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Chapter 17: Problem 63

The speed of a one-dimensional compressional wave traveling along a thin copper rod is 3.56 \(\mathrm{km} / \mathrm{s}\) . A copper bar is given a sharp compressional blow at one end. The sound of the blow, traveling through air at \(0^{\circ} \mathrm{C}\) , reaches the opposite end of the bar 6.40 \(\mathrm{ms}\) later than the sound transmitted through the metal of the bar. What is the length of the bar?

Short Answer

Expert verified
The length of the bar is approximately 7.32 meters.

Step by step solution

01

Understanding the Exercise

The problem provides the speed of sound in copper (\(v_{copper} = 3.56 \text{ km/s}\)) and states that the sound traveling through air reaches the opposite end of the copper bar 6.40 milliseconds (\(6.40 \text{ ms} = 6.40 \times 10^{-3} \text{ s}\)) later than the sound traveling through the copper. The speed of sound in air at 0掳C can be taken as approximately 331 m/s. We are asked to find the length of the copper bar.
02

Converting the Speed in Copper to Correct Units

First, we need the speed of sound in copper in meters per second. Multiply the given speed by 1000 to convert kilometers to meters: \(v_{copper} = 3.56 \text{ km/s} \times 1000 \text{ m/km} = 3560 \text{ m/s}\).
03

Establishing the Equations for Travel Times

Let the length of the bar be denoted by L. The time taken for the sound to travel through the copper is \(t_{copper} = \frac{L}{v_{copper}}\). The time taken for the sound to travel through the air is \(t_{air} = \frac{L}{v_{air}}\). The problem states that \(t_{air} = t_{copper} + 6.40 \times 10^{-3} \text{ s}\).
04

Setting up the Equation with the Given Time Delay

Substitute the expressions for \(t_{copper}\) and \(t_{air}\) into the equation that includes the time delay: \( \frac{L}{v_{air}} = \frac{L}{v_{copper}} + 6.40 \times 10^{-3}\).
05

Solving for the Length of the Bar

Multiply every term by \(v_{air} \times v_{copper}\) to clear the fractions: \((L \times v_{copper}) = (L \times v_{air}) + (6.40 \times 10^{-3} \text{ s} \times v_{air} \times v_{copper})\). Next, isolate L: \(L = \frac{6.40 \times 10^{-3} \text{ s} \times v_{air} \times v_{copper}}{v_{copper} - v_{air}}\). Substitute the known speeds: \(L = \frac{6.40 \times 10^{-3} \text{ s} \times 331 \text{ m/s} \times 3560 \text{ m/s}}{3560 \text{ m/s} - 331 \text{ m/s}}\).
06

Calculate the Length

Calculate the length of the bar: \(L = \frac{6.40 \times 10^{-3} \times 331 \times 3560}{3560 - 331}\) which simplifies to approximately 7.32 meters.

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Key Concepts

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

Waves in Physics
Understanding waves is crucial when studying physics, as they are disturbances that transfer energy from one point to another.

In physics, waves can be classified by how they move, and there are two main types:
  • Transverse waves: where the particles of the medium move perpendicular to the direction of the wave's travel.
  • Compressional or longitudinal waves: where particles move parallel to the wave's direction.
Sound waves are a typical example of longitudinal waves, where the particles in the medium (like air or a solid object) vibrate back and forth along the same direction the wave travels.

Studying waves helps us understand various phenomena, such as the echoes in a canyon, the color of the sky, and importantly, the topic of our exercise - the speed of sound through different media.
Compressional Wave
Compressional or longitudinal waves carry energy through a medium by compressing and rarefying the particles in the material. Imagine pushing and pulling a spring; the coils come together and then spread apart as the wave moves through.

The speed of these waves can vary vastly based on the medium they travel through. In solids, particles are closely packed, and the sound speed is typically higher compared to liquids and gases. In our exercise topic, the compressional wave is the sound wave that travels through the copper rod faster than through the air due to copper's dense atomic structure.

The ability of sound waves to travel through various mediums is a practical aspect to consider in various applications, including medical imaging, construction, and communications.
Sound Propagation
Sound propagation involves the movement of sound waves from their source to a listener. Sound travels as a wave of pressure variations through different kinds of mediums such as solids, liquids, and gases. The speed of sound is an essential aspect of how quickly this energy travels.

Factors that affect sound propagation include:
  • The medium's density: Sound travels faster in denser media because the particles are closer together and can transfer the vibrational energy more efficiently.
  • The medium's temperature: Usually, the speed of sound increases with the temperature due to increased energy and particle movement.
  • The medium's elasticity: Greater elasticity typically leads to faster sound speed, as the medium can quickly return to its original shape after being disturbed.
In our exercise, we calculate the bar's length by understanding the difference in sound propagation in air versus copper, demonstrating how sound speeds can aid in measurements and diagnostics.

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Most popular questions from this chapter

Suppose that you hear a clap of thunder 16.2 s after seeing the associated lightning stroke. The speed of sound waves in air is 343 \(\mathrm{m} / \mathrm{s}\) , and the speed of light is \(8.00 \times 10^{8} \mathrm{m} / \mathrm{s}\) . How fare you from the lightning stroke?

Assume that a loudspeaker broadcasts sound equally in all directions and produces sound with a level of 103 \(\mathrm{dB}\) at a distance of 1.60 \(\mathrm{m}\) from its center. (a) Find its sound power output. (b) If the salesperson claims to be giving you 150 \(\mathrm{W}\) per channel, he is referring to the electrical power input to the speaker. Find the efficiency of the speaker-that is, the fraction of input power that is converted into useful output power.

A bat, moving at 5.00 \(\mathrm{m} / \mathrm{s}\) , is chasing a flying insect (Fig. Pl7.7). If the bat emits a 40.0 \(\mathrm{kHz}\) chirp and receives back an echo at 40.4 \(\mathrm{kHz}\) , at what speed is the insect moving toward or away from the bat? (Take the speed of sound in air to be \(v=340 \mathrm{m} / \mathrm{s} . )\)

Find the speed of sound in mercury, which has a bulk modulus of approximately \(2.80 \times 10^{10} \mathrm{N} / \mathrm{m}^{2}\) and a density of 13600 \(\mathrm{kg} / \mathrm{m}^{3}\) .

A sinusoidal sound wave is described by the displacement wave function $$ s(x, t)=(2.00 \mu \mathrm{m}) \cos \left[\left(15.7 \mathrm{m}^{-1}\right) x-\left(858 \mathrm{s}^{-1}\right) t\right] $$ (a) Find the amplitude, wavelength, and speed of this wave. (b) Determine the instantaneous displacement from equilibrium of the elements of air at the position \(x=0.0500 \mathrm{m}\) at \(t=3.00 \mathrm{ms}\) . (c) Determine the maximum speed of the element's oscillatory motion.
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