/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 43 On the planet Arrakis a male orn... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

On the planet Arrakis a male ornithoid is flying toward his mate at 25.0 \(\mathrm{m} / \mathrm{s}\) while singing at a frequency of 1200 \(\mathrm{Hz}\) . If the stationary 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 on Arrakis is 775 m/s.

Step by step solution

01

Identify Known Values

First, identify the known values in the problem. The speed of the male ornithoid ( ) is 25.0 m/s (source moving towards stationary observer), the frequency emitted ( ) is 1200 Hz, and the frequency heard ( ) by the stationary female is 1240 Hz.
02

Use Doppler Effect Formula

The Doppler effect formula for a source moving towards a stationary observer is \( f' = f \times \frac{v + 0}{v - v_s} \), where \( f' \) is the detected frequency, \( f \) is the source frequency, \( v \) is the speed of sound, and \( v_s \) is the speed of the source.
03

Plug in Known Values

Substitute the values we know into the equation: \( 1240 = 1200 \times \frac{v}{v - 25 } \).
04

Solve for Speed of Sound

Rearrange the equation to solve for \( v \): \( 1240(v - 25) = 1200v \). Simplify to \( 1240v - 31000 = 1200v \). Then \( 40v = 31000 \), and solving for \( v \) gives \( v = \frac{31000}{40} = 775 \).
05

Final Answer

The speed of sound in the atmosphere of Arrakis is 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 91Ó°ÊÓ!

Key Concepts

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

Speed of Sound
The speed of sound is an important factor in understanding how sound waves travel through different mediums. On the planet described in the exercise, known as Arrakis, the speed of sound is key to solving the problem involving an ornithoid. The speed of sound varies depending on factors like the temperature, pressure, and medium through which it travels. In Earth's atmosphere, sound travels at approximately 343 meters per second at room temperature. However, on different planets or in unique environments, this rate can differ significantly. The exercise shows that the speed of sound on Arrakis is 775 meters per second. This faster speed could be due to a denser atmosphere or different atmospheric conditions compared to Earth. Understanding the speed of sound is crucial in physics problems that involve sound wave interactions and the Doppler Effect.
Frequency
Frequency is the number of waves that pass a given point per second, measured in Hertz (Hz). In the exercise, frequency is a central concept, with the male ornithoid emitting a sound at a frequency of 1200 Hz and the female hearing it at 1240 Hz due to the Doppler Effect. The difference in frequencies indicates that the sound waves are compressed as the source moves toward the observer. This is because the relative motion of the source affects how frequently the sound waves reach the observer.
  • Source frequency: The frequency of the sound emitted by the source, which in this case is 1200 Hz.
  • Observed frequency: The frequency at which the sound is heard by the observer, which is 1240 Hz in this scenario.
Frequency is an essential component in calculating and understanding wave behaviors and interactions in physics.
Physics Problem Solving
Physics problem-solving includes analyzing known values, understanding the concepts involved, and correctly applying formulas. Approaching the exercise on Arrakis involves this methodology.
  • Identify known values: Note down the values provided in the problem: the ornithoid's speed at 25.0 m/s, emitted frequency at 1200 Hz, and heard frequency at 1240 Hz.

  • Understanding the physics context: Comprehend how the Doppler Effect plays a part in observed frequency changes when the source approaches the observer. This understanding aids in employing the correct formula.

  • Application of correct formulas: Use the Doppler Effect formula to find the unknown, which in this problem is the speed of sound.
This structured approach is fundamental in physics problem-solving, facilitating a clear and logical path to finding the solution.
Equation Rearrangement
Rearranging equations is a critical skill in solving physics problems where variables need to be isolated. In the exercise, we start with the Doppler Effect equation: \[ f' = f \times \frac{v}{v - v_s} \]Given this formula, rearranging the terms to solve for the speed of sound "v" requires clear steps:
  • Substitute known values into the Doppler Effect equation.
  • Begin by multiplying each term to eliminate the fraction, giving: \[ 1240(v - 25) = 1200v \]
  • Simplify the equation through distribution: \[ 1240v - 31000 = 1200v \]
  • Combine like terms to facilitate solving for v:\[ 40v = 31000 \]
  • Finally, divide both sides by 40 to solve for "v": \[ v = \frac{31000}{40} = 775 \]
Rearranging equations allows the solution process to be methodical and accurate, which is vital in any physics calculations.

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

The sound source of a ship's sonar system operates at a frequency of 22.0 \(\mathrm{kHz}\) . The speed of sound in water (assumed to be at a uniform \(20^{\circ} \mathrm{C}\) ) is 1482 \(\mathrm{m} / \mathrm{s}\) . (a) What is the wavelength of the waves emitted by the source? (b) What is the difference in frequency between the directly radiated waves and the waves reflected from a whale traveling directly toward the ship at 4.95 \(\mathrm{m} / \mathrm{s}^{\prime \prime}\) The ship is at rest in the water.

A sound wave in air at \(20^{\circ} \mathrm{C}\) has a frequency of 150 \(\mathrm{Hz}\) and a displacement amplitude of \(5.00 \times 10^{-3} \mathrm{mm} .\) For this sound wave calculate the (a) pressure amplitude (in \(\mathrm{Pa} ) ;\) (b) intensity (in \(\mathrm{W} / \mathrm{m}^{2} ) ;(\mathrm{c})\) sound intensity level (in decibels).

cp Longitudinal Waves on a Spring. A long spring such as a Slinky"is 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 \(k^{\prime},\) the speed of longitudinal waves on the spring is \(v=L \sqrt{k^{\prime} / m}\) . (see Section 16.2\()\) (b) Evaluate \(v\) for a spring with \(m=0.250 \mathrm{kg}, L=2.00 \mathrm{m},\) and \(k^{\prime}=1.50 \mathrm{N} / \mathrm{m}\)

The intensity due to a number of independent sound sources is the sum of the individual intensities. (a) When four quadruplets cry simultaneously, how many decibels greater is the sound intensity level than when a single one cries? (b) To increase the sound intensity level again by the same number of decibels as in part (a), how many more crying babies are required?

You blow across the open mouth of an emply test tube and produce the fundamental standing wave of the air column inside the test tube. The speed of sound in air is 344 \(\mathrm{m} / \mathrm{s}\) and the test tube acts as a stopped pipe. (a) If the length of the air column in the test tube is \(14.0 \mathrm{cm},\) what is the frequency of this standing wave? (b) What is the frequency of the fundamental standing wave in the air column if the test tube is half filled with water?

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.