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Chapter 16: Problem 40

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 \(v=343 \mathrm{~m} / \mathrm{s}\).)

Short Answer

Expert verified
Answer: To find the change in frequency (|Δf|), first determine the frequencies when the train is approaching and receding using the Doppler effect formula. Then, calculate the difference between these two frequencies.

Step by step solution

01

Determine frequency when the train is approaching

Using the Doppler effect formula for the frequency of the sound when the train is approaching: $$ f_{\text{approaching}} = f \times \frac{v_\text{sound}}{v_\text{sound} - v}. $$ Here, \(f=3000\) Hz, \(v_\text{sound}=343\) m/s, and \(v=30\) m/s. Plugging in these values, we can calculate the frequency when the train is approaching the observer.
02

Determine frequency when the train is receding

Using the Doppler effect formula for the frequency of the sound when the train is receding: $$ f_{\text{receding}} = f \times \frac{v_\text{sound}}{v_\text{sound} + v}. $$ Here, \(f=3000\) Hz, \(v_\text{sound}=343\) m/s, and \(v=30\) m/s. Plugging in these values, we can calculate the frequency when the train is moving away from the observer.
03

Calculate the change in frequency

We can find the magnitude of the change in frequency \((|\Delta f|)\) by taking the difference between the frequencies when the train is approaching and receding: $$ |\Delta f| = |f_{\text{approaching}} - f_{\text{receding}}|. $$ Using the values from Step 1 and Step 2, we can now find the magnitude of the change in frequency as the train passes by.

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Key Concepts

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

Understanding Frequency Change
The change in frequency, often observed when a train passes by, is a classic example of the Doppler Effect. When an object emitting sound is moving towards an observer, the sound waves compress, increasing the frequency. This is why the sound of a train whistle seems higher as it approaches. Conversely, as the train moves away, the sound waves stretch out, leading to a decrease in frequency.

To calculate the frequency change, we use the Doppler Effect formulas. The approaching frequency is higher due to compressed sound waves, while the receding frequency is lower. The change in frequency, defined as \(|\Delta f|\), is obtained by subtracting these two frequencies. So, to understand the magnitude of the sound shift, you focus on the absolute difference between them.
The Role of Speed of Sound
The speed of sound is crucial in determining the frequency perceived. In this scenario, it is given as 343 m/s. This value allows us to account for how quickly sound waves travel through the air.

The speed of sound can vary based on factors such as temperature and air density. At sea level and around 20°C, it is approximately 343 m/s, which is the standard assumption in many physics problems.

When applying the Doppler Effect formula:
  • For approaching: \( f_{\text{approaching}} = f \times \frac{v_{\text{sound}}}{v_{\text{sound}} - v} \)
  • For receding: \( f_{\text{receding}} = f \times \frac{v_{\text{sound}}}{v_{\text{sound}} + v} \)
The speed of sound is the constant denominator that helps define the relationship between the source and observer.
Exploring Train Whistle Frequency
The frequency of the train whistle, given as 3000 Hz, acts as the fundamental tone that gets adjusted as the train moves. This initial frequency is what you would hear if the train was stationary.

The train's velocity affects how this frequency changes as perceived by an observer. When the train speed is 30 m/s, this contributes to the shift due to the Doppler Effect.

Understanding changes in the whistle frequency helps illustrate basic wave behavior:
  • Stationary source: The frequency remains at 3000 Hz.
  • Moving source: Frequency increases (approaching) or decreases (receding).
These perceptions are crucial for practical applications, such as railway safety and audio signal processing.

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Most popular questions from this chapter

In a suspense-thriller movie, two submarines, \(X\) and Y, approach each other, traveling at \(10.0 \mathrm{~m} / \mathrm{s}\) and \(15.0 \mathrm{~m} / \mathrm{s}\), respectively. Submarine X "pings" submarine Y by sending a sonar wave of frequency \(2000.0 \mathrm{~Hz}\). Assume that the sound travels at \(1500.0 \mathrm{~m} / \mathrm{s}\) in the water. a) Determine the frequency of the sonar wave detected by submarine Y. b) What is the frequency detected by submarine \(X\) for the sonar wave reflected off submarine Y? c) Suppose the submarines barely miss each other and begin to move away from each other. What frequency does submarine Y detect from the pings sent by X? How much is the Doppler shift?

Two 100.0-W speakers, A and B, are separated by a distance \(D=3.6 \mathrm{~m} .\) The speakers emit in-phase sound waves at a frequency \(f=10,000.0 \mathrm{~Hz}\). Point \(P_{1}\) is located at \(x_{1}=4.50 \mathrm{~m}\) and \(y_{1}=0 \mathrm{~m} ;\) point \(P_{2}\) is located at \(x_{2}=4.50 \mathrm{~m}\) and \(y_{2}=-\Delta y .\) Neglecting speaker \(\mathrm{B}\), what is the intensity, \(I_{\mathrm{A} 1}\) (in \(\mathrm{W} / \mathrm{m}^{2}\) ), of the sound at point \(P_{1}\) due to speaker \(\mathrm{A}\) ? Assume that the sound from the speaker is emitted uniformly in all directions. What is this intensity in terms of decibels (sound level, \(\beta_{\mathrm{A} 1}\) )? When both speakers are turned on, there is a maximum in their combined intensities at \(P_{1} .\) As one moves toward \(P_{2},\) this intensity reaches a single minimum and then becomes maximized again at \(P_{2}\). How far is \(P_{2}\) from \(P_{1},\) that is, what is \(\Delta y ?\) You may assume that \(L \gg \Delta y\) and that \(D \gg \Delta y\), which will allow you to simplify the algebra by using \(\sqrt{a \pm b} \approx a^{1 / 2} \pm \frac{b}{2 a^{1 / 2}}\) when \(a \gg b\).

Two trains are traveling toward each other in still air at \(25.0 \mathrm{~m} / \mathrm{s}\) relative to the ground. One train is blowing a whistle at \(300 .\) Hz. The speed of sound is \(343 \mathrm{~m} / \mathrm{s}\). a) What frequency is heard by a man on the ground facing the whistle-blowing train? b) What frequency is heard by a man on the other train?

A sound level of 50 decibels is a) 2.5 times as intense as a sound of 20 decibels. b) 6.25 times as intense as a sound of 20 decibels. c) 10 times as intense as a sound of 20 decibels. d) 100 times as intense as a sound of 20 decibels. e) 1000 times as intense as a sound of 20 decibels.

Two sources, \(A\) and \(B\), emit a sound of a certain wavelength. The sound emitted from both sources is detected at a point away from the sources. The sound from source \(\mathrm{A}\) is a distance \(d\) from the observation point, whereas the sound from source \(\mathrm{B}\) has to travel a distance of \(3 \lambda .\) What is the largest value of the wavelength, in terms of \(d\), for the maximum sound intensity to be detected at the observation point? If \(d=10.0 \mathrm{~m}\) and the speed of sound is \(340 \mathrm{~m} / \mathrm{s}\), what is the frequency of the emitted sound?
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