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Chapter 12: Problem 51

On the planet Arrakis, a male ornithoid is flying toward his stationary mate at \(25.0 \mathrm{~m} / \mathrm{s}\) while singing at a frequency of \(1200 \mathrm{~Hz}\). If the female hears a tone of \(1240 \mathrm{~Hz}\), what is the speed of sound in the atmosphere of Arrakis?

Short Answer

Expert verified
The speed of sound in the atmosphere of Arrakis is 775 m/s.

Step by step solution

01

Understand the Doppler Effect Formula

The Doppler effect describes how the frequency of a wave changes for an observer moving relative to the source of the wave. The formula when the source is moving towards a stationary observer is \( f' = \frac{f \cdot (v + v_o)}{v - v_s} \), where \( f' \) is the observed frequency, \( f \) is the emitted frequency, \( v \) is the speed of sound in the medium, \( v_o \) is the speed of the observer, and \( v_s \) is the speed of the source.
02

Plug in Known Values and Simplify

In this problem, the observer (female ornithoid) is stationary, so \( v_o = 0 \). The speed of the source (male ornithoid), \( v_s \), is 25 m/s, the emitted frequency \( f \) is 1200 Hz, and the observed frequency \( f' \) is 1240 Hz. Substituting these values into the formula yields: \( 1240 = \frac{1200 \cdot v}{v - 25} \).
03

Solve for Speed of Sound \( v \)

Rearrange and solve the equation: \( 1240(v - 25) = 1200v \). Expand and simplify: \( 1240v - 31000 = 1200v \). Moving terms to one side gives: \( 40v = 31000 \). Solving for \( v \) gives: \( v = \frac{31000}{40} = 775 \).
04

Conclude the Speed of Sound

The final calculation indicates that the speed of sound in the atmosphere of Arrakis is \( 775 \) m/s.

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Key Concepts

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

Frequency Change
The concept of frequency change is key to understanding the Doppler effect. This phenomenon occurs when a sound source moves relative to an observer, leading to a perceived shift in frequency. It's similar to the change in pitch you hear when an ambulance passes by. The frequency increases as the source approaches, then decreases as it moves away.

For example, in the provided problem, the male ornithoid is moving towards his mate while singing. The frequency initially emitted by him is 1200 Hz. However, because he's moving towards the stationary female, she hears it as 1240 Hz. This indicates a higher perceived pitch due to the relative motion.

Understanding this concept is crucial for solving physics problems related to sound waves and relative motion. It's all about how movement can alter what we 'hear' from a sound source.
Wave Motion
Wave motion is a fundamental concept in physics that describes how waves travel through different mediums. In the context of sound waves, it refers to the oscillation and travel of sound compressions through the air or another medium.

When a source of sound like the male ornithoid moves, it impacts the wave motion. In this scenario, the sound emitted by the ornithoid travels as a longitudinal wave, compressing and expanding the medium (air on Arrakis) as it moves.

The motion of these waves explains why we experience changes in frequency. As the source moves towards the observer, sound waves are compressed, resulting in a higher frequency. Conversely, if the source moves away, the waves spread out, leading to a lower frequency. This behavior is characteristic of wave motion and is central to comprehending the Doppler effect.
Speed of Sound
The speed of sound is a measure of how fast sound waves travel through a medium. It's dependent on several factors, including the type of medium and its temperature.

In the problem, solving for the speed of sound in the atmosphere of Arrakis involved using the rearranged Doppler effect equation. The speed of sound, denoted as \(v\), influenced how the frequency changed for the stationary female observing her mate's song.

The calculated speed of sound on Arrakis was found to be 775 m/s, which is notably higher than Earth's typical speed of sound in air at 343 m/s. This difference highlights how planetary atmospheres can vary significantly, impacting how sound is experienced and measured.
Physics Problem Solving
Effective problem solving in physics often requires a methodical approach.

In scenarios like this Doppler effect problem:
  • First, comprehend the formula or principle at play. Understand how variables like speed of sound and frequency relate to each other.
  • Next, identify and plug in known values while isolating the unknown, here being the speed of sound \(v\).
  • Solve the equation step-by-step, simplifying where possible to reach the solution.
  • Finally, review and confirm your results to ensure logical consistency and accuracy.
These steps ensure a clear, logical pathway to solving physics challenges, making complex ideas more manageable.

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Most popular questions from this chapter

Transverse waves on a string have wave speed \(8.00 \mathrm{~m} / \mathrm{s}\), amplitude \(0.0700 \mathrm{~m},\) and wavelength \(0.320 \mathrm{~m}\). These waves travel in the \(x\) direction, and at \(t=0\) the \(x=0\) end of the string is at \(y=0\) and moving downward. (a) Find the frequency, period, and wave number \(k=2 \pi / \lambda\) of these waves. (b) Write the equation for \(y(x, t)\) describing these waves. (c) Find the transverse displacement of a point on the string at \(x=0.360 \mathrm{~m}\) at time \(t=0.150 \mathrm{~s}\)

Transverse waves are traveling on a long string that is under a tension of \(4.00 \mathrm{~N}\). The equation describing these waves is $$y(x, t)=(1.25 \mathrm{~cm}) \sin \left[\left(415 \mathrm{~s}^{-1}\right) t-\left(44.9 \mathrm{~m}^{-1}\right) x\right]$$ (a) Find the speed of the wave using the equation. (b) Find the lincar mass density of this string.

One of the \(63.5-\mathrm{cm}\) -long strings of an ordinary guitar is tuned to produce the note \(\mathrm{B}_{3}\) (frequency \(245 \mathrm{~Hz}\) ) when vibrating in its fundamental mode. (a) Find the speed of transverse waves on this string. (b) If the tension in this string is increased by \(1.0 \%,\) what will be the new fundamental frequency of the string? (c) If the speed of sound in the surrounding air is \(344 \mathrm{~m} / \mathrm{s},\) find the frequency and wavelength of the sound wave produced in the air by the vibration of the \(\mathrm{B}_{3}\) string. How do these compare to the frequency and wavelength of the standing wave on the string?

A trumpet player is tuning his instrument by playing an A note simultaneously with the first-chair trumpeter, who has perfect pitch. The first-chair player's note is exactly \(440 \mathrm{~Hz}\), and 2.8 beats per second are heard. What are the two possible frequencies of the other player's note?

The range of audible frequencies is from about \(20 \mathrm{~Hz}\) to \(20.000 \mathrm{~Hz}\). What is the range of the wavelengths of audible sound in air? (b) Visible light. The range of visible light extends from \(400 \mathrm{nm}\) to \(700 \mathrm{nm}\). What is the range of visible frequencies of light? (c) Brain surgery. Surgeons can remove brain tumors by using a cavitron ultrasonic surgical aspirator, which produces sound waves of frequency \(23 \mathrm{kHz}\). What is the wavelength of these waves in air? (d) Sound in the body. What would be the wavelength of the sound in part (c) in bodily fluids in which the speed of sound is \(1480 \mathrm{~m} / \mathrm{s},\) but the frequency is unchanged?
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