Log out Go to App
Go to App

Learning Materials

Features

Discover

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.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

At a state championship high school football game, the intensity level of the shout of a single person in the stands at the center of the field is about \(50 \mathrm{~dB}\). What would be the intensity level at the center of the field if all 10,000 fans at the game shouted from roughly the same distance away from that center point?

A half-open pipe is constructed to produce a fundamental frequency of \(262 \mathrm{~Hz}\) when the air temperature is \(22^{\circ} \mathrm{C} .\) It is used in an overheated building when the temperature is \(35^{\circ} \mathrm{C} .\) Neglecting thermal expansion in the pipe, what frequency will be heard?

A bat flying toward a wall at a speed of \(7.0 \mathrm{~m} / \mathrm{s}\) emits an ultrasound wave with a frequency of \(30.0 \mathrm{kHz}\). What frequency does the reflected wave have when it reaches the flying bat?

A policeman with a very good ear and a good understanding of the Doppler effect stands on the shoulder of a freeway assisting a crew in a 40 -mph work zone. He notices a car approaching that is honking its horn. As the car gets closer, the policeman hears the sound of the horn as a distinct \(\mathrm{B} 4\) tone \((494 \mathrm{~Hz}) .\) The instant the car passes by, he hears the sound as a distinct \(\mathrm{A} 4\) tone \((440 \mathrm{~Hz}) .\) He immediately jumps on his motorcycle, stops the car, and gives the motorist a speeding ticket. Explain his reasoning.

A standing wave in a pipe with both ends open has a frequency of \(440 \mathrm{~Hz}\). The next higher harmonic has a frequency of \(660 \mathrm{~Hz}\) a) Determine the fundamental frequency. b) How long is the pipe?
See all solutions

Recommended explanations on Physics Textbooks

Famous Physicists

Read Explanation

Linear Momentum

Read Explanation

Nuclear Physics

Read Explanation

Magnetism

Read Explanation

Kinematics Physics

Read Explanation

Classical Mechanics

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free