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Concept Questions The table shows three situations in which the Doppler effect may arise. The first two columns indicate the velocities of the sound source and the observer, where the length of each arrow is proportional to the speed. For each situation, fill in the empty columns by deciding whether the wavelength of the sound and the frequency heard by the observer increase, decrease, or remain the same compared to the case when there is no Doppler effect. Provide a reason for each answer. $$ \begin{array}{|l|c|c|c|c|} \hline & \begin{array}{c} \text { Velocity of Sound } \\ \text { Source (Toward the } \\ \text { Observer) } \end{array} & \begin{array}{c} \text { Velocity of } \\ \text { Observer (Toward } \\ \text { the Source) } \end{array} & \text { Wavelength } & \begin{array}{c} \text { Frequency Heard by } \\ \text { Observer } \end{array} \\ \hline \text { (a) } & 0 \mathrm{~m} / \mathrm{s} & 0 \mathrm{~m} / \mathrm{s} & & \\ \hline \text { (b) } & \rightarrow & 0 \mathrm{~m} / \mathrm{s} & & \\ \hline \text { (c) } & \rightarrow & \leftarrow & & \\ \hline \end{array} $$ Problem The siren on an ambulance is emitting a sound whose frequency is \(2450 \mathrm{~Hz}\). The speed of sound is \(343 \mathrm{~m} / \mathrm{s}\). (a) If the ambulance is stationary and you (the "observer") are sitting in a parked car, what is the wavelength of the sound and the frequency heard by you? (b) Suppose the ambulance is moving toward you at a speed of \(26.8 \mathrm{~m} / \mathrm{s}\). Determine the wavelength of the sound and the frequency heard by you. (c) If the ambulance is moving toward you at a speed of \(26.8 \mathrm{~m} / \mathrm{s}\) and you are moving toward it at a speed of \(14.0 \mathrm{~m} / \mathrm{s}\), find the wavelength of the sound and the frequency that you hear. Be sure that your answers are consistent with your answers to the Concept Questions.

Short Answer

Expert verified
(a) Frequency: 2450 Hz, Wavelength: 0.140 m. (b) Frequency: 2675 Hz, Wavelength: 0.129 m. (c) Frequency: 2834 Hz, Wavelength: 0.121 m.

Step by step solution

01

Analyze Conceptual Situation (a)

For situation (a), both the sound source and the observer are stationary. Hence, there is no Doppler effect affecting the wavelength or frequency. Therefore, the wavelength remains constant at its normal value, and the frequency heard by the observer remains at the emitted frequency.
02

Analyze Conceptual Situation (b)

In situation (b), the sound source moves towards the observer while the observer remains stationary. The wavelength decreases because the sound waves are compressed. This leads to an increase in the frequency heard by the observer.
03

Analyze Conceptual Situation (c)

For situation (c), both the sound source and the observer are moving towards each other. The relative velocity between them increases the effect observed in (b), resulting in an even shorter wavelength and a higher frequency heard by the observer.
04

Solve Problem (a) - Stationary Conditions

With the ambulance stationary, the frequency heard is the same as the emitted frequency, 2450 Hz. The wavelength \(\lambda\) is calculated by \(\lambda = \frac{v}{f} = \frac{343}{2450} \approx 0.140 \text{ m}\).
05

Solve Problem (b) - Source Moving Toward Observer

When the ambulance moves toward you at 26.8 m/s, only the source is moving. The observed frequency \(f'\) is given by \(f' = f \frac{v}{v - v_s} = 2450 \frac{343}{343 - 26.8} \approx 2675 \text{ Hz}\). The new shorter wavelength \(\lambda'\) is \(\lambda' = \frac{v}{f'} \approx 0.129 \text{ m}\).
06

Solve Problem (c) - Both Moving Toward Each Other

With both the ambulance and observer moving towards each other, the relative velocity is the sum, \(v_{rel} = 26.8 + 14.0 = 40.8 \text{ m/s}\). The frequency heard is \(f' = f \frac{v + v_o}{v - v_s} = 2450 \frac{343 + 14.0}{343 - 26.8} \approx 2834 \text{ Hz}\). The wavelength \(\lambda'\) becomes \(\lambda' = \frac{v}{f'} \approx 0.121 \text{ m}\).

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Key Concepts

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

Sound Waves
Sound waves are vibrations that travel through the air or another medium. These waves move by creating areas of high and low pressure. When something like a bell rings, it causes the air molecules around it to vibrate.

These vibrations create waves that move outward from the source. Sound waves are longitudinal, meaning the movement of the molecules is parallel to the direction of the wave. Think of it like a slinky toy, where compressions and rarefactions move along the length of the toy.

One key thing to know about sound waves is that they carry energy, which is why you can hear the sound from far away. In summary, sound waves are how sound travels from the source to your ears.
Frequency
Frequency refers to the number of times a wave vibrates per second. In the context of sound, it is measured in hertz (Hz), where one hertz equals one vibration per second.

Frequency determines the pitch of the sound. A higher frequency means a higher-pitched sound, like that of a whistle or a bird chirping. On the other hand, a lower frequency produces a lower-pitched sound, like a drum or a cello.

In our case of the ambulance siren, the frequency tells us how high or low the sound of the siren is when it reaches the observer. Understanding frequency is crucial when dealing with concepts like the Doppler effect, as moving sources or observers can alter the frequency, hence bringing changes in the sound we hear.
Wavelength
Wavelength is the distance between consecutive points of a wave, such as from one crest to the next. In sound waves, it's the distance between compressions in the air.

