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On the planet Arrakis a male ornithoid is flying toward his mate at 25.0 m/s while singing at a frequency of 1200 Hz. If the stationary female hears a tone of 1240 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 formula for observed frequency when the source is moving towards a stationary observer is given by \( f' = \frac{f \cdot (v + v_o)}{v - v_s} \), where \( f' \) is the observed frequency, \( f \) is the source frequency, \( v \) is the speed of sound, \( v_o \) is the velocity of the observer (0 m/s in this case as the observer is stationary), and \( v_s \) is the velocity of the source (25 m/s here).
02

Substitute Known Values

Substitute the given values into the formula: the observed frequency \( f' = 1240 \) Hz, the source frequency \( f = 1200 \) Hz, and the velocity of the source \( v_s = 25 \) m/s. We will solve for \( v \), the speed of sound. The equation becomes \( 1240 = \frac{1200 \cdot v}{v - 25} \).
03

Solve for Speed of Sound

Rearrange the equation to solve for \( v \). Multiply both sides by \( v - 25 \) to eliminate the fraction: \( 1240(v - 25) = 1200v \). This simplifies to \( 1240v - 31000 = 1200v \). Simplifying further, we get \( 40v = 31000 \). Thus, \( v = \frac{31000}{40} = 775 \) m/s.

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Key Concepts

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

Observed Frequency
The observed frequency is the frequency of a wave as perceived by an observer. In the Doppler Effect scenario, this frequency changes depending on the motion of the source or the observer. Therefore, if the source is moving towards a stationary observer, the observed frequency will be higher than the source frequency.

In the exercise from Arrakis, the female ornithoid is stationary, and she perceives the frequency as 1240 Hz, which is higher than the source frequency of 1200 Hz. This increase demonstrates the principle that the frequency increases when the source approaches the observer. Understanding this principle helps in using the Doppler Effect effectively to calculate other parameters, like the speed of sound.
Source Frequency
The source frequency is the original frequency emitted by the source of a wave, without any alterations. In the Doppler Effect, this is the baseline frequency before any movement-related changes are accounted for.

In our scenario, the male ornithoid is producing a source frequency of 1200 Hz. This frequency is crucial to calculating the observed frequency once you include all parameters such as source speed and the speed of sound. It serves as one of the known variables to resolve the equation involved in analyzing the Doppler Effect.
  • Basic parameter needing no adjustments unless under influence.
  • Starting point for understanding Doppler Effect in motion scenarios.
Speed of Sound
The speed of sound is critical in understanding how frequencies shift with motion in the Doppler Effect. It refers to how fast sound waves propagate through a medium, impacting how quickly differences in frequency can be perceived.

In the exercise, we calculated the speed of sound on the planet Arrakis by rearranging and solving the Doppler Effect formula. By substituting the known observed frequency (1240 Hz) and source frequency (1200 Hz) with the source speed (25 m/s), we were able to determine that the speed of sound is 775 m/s. This step underlines the interdependence of frequency perceptions and sound speed in the medium where the sound travels.
Stationary Observer
A stationary observer does not move relative to the medium through which sound is propagating. In the Doppler Effect, this means any frequency change observed is solely due to the motion of the source.

The stationary observer in our problem is the female ornithoid. Her lack of motion simplifies the Doppler Effect formula, as the observer's velocity term ( $v_o)$ is zero. This stationary condition is essential for calculating the exact shift in observed frequency without additional variables. Understanding the role of a stationary observer is key to correctly applying the Doppler Effect to determine changes in perceived sound frequency.

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Most popular questions from this chapter

A loud factory machine produces sound having a displacement amplitude of 1.00 \(\mu\)m, but the frequency of this sound can be adjusted. In order to prevent ear damage to the workers, the maximum pressure amplitude of the sound waves is limited to 10.0 Pa. Under the conditions of this factory, the bulk modulus of air is 1.42 \(\times\) 10\(^5\) Pa. What is the highest-frequency sound to which this machine can be adjusted without exceeding the prescribed limit? Is this frequency audible to the workers?

An organ pipe has two successive harmonics with frequencies 1372 and 1764 Hz. (a) Is this an open or a stopped pipe? Explain. (b) What two harmonics are these? (c) What is the length of the pipe?

Two guitarists attempt to play the same note of wavelength 64.8 cm at the same time, but one of the instruments is slightly out of tune and plays a note of wavelength 65.2 cm instead. What is the frequency of the beats these musicians hear when they play together?

\(\textbf{Longitudinal Waves on a Spring.}\) A long spring such as a Slinky\(^{\mathrm{TM}}\) is often used to demonstrate longitudinal waves. (a) Show that if a spring that obeys Hooke’s law has mass \(m\), length \(L\), and force constant \(k'\), the speed of longitudinal waves on the spring is \(v = L\sqrt{ k'/m}\) (see Section 16.2). (b) Evaluate \(v\) for a spring with \(m =\) 0.250 kg, \(L =\) 2.00 m, and \(k' =\) 1.50 N\(/\)m. \(\textbf{ULTRASOUND IMAGING}\). A typical ultrasound transducer used for medical diagnosis produces a beam of ultrasound with a frequency of 1.0 MHz. The beam travels from the transducer through tissue and partially reflects when it encounters different structures in the tissue. The same transducer that produces the ultrasound also detects the reflections. The transducer emits a short pulse of ultrasound and waits to receive the reflected echoes before emitting the next pulse. By measuring the time between the initial pulse and the arrival of the reflected signal, we can use the speed of ultrasound in tissue, 1540 m/s, to determine the distance from the transducer to the structure that produced the reflection. As the ultrasound beam passes through tissue, the beam is attenuated through absorption. Thus deeper structures return weaker echoes. A typical attenuation in tissue is \(-\)100 dB/m \(\cdot\) MHz; in bone it is \(-\)500 dB/m \(\cdot\) MHz. In determining attenuation, we take the reference intensity to be the intensity produced by the transducer.

The shock-wave cone created by a space shuttle at one instant during its reentry into the atmosphere makes an angle of 58.0\(^\circ\) with its direction of motion. The speed of sound at this altitude is 331 m/s. (a) What is the Mach number of the shuttle at this instant, and (b) how fast (in m/s and in mi/h) is it traveling relative to the atmosphere? (c) What would be its Mach number and the angle of its shock-wave cone if it flew at the same speed but at low altitude where the speed of sound is 344 m/s ?

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