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Chapter 18: Problem 2

You are at a large outdoor concert, seated \(300 \mathrm{~m}\) from the speaker system. The concert is also being broadcast live via satellite (at the speed of light, \(3.0 \times 10^{8} \mathrm{~m} / \mathrm{s}\) ). Consider a listener \(5000 \mathrm{~km}\) away who receives the broadcast. Who hears the music first, you or the listener and by what time difference?

Short Answer

Expert verified
The listener hears the music 0.8573 seconds earlier.

Step by step solution

01

Calculate the Sound Travel Time to You

Sound travels at approximately 343 meters per second in air. Use the formula to calculate the time it takes for the sound to travel to you:\[ t_{\text{you}} = \frac{d_{\text{you}}}{v_{\text{sound}}} \]where \( d_{\text{you}} = 300 \) meters and \( v_{\text{sound}} = 343 \) meters per second.
02

Calculate the Broadcast Time to the Listener

Radio waves travel at the speed of light, which is \( 3.0 \times 10^8 \) meters per second. Calculate the distance in meters to the listener using the conversion 1 km = 1000 m:\[ d_{\text{listener}} = 5000 \times 1000 = 5,000,000 \text{ meters} \]Now, use the formula to calculate the time it takes for the broadcast to reach the listener:\[ t_{\text{listener}} = \frac{d_{\text{listener}}}{v_{\text{light}}} \]where \( d_{\text{listener}} = 5,000,000 \) meters and \( v_{\text{light}} = 3.0 \times 10^8 \) meters per second.
03

Compare the Two Times

Compare the time it takes for the sound to reach you and the time it takes for the broadcast to reach the listener. The difference in these times will tell us who hears the music first and by how much time.
04

Calculation

First, calculate your time:\[ t_{\text{you}} = \frac{300}{343} \approx 0.874 \text{ seconds} \]Next, calculate the listener's time:\[ t_{\text{listener}} = \frac{5,000,000}{3.0 \times 10^8} \approx 0.0167 \text{ seconds} \]Now, subtract the listener's time from your time to find the difference in time:\[ \Delta t = t_{\text{you}} - t_{\text{listener}} = 0.874 - 0.0167 \approx 0.8573 \text{ seconds} \]
05

Conclusion

The listener hears the music faster by approximately 0.8573 seconds.

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Key Concepts

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

Speed of Sound
In this exercise, we need to understand how sound waves travel. Sound travels through the air at a speed of approximately 343 meters per second (m/s). This speed can vary depending on conditions like temperature, humidity, and air pressure, but for simplicity, we'll use the standard speed.
When you hear music at a concert, sound waves emanate from the speaker systems and move through the air to reach your ears.
If you are seated 300 meters away from the speakers, we can calculate the time it takes for the sound to reach you using the formula:
\[ t_{\text{you}} = \frac{d_{\text{you}}}{v_{\text{sound}}} \]
Substitute your distance to the speakers and the speed of sound:
\[ t_{\text{you}} = \frac{300}{343} \]
This calculation gives approximately 0.874 seconds. So, it takes 0.874 seconds for the sound to travel 300 meters to your ears.
Speed of Light
In contrast to sound, light travels incredibly fast. The speed of light in a vacuum is about 300,000,000 meters per second (or 3.0 脳 10^8 m/s).
Light, including radio waves used for broadcasting, travels much faster than sound.
For listeners who are not at the concert but instead are receiving the broadcast live via satellite, the sound gets converted into radio waves, which travel at the speed of light.
Let's calculate the time it takes for the broadcast to reach a listener 5000 kilometers away:
1 km = 1000 meters, so 5000 km is equal to 5,000,000 meters.
We use the formula:
\[ t_{\text{listener}} = \frac{d_{\text{listener}}}{v_{\text{light}}} \]
Substituting the listener's distance and the speed of light:
\[ t_{\text{listener}} = \frac{5,000,000}{3.0 \times 10^8} \]
This calculation results in approximately 0.0167 seconds.
As you can see, light travels this distance almost instantaneously compared to sound.
Time Calculation
Now that we've calculated the times for both sound and light travel, let's compare them. We find that:
\[ t_{\text{you}} = 0.874 \text{ seconds} \] \[ t_{\text{listener}} = 0.0167 \text{ seconds} \]
We subtract the listener鈥檚 time from your time to find the time difference:
\[ \text{Difference} = t_{\text{you}} - t_{\text{listener}} = 0.874 - 0.0167 \text{ seconds} \] \[ \text{Difference} = 0.8573 \text{ seconds} \]
This means the listener receiving the broadcast 5000 kilometers away hears the music about 0.8573 seconds faster than you.
While these calculations may seem small, they highlight the significant differences in travel times for sound and light.
In practice, this is why lightning appears before we hear thunder, and why events broadcast live can be experienced almost simultaneously around the world.
Understanding these differences helps grasp crucial concepts in physics and their real-world impacts.

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Most popular questions from this chapter

Two loudspeakers are located \(3.55 \mathrm{~m}\) apart on an outdoor stage. A listener is \(18.3 \mathrm{~m}\) 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 are the three lowest frequencies at which the listener will hear a minimum signal because of destructive interference? (b) What are the three lowest frequencies at which the listener will hear a maximum signal?

The period of a pulsating variable star may be estimated by considering the star to be executing radial longitudinal pulsations in the fundamental standing wave mode. That is, the star's radius varies periodically with time, with a displacement antinode at the star's surface. (a) Would you expect the center of the star to be a displacement node or antinode? (b) By analogy with a pipe with one open end, show that the period of pulsation \(T\) is given by $$T=\frac{4 R}{\langle v\rangle}$$ where \(R\) is the equilibrium radius of the star and \(\langle v\) ) is the average sound speed in the material of the star. (c) Typical white dwarf stars are composed of material with a bulk modulus of \(1.33 \times 10^{22} \mathrm{~Pa}\) and a density of \(10^{10} \mathrm{~kg} / \mathrm{m}^{3}\). They have radii equal to \(9.0 \times 10^{-3}\) solar radius. What is the approximate pulsation period of a white dwarf?

A tube \(1.20 \mathrm{~m}\) long is closed at one end. A stretched wire is placed near the open end. The wire is \(0.330 \mathrm{~m}\) long and has a mass of \(9.60 \mathrm{~g} .\) It is fixed at both ends and oscillates in its fundamental mode. By resonance, it sets the air column in the tube into oscillation at that column's fundamental frequency. Find (a) that frequency and (b) the tension in the wire.

A source emits sound waves isotropically. The intensity of the waves \(2.50 \mathrm{~m}\) from the source is \(1.91 \times 10^{-4} \mathrm{~W} / \mathrm{m}^{2}\) Assuming that the energy of the waves is conserved, find the power of the source.

A plane flies at \(1.25\) times the speed of sound. Its sonic boom reaches a man on the ground \(1.00\) min after the plane passes directly overhead. What is the altitude of the plane? Assume the speed of sound to be \(330 \mathrm{~m} / \mathrm{s}\).
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