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A railroad train is traveling at \(30.0 \mathrm{~m} / \mathrm{s}\) in still air. The frequency of the note emitted by the train whistle is \(262 \mathrm{~Hz}\). What frequency is heard by a passenger on a train moving in the opposite direction to the first at \(18.0 \mathrm{~m} / \mathrm{s}\) and (a) approaching the first; and (b) receding from the first?

Short Answer

Expert verified
For approaching: 295.92 Hz. For receding: 234.21 Hz.

Step by step solution

01

Understanding the Doppler Effect Formula

To find the frequency heard by a passenger, we use the Doppler Effect formula: \( f' = f \frac{v + v_0}{v - v_s} \), where \( f \) is the source frequency, \( v \) is the speed of sound in air (approximately 343 m/s at 20°C), \( v_0 \) is the speed of the observer relative to the medium, and \( v_s \) is the speed of the source relative to the medium.
02

Factors for case (a) - Approaching Trains

For the case where the trains are approaching each other: \( v_0 = 18 \, \mathrm{m/s} \) (observer speed), \( v_s = 30 \, \mathrm{m/s} \) (source speed), and we use the Doppler formula with observer and source speeds being towards each other.
03

Calculation for Approaching Trains

Substitute the values into the formula for case (a): \( f' = 262 \frac{343 + 18}{343 - 30} \). Calculate the fraction and the result to find \( f' \).
04

Result for Approaching Trains

The frequency heard by the passenger when the trains are approaching is approximately \( 295.92 \mathrm{~Hz} \).
05

Factors for case (b) - Receding Trains

For the case where the trains are moving away from each other: Use the formula \( f' = 262 \frac{343 - 18}{343 + 30} \).
06

Calculation for Receding Trains

Substitute the values into the formula for case (b) and calculate \( f' = 262 \frac{343 - 18}{343 + 30} \). Calculate the fraction and the result.
07

Result for Receding Trains

The frequency heard by the passenger when the trains are receding is approximately \( 234.21 \mathrm{~Hz} \).

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Key Concepts

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

Understanding Acoustics and the Doppler Effect
Acoustics is the science of sound, exploring how sound behaves in different environments. In our daily lives, we experience sound through many contexts, like music or traffic noises. When we analyze sound from scientific angles, we often delve into acoustics, where the Doppler Effect plays a vital role.
This effect describes how the frequency of sound changes for an observer when the source of the sound is in motion relative to them. It can easily be noticed when a vehicle whizzing by changes the pitch of the sound it makes. This important phenomenon is named after Christian Doppler, who first proposed it.
In the case of our exercise, where two trains are involved, we're focusing on how the motion of the trains affects the sound's frequency. The Doppler Effect is not just about sound; it also applies to light and other waves, but in acoustics, it helps us understand experiences like changes in pitch as a train horn approaches or recedes.
Frequency Calculation for Moving Sources and Observers
Frequency calculation becomes critical when sources and observers are moving, as is often the case in real-world scenarios. Frequency is the number of wave cycles passing a point per second, measured in Hertz (Hz).
In the Doppler Effect, when the source of a sound and the observer are moving relative to each other, the frequency the observer hears is different from what is emitted.
Using the formula:
  • \( f' = f \frac{v + v_0}{v - v_s} \)
  • \( f' \) represents the frequency heard by the observer,
  • \( f \) is the original frequency emitted,
  • \( v \) is the speed of sound in air,
  • \( v_0 \) is the speed of the observer,
  • \( v_s \) is the speed of the source.
To arrive at the correct frequency, we substitute these values into the formula. It's fascinating when you apply this knowledge to situations, such as this train example, and see how speed impacts sound perception.
Approaching Physics Problem-Solving
Physics problem-solving involves breaking down the problem into manageable parts, just like how you tackle any puzzle. Applying the Doppler Effect formula is a good example of using physics to solve real-world problems.
When you face a problem, start by identifying what's given and what needs to be found. In our exercise, understanding that there are two scenarios (approaching and receding) helps in setting up the equation correctly. Each train's motion affects how the sound is perceived, necessitating different calculations.
After setting up the formula with the correct parameters, solving goes beyond basic arithmetic; it requires understanding why each component is used and what it represents. By following each step methodically, you find the solution and learn more about the principles governing these phenomena.
Analyzing Train Motion and Its Effects on Sound
Train motion analysis is intriguing when coupled with sound mechanics, like the Doppler Effect. Trains are often used in textbook problems because they offer a clear real-life example of how motion impacts sound waves. When trains move towards each other, the sound waves compress, leading to a higher pitch, which is a direct result of the Doppler Effect.
Conversely, when they move apart, the sound waves elongate, causing a drop in pitch. In our problem, the first setup involved trains approaching: both speeds add up, enhancing the frequency. For receding trains, speeds counteract, and thus, the perceived frequency drops.
  • Approaching trains: Frequencies increase due to compressed wavefronts.
  • Receding trains: Frequencies decrease as waves stretch apart.
Understanding these principles aids in solving complex physics problems and demonstrates the interconnectedness of motion and sound perception.

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Most popular questions from this chapter

Standing sound waves are produced in a pipe that is \(1.20 \mathrm{~m}\) long. For the fundamental frequency and the first two overtones, determine the locations along the pipe (measured from the left end) of the displacement nodes if (a) the pipe is open at both ends; (b) the pipe is closed at the left end and open at the right end.

While sitting in your car by the side of a country road, you see your friend, who happens to have an identical car with an identical horn, approaching you. You blow your horn, which has a frequency of \(260 \mathrm{~Hz}\); your friend begins to blow his horn as well, and you hear a beat frequency of \(6.0 \mathrm{~Hz}\). How fast is your friend approaching you?

(a) Whale communication. Blue whales apparently communicate with each other using sound of frequency \(17 \mathrm{~Hz}\), which can be heard nearly \(1000 \mathrm{~km}\) away in the ocean. What is the wavelength of such a sound in seawater, where the speed of sound is \(1531 \mathrm{~m} / \mathrm{s} ?\) (b) Dolphin clicks. One type of sound that dolphins emit is a sharp click of wavelength \(1.5 \mathrm{~cm}\) in the ocean. What is the frequency of such clicks? (c) Dog whistles. One brand of dog whistles claims a frequency of \(25 \mathrm{kHz}\) for its product. What is the wavelength of this sound? (d) Bats. While bats emit a wide varicty of sounds, one type emits pulses of sound having a frequency between \(39 \mathrm{kHz}\) and \(78 \mathrm{kHz}\). What is the range of wavelengths of this sound? (e) Sonograms. Ultrasound is used to view the interior of the body, much as X-rays are utilized. For sharp imagery, the wavelength of the sound should be around one-fourth (or less) the size of the objects to be viewed. Approximately what frequency of sound is needed to produce a clear image of a tumor that is \(1.0 \mathrm{~mm}\) across if the speed of sound in the tissue is \(1550 \mathrm{~m} / \mathrm{s} ?\)

How fast (as a percentage of light speed) would a star have to be moving so that the frequency of the light we receive from it is \(10.0 \%\) higher than the frequency of the light it is emitting? Would it be moving away from us or toward us? (Assume it is moving either directly away from us or directly toward us.)

A bat flies toward a wall, emitting a steady sound of frequency \(2000 \mathrm{~Hz}\). The bat hears its own sound, plus the sound reflected by the wall. How fast should the bat fly in order to hear a beat frequency of \(10.0 \mathrm{~Hz}\) ? (Hint: Break this problem into two parts, first with the bat as the source and the wall as the listener and then with the wall as the source and the bat as the listener.)

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