Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

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.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

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.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

While sitting in your car by the side of a country road, you are approached by your friend, who happens to be in an identical car. You blow your car's horn, which has a frequency of 260 Hz. Your friend blows his car's horn, which is identical to yours, and you hear a beat frequency of 6.0 Hz. How fast is your friend approaching you?

A car alarm is emitting sound waves of frequency 520 Hz. You are on a motorcycle, traveling directly away from the parked car. How fast must you be traveling if you detect a frequency of 490 Hz?

A violinist is tuning her instrument to concert A (440 Hz). She plays the note while listening to an electronically generated tone of exactly that frequency and hears a beat frequency of 3 Hz, which increases to 4 Hz when she tightens her violin string slightly. (a) What was the frequency of the note played by her violin when she heard the 3-Hz beats? (b) To get her violin perfectly tuned to concert A, should she tighten or loosen her string from what it was when she heard the 3-Hz beats?

A swimming duck paddles the water with its feet once every 1.6 s, producing surface waves with this period. The duck is moving at constant speed in a pond where the speed of surface waves is 0.32 m/s, and the crests of the waves ahead of the duck are spaced 0.12 m apart. (a) What is the duck's speed? (b) How far apart are the crests behind the duck?

(a) By what factor must the sound intensity be increased to raise the sound intensity level by 13.0 dB? (b) Explain why you don't need to know the original sound intensity.

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free