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Chapter 16: Problem 12

A classic demonstration of the physics of sound involves an alarm clock in a bell vacuum jar. The demonstration starts with air in the vacuum jar at normal atmospheric pressure and then the jar is evacuated to lower and lower pressures. Describe the expected outcome.

Short Answer

Expert verified
Answer: As the pressure inside the vacuum jar decreases, the sound of the alarm clock will become quieter. Once the jar is fully evacuated and a vacuum is created, we will no longer be able to hear the sound of the alarm clock, as there is no medium present to transmit the sound waves.

Step by step solution

01

Understand the propagation of sound

Sound travels through the vibration of particles in a medium, such as air, water, or solid materials. In this case, the alarm clock's sound waves are generated by the rapid vibration of its internal mechanism and require a medium to propagate. At normal atmospheric pressure, the air inside the vacuum jar will carry these vibrations, allowing us to hear the alarm clock.
02

Explain the effect of evacuating the jar

As the jar is evacuated, the air pressure inside the jar decreases. This means that there are fewer air molecules present for the sound waves to propagate. At lower pressures, the air molecules are spaced farther apart, making it more difficult for them to transfer the vibrations to each other. As a result, the sound propagation will become less effective.
03

Describe the outcome at lower pressure

As the pressure inside the vacuum jar continues to decrease, the sound from the alarm clock will gradually become quieter. At extremely low pressures, there may not be enough air molecules to effectively transmit the sound of the alarm clock. In this case, we may no longer be able to hear the alarm clock even though it is still functioning properly.
04

Explain the concept of a vacuum

A vacuum is defined as a space devoid of matter, which means there are no particles present to transmit sound waves. When the vacuum jar is fully evacuated to create a vacuum, there will be no air molecules left inside the jar. At this stage, the sound of the alarm clock cannot propagate through the vacuum, and we will no longer be able to hear it.
05

Summarize the expected outcome

In conclusion, as the vacuum jar is gradually evacuated and the pressure inside the jar lowers, the sound of the alarm clock will become quieter. Once the jar is fully evacuated and a vacuum is created inside it, we will no longer be able to hear the sound of the alarm clock as there is no medium present to transmit the sound waves.

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Most popular questions from this chapter

Two people are talking at a distance of \(3.0 \mathrm{~m}\) from where you are, and you measure the sound intensity as \(1.1 \cdot 10^{-7} \mathrm{~W} / \mathrm{m}^{2}\). Another student is \(4.0 \mathrm{~m}\) away from the talkers. What sound intensity does the other student measure?

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) \(55 \mathrm{~dB}\)

A train whistle emits a sound at a frequency \(f=3000 .\) Hz when stationary. You are standing near the tracks when the train goes by at a speed of \(v=30.0 \mathrm{~m} / \mathrm{s}\). What is the magnitude of the change in the frequency \((|\Delta f|)\) of the whistle as the train passes? (Assume that the speed of sound is \(v=343 \mathrm{~m} / \mathrm{s}\).)

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?

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?
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