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Chapter 16: Problem 50

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 tirst at 18.0 \(\mathrm{m} / \mathrm{s}\) and (a) approaching the first and (b) receding from the first?

Short Answer

Expert verified
(a) Approximately 303 Hz, (b) approximately 272 Hz.

Step by step solution

01

Define the Problem

We need to calculate the frequency heard by a passenger on a second train. First, identify the given values: Original frequency \( f = 262 \, \text{Hz} \), speed of sound in air \( v = 343 \, \text{m/s} \), speed of the first train \( v_s = 30.0 \, \text{m/s} \), and speed of the second train \( v_o = 18.0 \, \text{m/s} \).
02

Apply the Doppler Effect Formula for Approaching Trains

For part (a), as the second train is approaching the first train, the formula is:\[f' = \left(\frac{v + v_o}{v - v_s}\right) f\]Substitute the values:\[f' = \left(\frac{343 + 18.0}{343 - 30.0}\right) \times 262\]
03

Calculate the Frequency for Approaching Trains

Calculate the frequency:\[f' = \left(\frac{361}{313}\right) \times 262 \]\[f' \approx 302.7 \, \text{Hz}\]So, the frequency heard by the passenger when approaching is approximately 303 Hz.
04

Apply the Doppler Effect Formula for Receding Trains

For part (b), as the second train is moving away, the formula is:\[f' = \left(\frac{v - v_o}{v - v_s}\right) f\]Substitute the values:\[f' = \left(\frac{343 - 18.0}{343 - 30.0}\right) \times 262\]
05

Calculate the Frequency for Receding Trains

Calculate the frequency:\[f' = \left(\frac{325}{313}\right) \times 262 \]\[f' \approx 271.5 \, \text{Hz}\]So, the frequency heard by the passenger when receding is approximately 272 Hz.

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Key Concepts

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

Understanding Frequency
Frequency is a fundamental concept in wave physics. It refers to the number of occurrences of a repeating event per unit time. In the context of sound waves, frequency determines the pitch of the sound. For instance, a high frequency corresponds to a high-pitched sound, like a soprano's singing voice or a bird's chirp.
  • In our train whistle problem, the original frequency of the train whistle is given as 262 Hz. This means the sound wave repeats 262 times per second.
  • This frequency is altered by the relative motion of the trains due to what is known as the Doppler Effect.
Understanding frequency is crucial for solving problems involving sound waves in motion, like the train whistle scenario.
The Role of the Train Whistle
A train whistle is a device designed to create a loud sound to alert people and vehicles when a train is approaching. This sound is a result of sound waves being generated by the whistle and shot out into the air.
  • In this scenario, the whistle produces a steady frequency of 262 Hz when the train is stationary or heard at rest relative to the listener.
  • However, as the train moves, the sound frequency perceived by an observer changes. This is due to the relative motion between the sound source (train) and the observer (passenger on another train).
Thus, the train whistle serves as a practical example of the Doppler Effect at work, altering the perceived frequency depending on whether the train is approaching or receding.
Velocity and Their Influence in the Scenario
Velocity plays a significant role in determining how we perceive the frequency of sound waves from moving sources like trains. Both the velocity of the sound wave and the velocities of the observer and the source are crucial here.
  • In the original problem, the speed of sound in air is assumed to be 343 m/s.
  • The first train moves at a velocity of 30.0 m/s while the second train has a velocity of 18.0 m/s, each contributing differently when trains are approaching or receding.
For this scenario:
- When the two trains are moving towards each other, the frequencies heard become higher.
- Conversely, when they move away, the perceived frequency lowers.
This change occurs because their relative velocities alter the compression and elongation of sound waves, effectively squeezing or stretching them.
The Nature of Sound Waves
Sound waves are vibrations that travel through the air (or other mediums) and are perceived by our ears as sound. They can be described as mechanical waves, meaning they require a medium like air or water to travel.
  • They propagate in the form of longitudinal waves, where particle displacement is parallel to the direction of wave propagation.
  • The speed at which sound waves move through air is determined by various factors including temperature and pressure, but is assumed to be approximately 343 m/s at room temperature.
In practical applications such as the problem with the trains, sound waves help illustrate how the speed and direction of an object affect the frequency detected by an observer. The Doppler Effect leverages this principle by altering the frequency due to the motion of the sound source relative to the listener.

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Most popular questions from this chapter

CP You have a stopped pipe of adjustable length close to a taut \(85.0-\mathrm{cm}, 7.25-\mathrm{g}\) wire under a tension of 4110 \(\mathrm{N}\) . You want to adjust the length of the pipe so that, when it produces sound at its fundamental frequency, this sound causes the wire to vibrate in its second overtone with very large amplitude. How long should the pipe be?

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

CP Two identical taut strings under the same tension \(F\) produce a note of the same fundamental frequency \(f_{0}\) . The tension in one of them is now increased by a very small amount \(\Delta F .\) (a) If they are played together in their fundamental, show that the frequency of the beat produced is \(f_{\text { beat }}=f_{0}(\Delta F / 2 F)\) . (b) Two identical violin strings, when in tune and stretched with the same tension, have a fundamental frequency of 440.0 Hz. One of the strings is retuned by increasing its tension. When this is done, 1.5 beats per second are heard when both strings are plucked simultaneously at their centers. By what percentage was the string tension changed?

Wagnerian Opera. A man marries a great Wagnerian soprano but, alas, he discovers he cannot stand Wagnerian opera. In order to save his eardrums, the unhappy man decides he must silence his larklike wife for good. His plan is to tie her to the front of his car and send car and soprano speeding toward a brick wall. This soprano is quite shrewd, however, having studied physics in her student days at the music conservatory. She realizes that this wall has a resonant frequency of 600 \(\mathrm{Hz}\) , which means that if a continuous sound wave of this frequency hits the wall, it will fall down, and she will be saved to sing more Isoldes. The car is heading toward the wall at a high speed of 30 \(\mathrm{m} / \mathrm{s}\) . (a) At what frequency must the soprano sing so that the will will crumble? (b) What frequency will the soprano hear reflected from the wall just before it crumbles?

Two train whistles, \(A\) and \(B\) , each have a frequency of 392 \(\mathrm{Hz} . A\) is stationary and \(B\) is moving toward the right (away from \(A\) ) at a speed of 35.0 \(\mathrm{m} / \mathrm{s}\) . A listener is between the two whistles and is moving toward the right with a speed of 15.0 \(\mathrm{m} / \mathrm{s}\) (Fig. E16.45). No wind is blowing. (a) What is the frequency from A as heard by the listener? (b) What is the frequency from \(B\) as heard by the listener? (c) What is the beat frequency detected by the listener?
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