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Chapter 17: Problem 34

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 \(60 \mathrm{~ms}\) 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 22.04 meters.

Step by step solution

01

Understand the Problem

We need to find the length of the rod, knowing that sound travels through the rod and air, and that the time interval between hearing these two sounds is 60 ms. Given is the speed of sound in air (343 m/s) and the fact that sound in the rod travels 15 times faster.
02

Determine Speed of Sound in Rod

The speed of sound in the rod is 15 times the speed of sound in air. Thus, \( v_{\text{rod}} = 15 \times 343 = 5145 \, \mathrm{m/s} \).
03

Set Up Time Equations for Both Paths

Let's denote the length of the rod as \( L \). The time, \( t_{\text{rod}} \), for sound traveling through the rod is \( t_{\text{rod}} = \frac{L}{v_{\text{rod}}} \). The time, \( t_{\text{air}} \), for sound traveling through the air is \( t_{\text{air}} = \frac{L}{v_{\text{air}}} \), where \( v_{\text{air}} = 343 \, \mathrm{m/s} \).
04

Use the Given Time Interval Information

The woman hears the sound through the rod first and then after 60 ms through the air. Therefore, \( t_{\text{air}} - t_{\text{rod}} = 60 \, \mathrm{ms} = 0.060 \, \mathrm{s} \).
05

Substitute and Solve for L

Substitute the expressions from the time equations into the interval equation:\( \frac{L}{343} - \frac{L}{5145} = 0.060 \). Solve for \( L \):First, find a common denominator and solve:\[\frac{5145L}{343 \times 5145} - \frac{343L}{343 \times 5145} = 0.060 \\rightarrow \frac{(5145 - 343)L}{1763205} = 0.060 \\rightarrow 4802L = 0.060 \times 1763205 \\rightarrow 4802L = 105792.3 \\rightarrow L = \frac{105792.3}{4802} \approx 22.04 \, \text{m}\]
06

Verify the Solution

The calculations show that the rod length is approximately 22.04 meters. Double-check your arithmetic and ensure all steps logically follow.

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Key Concepts

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

Speed of Sound
Sound travels at different speeds depending on the medium it moves through. We often think about how fast sound travels in air, a speed known as the "speed of sound in air." For average conditions, this speed is around 343 meters per second (m/s). In different materials, the speed of sound can change considerably. In solids, sound waves travel faster than in gases. This is due to the tightly packed particles present in solids that transmit vibrations more effectively.

In the example problem, the speed of sound in the rod is 15 times greater than in the air, specifically calculated to be 5145 m/s. This means sound waves travel much quicker through the dense medium of the rod than the air surrounding it.

Key points to understand include:
  • The density and elasticity of a medium affect the speed of sound.
  • Solids allow faster sound travel than air due to their tightly bound particles.
  • In our scenario, understanding this concept helps compute the time difference in hearing the sound through different mediums.
Understanding these concepts is crucial for solving problems related to sound propagation.
Wave Propagation in Solids
Wave propagation refers to how sound waves move through a medium, such as solids, liquids, or gases. In solids, these waves are primarily longitudinal, meaning particles in the solid vibrate parallel to the direction of the wave travel.

When a sound wave is introduced, such as a hammer striking a rod, these particles transfer energy from one particle to another efficiently, allowing sound to move quickly.

Here's why sound waves travel faster in solids:
  • Particles in solids are more tightly bound.
  • The stiffness (elasticity) of the material helps in fast energy transfer.
  • Consequently, the speed of sound in solids is usually greater than in fluids.
In the context of the problem, understanding wave propagation in solids allowed us to calculate the speed of sound in the rod and use it to determine the length of the rod based on the time interval of sound waves traveling through different pathways.
Problem Solving in Physics
Solving physics problems often requires a structured approach. The original exercise involves calculating the length of a rod using a known speed of sound in different mediums.

Here's a general breakdown of the problem-solving process in physics:
  • Understand the Problem: Grasp the scenario and identify what you need to find, like the length of the rod.
  • Gather Information: Take note of the given values, such as the speeds of sound and the time difference.
  • Formulate Equations: Use your understanding of physics principles to write equations. Here, one equation represents sound travel through the rod and another through air.
  • Substitute and Solve: Replace known values into your equations and do the necessary algebra to find unknowns.
  • Verify and Review: Check if your solution makes sense and correlates with the problem statement.
By breaking down problems like this into manageable steps, and understanding fundamental concepts, students can effectively tackle physics challenges.

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Most popular questions from this chapter

A bat is flitting about in a cave, navigating via ultrasonic bleeps. Assume that the sound emission frequency of the bat is \(39000 \mathrm{~Hz}\). During one fast swoop directly toward a flat wall surface, the bat is moving at \(0.020\) times the speed of sound in air. What frequency does the bat hear reflected off the wall?

The source of a sound wave has a power of \(3.00 \mu \mathrm{W}\). If it is a point source, (a) what is the intensity \(4.20 \mathrm{~m}\) away and (b) what is the sound level in decibels at that distance?

Two trains are traveling toward each other at \(35.7 \mathrm{~m} / \mathrm{s}\) relative to the ground. One train is blowing a whistle at \(850 \mathrm{~Hz}\). (a) What frequency is heard on the other train in still air? (b) What frequency is heard on the other train if the wind is blowing at \(35.7 \mathrm{~m} / \mathrm{s}\) toward the whistle and away from the listener? (c) What frequency is heard if the wind direction is reversed?

Two loud speakers are located \(3.35 \mathrm{~m}\) apart on an outdoor stage. A listener is \(17.5 \mathrm{~m}\) from one and \(19.5 \mathrm{~m}\) from the other. During the sound check, a signal generator drives the two speakers in phase with the same amplitude and frequency. The transmitted frequency is swept through the audible range ( \(20 \mathrm{~Hz}\) to 20 \(\mathrm{kHz}\) ). (a) What is the lowest frequency \(f_{\min , 1}\) that gives minimum signal (destructive interference) at the listener's location? By what number must \(f_{\min , 1}\) be multiplied to get (b) the second lowest frequency \(f_{\min , 2}\) that gives minimum signal and (c) the third lowest frequency \(f_{\min , 3}\) that gives minimum signal? (d) What is the lowest frequency \(f_{\text {max }, 1}\) that gives maximum signal (constructive interference) at the listener's location? By what number must \(f_{\max , 1}\) be multiplied to get (e) the second lowest frequency \(f_{\max , 2}\) that gives maximum signal and (f) the third lowest frequency \(f_{\max , 3}\) that gives maximum signal?

Approximately a third of people with normal hearing have ears that continuously emit a low-intensity sound outward through the ear canal. A person with such spontaneous otoacoustic emission is rarely aware of the sound, except perhaps in a noise-free environment, but occasionally the emission is loud enough to be heard by someone else nearby. In one observation, the sound wave had a frequency of \(1200 \mathrm{~Hz}\) and a pressure amplitude of \(2.50 \times 10^{-3} \mathrm{~Pa}\). What were (a) the displacement amplitude and (b) the intensity of the wave emitted by the ear?
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