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Chapter 17: Problem 84

The speed of sound in a certain metal is \(v_{m}\). One end of a long pipe of that metal of length \(L\) is struck a hard blow. A listener at the other end hears two sounds, one from the wave that travels along the pipe's metal wall and the other from the wave that travels through the air inside the pipe. (a) If \(v\) is the speed of sound in air, what is the time interval \(\Delta t\) between the arrivals of the two sounds at the listener's ear? (b) If \(\Delta t=1.00 \mathrm{~s}\) and the metal is steel, what is the length \(L ?\)

Short Answer

Expert verified
(a) \(\Delta t = \left( \frac{1}{v} - \frac{1}{v_{m}} \right) L\); (b) \(L = v_{m} v \Delta t (v_{m} - v)^{-1}\).

Step by step solution

01

Understanding the Problem

We are asked to find the time interval \(\Delta t\) between the arrival of sound waves traveling through metal and air. We need to calculate the arrival times for each path and then find their difference.
02

Calculate Time for Sound in Metal

The time taken for sound to travel through the metal \(t_{m}\) is given by the formula \(t_{m} = \frac{L}{v_{m}}\). Here, \(v_{m}\) is the speed of sound in metal, and \(L\) is the length of the pipe.
03

Calculate Time for Sound in Air

Similarly, the time taken for sound to travel through air \(t_{a}\) is given by \(t_{a} = \frac{L}{v}\), where \(v\) is the speed of sound in air.
04

Find the Time Interval

The time interval \(\Delta t\) between the arrivals is the difference of the two times: \(\Delta t = t_{a} - t_{m} = \frac{L}{v} - \frac{L}{v_{m}}\).
05

Solve for Length L

Rearrange the equation \(\Delta t = \frac{L}{v} - \frac{L}{v_{m}}\) to solve for \(L\). Substituting \(\Delta t = 1.00 \text{ s}\) and the given values for the speed of sound in steel and air, solve the equation to find \(L\).

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Key Concepts

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

Time Interval
When sound travels through different mediums, it doesn't always reach us simultaneously. This gives rise to a **time interval** between the arrival of sounds. The time interval, denoted by \( \Delta t \), is essentially the difference in time it takes for the sound to travel through two mediums. In this exercise, sound travels through a metal pipe, which is a faster medium, and through air inside the pipe.To calculate this interval, we follow these steps:
  • Calculate the time \( t_{m} \) taken by the sound to travel through metal using \( t_{m} = \frac{L}{v_{m}} \).
  • Calculate the time \( t_{a} \) taken by the sound to travel through air using \( t_{a} = \frac{L}{v} \).
  • The time interval is then \( \Delta t = t_{a} - t_{m} \), which represents the delay or lag of the sound through air relative to the sound through metal.
Understanding these calculations helps us determine how swiftly sound travels in different environments.
Wave Propagation
Wave propagation refers to the movement of waves through a medium. In physics, sound waves are a classic example of this process, where the wave moves particles in a medium, allowing the sound to travel from one point to another. In our problem, we see wave propagation in two different phases:
  • **Through Metal**: The speed of sound in metal is greater than in air. This is because the particles in metal are packed closer together, providing a more efficient medium for the wave to propagate quickly and with less resistance.
  • **Through Air**: Air particles are much less dense compared to metal, making wave propagation slower. The speed of sound in air is affected by factors like temperature and air pressure.
By examining wave propagation through different mediums, we get a better understanding of how sound behaves under various conditions. This knowledge is crucial in fields such as engineering and acoustics.
Physics Problem Solving
Physics problem solving is an essential skill that involves understanding, breaking down, and solving given problems. The key steps include identifying the knowns and unknowns, applying the appropriate formulas, and conducting precise calculations.For example, to solve the exercise provided:
  • **Understand the Problem**: Recognize that we have sound waves traveling through a pipe's metal and air, and we need to determine their time interval \( \Delta t \).
  • **Apply Formulas**: Use the formulas for time, \( t_{m} = \frac{L}{v_{m}} \) and \( t_{a} = \frac{L}{v} \), to find the travel times in each medium.
  • **Find the Solution**: Calculate the time interval \( \Delta t = \frac{L}{v} - \frac{L}{v_{m}} \) and solve any necessary steps, such as rearranging to find the pipe's length with known values.
Effective problem solving in physics involves logical thinking and careful analysis, providing a structured approach to tackle various physics-related challenges.

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Most popular questions from this chapter

Hot chocolate effect. Tap a metal spoon inside a mug of water and note the frequency \(f_{i}\) you hear. Then add a spoonful of powder (say, chocolate mix or instant coffee) and tap again as you stir the powder. The frequency you hear has a lower value \(f_{s}\) because the tiny air bubbles released by the powder change the water's bulk modulus. As the bubbles reach the water surface and disappear, the frequency gradually shifts back to its initial value. During the effect, the bubbles don't appreciably change the water's density or volume or the sound's wavelength.

A sound source moves along an \(x\) axis, between detectors \(A\) and \(B .\) The wavelength of the sound detected at \(A\) is \(0.500\) that of the sound detected at \(B .\) What is the ratio \(v_{s} / v\) of the speed of the source to the speed of sound?

Earthquakes generate sound waves inside Earth. Unlike a gas, Earth can experience both transverse (S) and longitudinal (P) sound waves. Typically, the speed of S waves is about \(4.5 \mathrm{~km} / \mathrm{s}\), and that of \(\mathrm{P}\) waves \(8.0 \mathrm{~km} / \mathrm{s}\). A seismograph records \(\mathrm{P}\) and \(\mathrm{S}\) waves from an earthquake. The first \(\mathrm{P}\) waves arrive \(3.0 \mathrm{~min}\) before the first S waves. If the waves travel in a straight line, how far away did the earthquake occur?

A listener at rest (with respect to the air and the ground) hears a signal of frequency \(f_{1}\) from a source moving toward him with a velocity of \(15 \mathrm{~m} / \mathrm{s}\), due east. If the listener then moves toward the approaching source with a velocity of \(25 \mathrm{~m} / \mathrm{s}\), due west, he hears a frequency \(f_{2}\) that differs from \(f_{1}\) by \(37 \mathrm{~Hz}\). What is the frequency of the source? (Take the speed of sound in air to be \(340 \mathrm{~m} / \mathrm{s}\).)

A point source that is stationary on an \(x\) axis emits a sinusoidal sound wave at a frequency of \(686 \mathrm{~Hz}\) and speed \(343 \mathrm{~m} / \mathrm{s}\). The wave travels radially outward from the source, causing air molecules to oscillate radially inward and outward. Let us define a wavefront as a line that connects points where the air molecules have the maximum, radially outward displacement. At any given instant, the wavefronts are concentric circles that are centered on the source. (a) Along \(x\), what is the adjacent wavefront separation? Next, the source moves along \(x\) at a speed of \(110 \mathrm{~m} / \mathrm{s}\). Along \(x\), what are the wavefront separations (b) in front of and (c) behind the source?
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