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A bird is flying directly toward a stationary bird-watcher and emits a frequency of \(1250 \mathrm{Hz}\). The bird-watcher, however, hears a frequency of \(1290 \mathrm{Hz} .\) What is the speed of the bird, expressed as a percentage of the speed of sound?

Short Answer

Expert verified
The bird's speed is approximately 3.11% of the speed of sound.

Step by step solution

01

Understand the Doppler Effect Formula

The Doppler Effect formula for sound when the source is moving towards the observer is expressed as \( f' = \frac{f}{1 - \frac{v_s}{v}} \), where \( f' \) is the observed frequency (1290 Hz), \( f \) is the emitted frequency (1250 Hz), \( v_s \) is the speed of the source (bird), and \( v \) is the speed of sound in air (approximately 343 m/s).
02

Rearrange the Formula

To find the speed of the source, \( v_s \), rearrange the formula: \( v_s = v \left( 1 - \frac{f}{f'} \right) \). Plug in the values for \( f \) and \( f' \) to solve for \( v_s \).
03

Plug in Known Values

Substitute the known values into the rearranged formula: \( v_s = 343 \left( 1 - \frac{1250}{1290} \right) \). Evaluate the fraction \( \frac{1250}{1290} \) and subtract from 1.
04

Solve for the Speed of the Bird

Calculate the expression \( v_s = 343 \left( 1 - \frac{1250}{1290} \right) \). First find the fraction \( \frac{1250}{1290} \approx 0.968992 \). Then compute the expression: \( 1 - 0.968992 = 0.031008 \). Multiply by 343 m/s: \( v_s \approx 343 \times 0.031008 = 10.64774 \) m/s.
05

Express Bird's Speed as a Percentage of Speed of Sound

To express the bird's speed as a percentage of the speed of sound, use the formula \( \frac{v_s}{v} \times 100\% \). Calculate: \( \frac{10.64774}{343} \times 100\% \approx 3.106\%. \)

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Key Concepts

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

Frequency
Frequency is a fundamental concept in physics, particularly in the study of waves and sound. In simple terms, frequency refers to the number of oscillations or cycles a wave completes in one second. It is measured in Hertz (Hz).
For sound, frequency determines the pitch of the sound we hear - higher frequency means a higher pitch, and lower frequency means a lower pitch.
In the provided exercise, the bird emits a sound at a frequency of 1250 Hz. However, due to the Doppler Effect, the bird-watcher perceives it as 1290 Hz. This change is because the bird is moving towards the observer.
To understand frequency changes better, think of how a sound might seem higher or lower depending on the movement of the source emitting it. Also, remember that frequency changes can provide useful information in many real-world applications, from radar to medical imaging.
Speed of Sound
The speed of sound is the speed at which sound waves travel through a medium. In air, at room temperature, it is approximately 343 meters per second (m/s).
The speed can vary depending on factors like temperature and air density, but for most calculations, using 343 m/s is accurate enough.
Understanding the speed of sound is critical in applying the Doppler Effect formula correctly. This speed acts as the baseline for calculating how the observed frequency changes relative to the speed of the source.
  • If the source moves towards the observer, the observed frequency increases.
  • If the source moves away, the observed frequency decreases.
In the exercise, this constant speed of sound helps us find the bird's speed as it flies towards the bird-watcher.
Observed Frequency
The observed frequency is what the listener actually hears, which can differ from the emitted frequency due to motion between the source and the observer.
In the context of the Doppler Effect, the observed frequency changes when there is relative motion. For the exercise, the bird-watcher hears an observed frequency of 1290 Hz, which is higher than the emitted frequency of the bird because the bird approaches the bird-watcher.
The phenomenon can be explained with the formula: \[ f' = \frac{f}{1 - \frac{v_s}{v}} \]where:
  • \( f' \) is the observed frequency (1290 Hz)
  • \( f \) is the emitted frequency (1250 Hz)
  • \( v_s \) is the speed of the source (the bird).
  • \( v \) is the speed of sound (343 m/s)
Using this understanding helps us rearrange the formula to find the speed of the bird, given changes in observed frequency, showcasing how crucial these principles are in solving real-world challenges.

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Most popular questions from this chapter

Suppose that a public address system emits sound uniformly in all directions and that there are no reflections. The intensity at a location \(22 \mathrm{m}\) away from the sound source is \(3.0 \times 10^{-4} \mathrm{W} / \mathrm{m}^{2} .\) What is the intensity at a spot that is \(78 \mathrm{m}\) away?

A Mysterious Underwater Object. You and your team are on a reconnaissance mission in a submarine exploring a mysterious object in the cold waters of the Weddell Sea, off the coast of Antarctica. The sonar indicates that the object, which had otherwise been moving erratically, has changed course and is now on a direct collision course with your sub. The captain issues an "all stop" order, bringing the sub to a halt relative to the water. The sonar operator "pings" the object, which amounts to sending a short blast of sound in the direction of the object. The emitted sound wave has a frequency of \(1550 \mathrm{Hz}\) and a speed of \(1552 \mathrm{m} / \mathrm{s}\) (the speed of sound in seawater). The sound reflects from the object and returns \(2.582 \mathrm{s}\) after it was emitted from your sub, and its frequency has shifted to \(1598 \mathrm{Hz}\). (a) How far from the sub was the object when the sound reflected from it? (b) What is the object's speed? (c) How long after you receive the return signal will it take the object to reach your submarine?

A car driving along a highway at a speed of \(23 \mathrm{m} / \mathrm{s}\) strays onto the shoulder. Evenly spaced parallel grooves called "rumble strips" are carved into the pavement of the shoulder. Rolling over the rumble strips causes the car's wheels to oscillate up and down at a frequency of 82 Hz. How far apart are the centers of adjacent rumble-strip grooves?

A wave traveling in the \(+x\) direction has an amplitude of \(0.35 \mathrm{m}\), a speed of \(5.2 \mathrm{m} / \mathrm{s},\) and a frequency of \(14 \mathrm{Hz}\). Write the equation of the wave in the form given by either Equation 16.3 or 16.4

The volume control on a surround-sound amplifier is adjusted so the sound intensity level at the listening position increases from 23 to 61 dB. What is the ratio of the final sound intensity to the original sound intensity?

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