The wavelength affects the sound's frequency. A shorter wavelength results in a higher frequency, while a longer wavelength results in a lower frequency.

For an ambulance siren, as it approaches a listener, the sound waves get compressed, leading to a shorter wavelength and an increased frequency heard by the observer. Conversely, when the ambulance moves away, the wavelengths stretch out, resulting in a lower frequency. This is a visual and auditory demonstration of the Doppler effect in action.
Ambulance Siren
An ambulance siren is an excellent real-life example of the Doppler effect. Normally, the siren emits sound waves at a specific frequency. But when the ambulance moves, it affects the sound waves and, consequently, how we perceive them.

As the ambulance approaches, sound waves are compressed, making them hit our ears more frequently, increasing the frequency of the sound we interpret. When the ambulance moves away, the opposite happens; the sound waves stretch, resulting in a lower frequency. This is why the pitch of an ambulance siren seems to change as it passes by.

This Doppler effect is crucial for understanding the change in sound frequency that occurs due to the motion of the source or the observer.
Velocity of Sound
The velocity of sound is the speed at which sound waves travel through a medium. In air, this speed is typically around 343 meters per second, but this can vary depending on factors like temperature and humidity.

Knowing the velocity of sound is critical for solving problems involving the Doppler effect. It helps us calculate how the frequency and wavelength of sound change as the source or observer moves.

For instance, in the case of a moving ambulance, understanding the velocity of sound allows us to determine the new frequency and wavelength when the Doppler effect comes into play. The velocity acts as a reference against which other velocities (like that of the source or the observer) are compared in these calculations.

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Most popular questions from this chapter

Concept Questions A uniform rope of mass \(m\) and length \(L\) is hanging straight down from the ceiling. (a) Is the tension in the rope greater near the top or near the bottom of the rope, or is the tension the same everywhere along the rope? Why? (b) A small-amplitude transverse wave is sent up the rope from the bottom end. Is the speed of the wave greater near the bottom or near the top of the rope, or is the speed of the wave the same everywhere along the rope? Explain. (c) Consider a section of the rope between the bottom end and a point that is a distance \(y\) meters above the bottom. What is the weight of this section? Express your answer in terms of \(m, L, y,\) and \(g\) (the acceleration due to gravity). Problem (a) For the rope described in the Concept Questions, derive an expression that gives the speed of the wave on the rope in terms of the distance \(y\) above the bottom end and the acceleration \(g\) due to gravity. (b) Use the expression that you have derived to calculate the speeds at distances of \(0.50 \mathrm{~m}\) and \(2.0 \mathrm{~m}\) above the bottom end of the rope. Be sure that your answers are consistent with your answer to Concept Question (b).

Concept Questions Multiple-Concept 11 provides a model for solving this type of problem. A wireless transmitting microphone is mounted on a small platform, which can roll down an incline, away from a speaker that is mounted at the top of the incline. The speaker broadcasts a fixed-frequency tone. (a) The platform is positioned in front of the speaker and released from rest. Describe how the velocity of the platform changes and why. (b) How is the changing velocity related to the acceleration of the platform? (c) Describe how the frequency detected by the microphone changes. Explain why the frequency changes as you have described. (d) Which equation given in the chapter applies to this situation? Justify your answer. Problem The speaker broadcasts a tone that has a frequency of \(1.000 \times 10^{4} \mathrm{~Hz}\), and the speed of sound is \(343 \mathrm{~m} / \mathrm{s}\). At a time of \(1.5 \mathrm{~s}\) following the release of the platform, the microphone detects a frequency of \(9939 \mathrm{~Hz}\). At a time of \(3.5 \mathrm{~s}\) following the release of the platform, the microphone detects a frequency of \(9857 \mathrm{~Hz}\). What is the acceleration (assumed constant) of the platform?

Two submarines are underwater and approaching each other head-on. Sub A has a speed of \(12 \mathrm{~m} / \mathrm{s}\) and sub \(\mathrm{B}\) has a speed of \(8 \mathrm{~m} / \mathrm{s}\). Sub A sends out a 1550 -Hz sonar wave that travels at a speed of \(1522 \mathrm{~m} / \mathrm{s}\). (a) What is the frequency detected by sub \(\mathrm{B}\) ? (b) Part of the sonar wave is reflected from \(\mathrm{B}\) and returns to \(\mathrm{A}\). What frequency does A detect for this reflected wave?

A monatomic ideal gas \((\gamma=1.67)\) is contained within a box whose volume is \(2.5 \mathrm{~m}^{3}\). The pressure of the gas is \(3.5 \times 10^{5} \mathrm{~Pa}\). The total mass of the gas is \(2.3 \mathrm{~kg}\). Find the speed of sound in the gas.

The tension in a string is \(15 \mathrm{~N},\) and its linear density is \(0.85 \mathrm{~kg} / \mathrm{m} .\) A wave on the string travels toward the \(-x\) direction; it has an amplitude of \(3.6 \mathrm{~cm}\) and a frequency of \(12 \mathrm{~Hz}\) What are the (a) speed and (b) wavelength of the wave? (c) Write down a mathematical expression (like Equation 16.3 or 16.4 ) for the wave, substituting numbers for the variables \(A, f,\) and \(\lambda\)

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