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A train on one track moves in the same direction as a second train on the adjacent track. The first train, which is ahead of the second train and moves with a speed of \(36.8 \mathrm{m} / \mathrm{s},\) blows a horn whose frequency is \(124 \mathrm{Hz}\). If the frequency heard on the second train is \(135 \mathrm{Hz},\) what is its speed?

Short Answer

Expert verified
The speed of the second train is approximately 71.3 m/s.

Step by step solution

01

Understanding the Doppler Effect

When a sound source and a listener are moving towards each other, the frequency heard by the listener is higher than the actual frequency. This is due to the Doppler Effect, which is the change in frequency of a wave in relation to an observer moving relative to the wave source.
02

Determine the Formula for the Situation

The Doppler Effect formula when both source and observer are in motion is given by: \[ f' = f \left(\frac{v + v_0}{v + v_s}\right) \] where \(f'\) is the observed frequency, \(f\) is the emitted frequency, \(v\) is the speed of sound, \(v_0\) is the speed of the observer, and \(v_s\) is the speed of the source.
03

Identify Known Values

Given: \( f = 124 \, \mathrm{Hz} \), \( f' = 135 \, \mathrm{Hz} \), \( v_s = 36.8 \, \mathrm{m/s} \). The speed of sound in air, \( v \), is approximately \( 343 \, \mathrm{m/s} \). We need to find \( v_0 \), the speed of the second train.
04

Solving for the Second Train's Speed

Plug the known values into the formula and solve for \( v_0 \): 1. Rearrange the formula to solve for \(v_0\): \[ f' = f \left(\frac{v + v_0}{v + v_s}\right) \] \[ 135 = 124 \left(\frac{343 + v_0}{343 + 36.8}\right) \]2. Simplify and solve: \[ \frac{135}{124} = \frac{343 + v_0}{379.8} \]3. Multiply both sides by \( 379.8 \): \[ 135 \times 379.8 = 124 \times (343 + v_0) \]4. Calculate: \[ 51273 = 42432 + 124 \times v_0 \]5. Solve for \( v_0 \): \[ 124 \times v_0 = 51273 - 42432 \] \[ v_0 = \frac{8841}{124} \approx 71.3 \, \mathrm{m/s} \]
05

Result Interpretation

The calculated speed indicates that the second train is moving at approximately \(71.3 \, \mathrm{m/s}\) to hear the frequency as \(135 \, \mathrm{Hz}\), meaning it is moving faster than the first train, approaching the sound source.

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Key Concepts

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

Frequency Change
When dealing with sound waves, understanding frequency change is crucial especially in the context of the Doppler Effect. The frequency observed by a listener differs from the source frequency due to relative motion. This phenomenon is essentially because of varying distances covered by successive sound waves.
For instance, when a sound source moves closer to an observer, the waves are compressed, leading to a higher frequency or pitch. Conversely, if the source moves away, the waves are stretched, causing a lower frequency. This is analogous to the perception of pitch for an approaching versus receding siren.
In the exercise, the train blowing a horn has an emitted frequency of 124 Hz. The observer on the second train hears this frequency as 135 Hz, a higher value due to both trains moving towards one another. The central concept here is the change in frequency due to relative motion.
Sound Wave
Sound waves are disturbances that travel through a medium, such as air, in the form of longitudinal waves, where particles oscillate parallel to the wave direction. These vibrations cause areas of compression and rarefaction, enabling sound to propagate.
Each sound wave has characteristics such as frequency, wavelength, and speed, all interdependent. In this scenario, the frequency is altered by the movement of trains, highlighting how motion affects wave properties despite the speed of sound in air remaining relatively constant at around 343 m/s.
It's essential to understand that as long as the source or observer moves, the perceived frequency changes, not the wave speed. In our problem, the Doppler Effect explains the change in frequency while the sound wave's speed remains constant.
Relative Motion
Relative motion, a fundamental concept in physics, describes how one body moves in relation to another. In the Doppler Effect, it's the relative speeds of the source and observer that provoke frequency changes. Relative motion is integral in determining the perceived direction and magnitude of frequency shifts.
In the Doppler Effect formula, each variable accounts for these relative movements. For example, in our exercise, the speed of the source train effectively alters the frequency heard by the observer on the second train, due to them both moving in the same direction, albeit at different speeds.
By determining the observer's speed, which was approximately 71.3 m/s in the problem, we see how relative motion affects the sound heard, exemplifying why shifts occur as objects approach one another. The faster the listener approaches the sound source, the greater the increase in frequency.

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Most popular questions from this chapter

Waves on a particular string travel with a speed of \(16 \mathrm{m} / \mathrm{s}\). By what factor should the tension in this string be changed to produce waves with a speed of \(32 \mathrm{m} / \mathrm{s} ?\)

Dolphin Ultrasound Dolphins of the open ocean are classified as Type II Odontocetes (toothed whales). These animals use ultrasonic "clicks" with a frequency of about \(55 \mathrm{kHz}\) to navigate and find prey. (a) Suppose a dolphin sends out a series of clicks that are reflected back from the bottom of the ocean \(75 \mathrm{m}\) below. How much time elapses before the dolphin hears the echoes of the clicks? (The speed of sound in seawater is approximately \(1530 \mathrm{m} / \mathrm{s} .\) ) (b) What is the wavelength of \(55-\mathrm{kHz}\) sound in the occan?

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