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Chapter 16: Problem 37

An explosion occurs at the end of a pier. The sound reaches the other end of the pier by traveling through three media: air, fresh water, and a slender metal handrail. The speeds of sound in air, water, and the handrail are \(343,1482,\) and \(5040 \mathrm{~m} / \mathrm{s},\) respectively. The sound travels a distance of \(125 \mathrm{~m}\) in each medium. (a) Through which medium does the sound arrive first, second, and third? (b) After the first sound arrives, how much later do the second and third sounds arrive?

Short Answer

Expert verified
The sound arrives first through the metal handrail, then water, and finally air. The second sound arrives 0.059 s later, and the third sound arrives 0.339 s later.

Step by step solution

01

Calculate the Time for Sound Through Air

Use the formula for time, \[\text{Time} (t) = \frac{\text{Distance} (d)}{\text{Speed} (v)}\]Insert the given values: distance \(d = 125 \text{ m}\), and speed of sound in air \(v = 343 \text{ m/s}\):\[t_{\text{air}} = \frac{125 \text{ m}}{343 \text{ m/s}} \approx 0.364 \text{ seconds}\]
02

Calculate the Time for Sound Through Water

Use the same formula for time:\[t_{\text{water}} = \frac{125 \text{ m}}{1482 \text{ m/s}} \approx 0.084 \text{ seconds}\]
03

Calculate the Time for Sound Through Metal Handrail

Use the same formula for time:\[t_{\text{metal}} = \frac{125 \text{ m}}{5040 \text{ m/s}} \approx 0.025 \text{ seconds}\]
04

Determine the Arrival Order

Compare the calculated times for each medium:- Metal handrail: \(0.025 \text{ s}\)- Water: \(0.084 \text{ s}\)- Air: \(0.364 \text{ s}\)The sound first arrives through the metal handrail, then through water, and finally through air.
05

Calculate Time Delay After First Arrival

Calculate the differences between the times for sound to arrive through each medium compared to the metal handrail (first arrival):- Time delay for water: \[t_{\text{water delay}} = t_{\text{water}} - t_{\text{metal}} = 0.084 \text{ s} - 0.025 \text{ s} = 0.059 \text{ s}\]- Time delay for air:\[t_{\text{air delay}} = t_{\text{air}} - t_{\text{metal}} = 0.364 \text{ s} - 0.025 \text{ s} = 0.339 \text{ s}\]

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Key Concepts

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

Speed of Sound
The speed of sound refers to the rate at which sound waves travel through a medium. Sound propagates by molecular collisions, so the speed of sound depends on how quickly these collisions occur, which is influenced by the medium's properties.
Factors that affect the speed of sound include:
  • Density: Generally, sound travels faster in solids than in liquids, and faster in liquids than in gases.
  • Elasticity: More elastic media allow sound waves to travel faster.
  • Temperature: Higher temperatures usually increase the speed of sound in gases, as molecules move more swiftly.
Knowing the speed of sound in different environments is crucial for calculations involving distance and time, especially when dealing with events like explosions or using technologies like sonar.
Medium of Transmission
The medium of transmission is the material through which sound waves travel. It includes the usual states of matter: solid, liquid, and gas. For example, in the given exercise, sound travels through air, fresh water, and a slender metal handrail. Each of these media has unique properties affecting sound transmission.
Consider the mediums from the problem:
  • Air: Sound speed is slowest here at 343 m/s due to lower density and elasticity.
  • Fresh Water: Sound travels faster at 1482 m/s. Water's molecules are closer together, facilitating quicker transmission than in air.
  • Metal Handrail: The fastest medium in the scenario, with a sound speed of 5040 m/s. Metal is dense and very elastic, enabling rapid sound conduction.
Understanding mediums helps in predicting how sound will behave, crucial for applications such as underwater communication and structural health monitoring.
Time Calculation
To determine the time it takes for sound to travel through a medium, you can use the simple formula: \[ t = \frac{d}{v} \] where:
  • \(t\) is the time,
  • \(d\) is the distance,
  • \(v\) is the speed of sound in the medium.
In our scenario:
  • Air: \( t_{\text{air}} = \frac{125 \text{ m}}{343 \text{ m/s}} \approx 0.364 \text{ s} \)
  • Water: \( t_{\text{water}} = \frac{125 \text{ m}}{1482 \text{ m/s}} \approx 0.084 \text{ s} \)
  • Metal: \( t_{\text{metal}} = \frac{125 \text{ m}}{5040 \text{ m/s}} \approx 0.025 \text{ s} \)
By calculating these times, we can understand the delay in sound reaching different points or observers, which is essential in sound recording, sonar detection, and other audio-related fields.
Sound Travel Through Different Media
When sound travels through different media, its velocity changes due to differences in each medium's density and elasticity. This behavior explains why sound arrives at different times once it travels through air, water, and metals, as shown in the exercise.

In the scenario with the explosion:
  • The fastest arrival was through the metal handrail because metals are dense and elastic, making sound travel swiftly.
  • The next arrival was through water, as it offers less resistance than air, although more than metal.
  • Lastly, sound reached through air, the slowest medium due to its lower density and elasticity.
This helps us understand practical phenomena like echoes in different environments and clarifies why underwater sounds might seem closer. Applications extend to designing concert halls for better acoustics or engineering communication devices that work in specific environments.

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Most popular questions from this chapter

The sound intensity level at a rock concert is \(115 \mathrm{~dB}\), while that at a jazz fest is \(95 \mathrm{~dB}\). Determine the ratio of the sound intensity at the rock concert to that at the jazz fest.

A rocket, starting from rest, travels straight up with an acceleration of \(58.0 \mathrm{~m} / \mathrm{s}^{2}\). When the rocket is at a height of \(562 \mathrm{~m}\), it produces sound that eventually reaches a ground-based monitoring station directly below. The sound is emitted uniformly in all directions. The monitoring station measures a sound intensity \(I .\) Later, the station measures an intensity \(\frac{1}{3} I\). Assuming that the speed of sound is \(343 \mathrm{~m} / \mathrm{s},\) find the time that has elapsed between the two measurements.

A wave traveling in the \(+x\) direction has an amplitude of \(0.35 \mathrm{~m},\) a speed of \(5.2 \mathrm{~m} /\) \(\mathrm{s},\) and a frequency of \(14 \mathrm{~Hz}\). Write the equation of the wave in the form given by either Equation 16.3 or 16.4

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.

So Concept Questions Example 4 in the text discusses an ultrasonic ruler that displays the distance between the ruler and an object, such as a wall. The ruler sends out a pulse of ultrasonic sound and measures the time it takes for the pulse to reflect from the object and return. The ruler uses this time, along with a preset value for the speed of sound in air, to determine the distance. Suppose you use this ruler underwater, rather than in air. (a) Is the speed of sound in water greater than, less than, or equal to the speed of sound in air? (b) Is the reading on the ruler greater than, less than, or equal to the actual distance? Provide reasons for your answers. Problem The actual distance from the ultrasonic ruler to an object is \(25.0 \mathrm{~m} .\) The adiabatic bulk modulus and density of seawater are \(B_{\mathrm{ad}}=2.37 \times 10^{9} \mathrm{~Pa}\) and \(\rho=1025\) \(\mathrm{kg} / \mathrm{m}^{3}\), respectively. Assume that the ruler uses a preset value of \(343 \mathrm{~m} / \mathrm{s}\) for the speed of sound in air, and determine the distance reading on its display. Verify that your answer is consistent with your answers to the Concept Questions.
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