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Chapter 17: Problem 6

A man strikes one end of a thin rod with a hammer. The speed of sound in the rod is 15 times the speed of sound in air. A woman, at the other end with her ear close to the rod, hears the sound of the blow twice with a 0.12 s interval between; one sound comes through the rod and the other comes through the air alongside the rod. If the speed of sound in air is \(343 \mathrm{~m} / \mathrm{s},\) what is the length of the rod?

Short Answer

Expert verified
The length of the rod is approximately 41.16 meters.

Step by step solution

01

Understand Given Values

We know the speed of sound in air is \(343\, \text{m/s}\). The speed of sound in the rod is given as 15 times the speed of sound in air, which calculates to \(15 \times 343 = 5145\, \text{m/s}\). The difference in arrival time of the sounds through the rod and air is 0.12 seconds.
02

Calculate Time Taken Through Rod

Let the length of the rod be \(L\). The time taken for the sound to travel through the rod will be \( \frac{L}{5145} \).
03

Calculate Time Taken Through Air

The time taken for the sound to travel through the air will be \( \frac{L}{343} \).
04

Set Up the Equation Using Time Difference

According to the problem, the difference in time between the sound traveling through air and the rod is 0.12 seconds. Therefore, we have the equation: \( \frac{L}{343} - \frac{L}{5145} = 0.12 \).
05

Solve for Length of the Rod

Solve the equation for \(L\): \(\frac{L}{343} - \frac{L}{5145} = 0.12\)This simplifies to: \(L \left( \frac{1}{343} - \frac{1}{5145} \right) = 0.12\)Calculate the expression in the parenthesis:\(\frac{1}{343} - \frac{1}{5145} \approx 0.0029152\)Now solve for \(L\):\(L = \frac{0.12}{0.0029152} \approx 41.16\, \text{meters}\)

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Key Concepts

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

Physics Problem Solving
Physics problem solving involves breaking down complex situations into understandable parts. In this problem, we must determine the length of a rod using the given speed of sound in air and the rod, as well as the time difference in sound travel.
The problem begins by establishing known variables:
  • Speed of sound in air: 343 m/s
  • Speed of sound in the rod: 15 times that in air, calculated as 5145 m/s
  • Time difference of sound arrival at the other end of the rod: 0.12 seconds
Understanding these is critical. Next, we determine how the sounds travel through different mediums, leading to a key equation relating these times to the rod length.
Wave Propagation
Wave propagation is the transmission of waves through a medium, such as the sound waves traveling through air and a rod in this scenario. Sound, a type of mechanical wave, requires a medium to travel, and its speed varies depending on that medium.
In this problem: - Sound travels faster in solid materials, like the rod, compared to air. This is due to the molecules being closely packed, allowing for quicker vibration transmission. - The speed of sound in the rod (5145 m/s) is significantly faster than in air (343 m/s). This speed discrepancy causes a difference in sound arrival times at the opposite end of the rod. Recognizing how wave propagation operates helps frame the problem: despite originating from the same source, sounds arrive at different times due to propagation differences in their respective mediums.
Time Difference Calculation
Calculating the time difference is a pivotal step in solving the length of the rod. The sound reaches the observer through two different paths鈥攖hrough air and through the rod鈥攚ith a known time difference of 0.12 seconds.
The equations for travel time are:
  • Through the rod: \( \frac{L}{5145} \)
  • Through air: \( \frac{L}{343} \)
Given the time difference, we set up the equation:\[ \frac{L}{343} - \frac{L}{5145} = 0.12 \]This equation is solved by isolating \(L\). First, calculate the coefficients:\[ \frac{1}{343} - \frac{1}{5145} \approx 0.0029152 \]Then, simply solve for \(L\):\[ L = \frac{0.12}{0.0029152} \approx 41.16 \text{ meters} \]This process relies on understanding how differences in propagation speed impact travel time, allowing precise time difference calculations to infer the rod's length.

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Most popular questions from this chapter

At a certain point, two waves produce pressure variations given by \(\Delta p_{1}=\Delta p_{m} \sin \omega t\) and \(\Delta p_{2}=\Delta p_{m} \sin (\omega t-\phi) .\) At this point, what is the ratio \(\Delta p_{r} / \Delta p_{m},\) where \(\Delta p_{r}\) is the pressure amplitude of the resultant wave, if \(\phi\) is (a) \(0,\) (b) \(\pi / 2,\) (c) \(\pi / 3\), and (d) \(\pi / 4 ?\)

You are standing at a distance \(D\) from an isotropic point source of sound. You walk \(50.0 \mathrm{~m}\) toward the source and observe that the intensity of the sound has doubled. Calculate the distance \(D\).

A source \(S\) and a detector \(D\) of radio waves are a distance \(d\) apart on level ground (Fig. 17-52). Radio waves of wavelength \(\lambda\) reach \(\mathrm{D}\) either along a straight path or by reflecting (bouncing) from a certain layer in the atmosphere. When the layer is at height \(H,\) the two waves reaching \(\mathrm{D}\) are exactly in phase. If the layer gradually rises, the phase difference between the two waves gradually shifts, until they are exactly out of phase when the layer is at height \(H+h .\) Express \(\lambda\) in terms of \(d, h,\) and \(H\).

A sound wave of frequency \(300 \mathrm{~Hz}\) has an intensity of \(1.00 \mu \mathrm{W} / \mathrm{m}^{2} .\) What is the amplitude of the air oscillations caused by this wave?

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 \(340 \mathrm{~m} / \mathrm{s}\).)
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