective length of the bottle.
03

Calculate the new resonance frequency

As water is added to the bottle, the effective length L of the air column inside the bottle decreases. Now, when the equation f = v / (4 * L) is evaluated for this new length, a higher resonance frequency is observed. This new resonance frequency corresponds to a higher tone, and therefore, the pitch of the sound increases.
04

Understand the relationship between pitch and resonance frequency

The pitch of a sound is determined by its frequency. Higher frequency waves result in higher pitch tones, while lower frequency waves produce lower pitch tones. As the resonance frequency increases due to the added water in the bottle, the pitch of the produced tone also increases. In conclusion, when water is added to a soda bottle and air is blown across its mouth, the pitch of the resulting tone increases because the effective length of the air column inside the bottle decreases, which in turn increases the resonance frequency of the bottle. Higher resonance frequencies lead to higher pitch tones, thus explaining the observed change in pitch.

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Most popular questions from this chapter

You are playing a note that has a fundamental frequency of \(400 .\) Hz on a guitar string of length \(50.0 \mathrm{~cm}\). At the same time, your friend plays a fundamental note on an open organ pipe, and 4 beats per seconds are heard. The mass per unit length of the string is \(2.00 \mathrm{~g} / \mathrm{m}\). Assume the velocity of sound is \(343 \mathrm{~m} / \mathrm{s}\). a) What are the possible frequencies of the open organ pipe? b) When the guitar string is tightened, the beat frequency decreases. Find the original tension in the string. c) What is the length of the organ pipe?

A police siren contains at least two frequencies, producing the wavering sound (beats). Explain how the siren sound changes as a police car approaches, passes, and moves away from a pedestrian.

Two identical half-open pipes each have a fundamental frequency of \(500 .\) Hz. What percentage change in the length of one of the pipes will cause a beat frequency of \(10.0 \mathrm{~Hz}\) when they are sounded simultaneously?

You are standing between two speakers that are separated by \(80.0 \mathrm{~m}\). Both speakers are playing a pure tone of \(286 \mathrm{~Hz}\). You begin running directly toward one of the speakers, and you measure a beat frequency of \(10.0 \mathrm{~Hz}\). How fast are you running?

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.
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