/*! 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 92 You are flying in an ultralight ... [FREE SOLUTION] | 91影视

91影视

You are flying in an ultralight aircraft at a speed of 39 m/s. An eagle, whose speed is 18 m/s, is flying directly toward you. Each of the given speeds is relative to the ground. The eagle emits a shrill cry whose frequency is 3400 Hz. The speed of sound is 330 m/s. What frequency do you hear?

Short Answer

Expert verified
The frequency heard by the pilot is approximately 4021 Hz.

Step by step solution

01

Understand the Problem

We need to determine the frequency of the eagle's cry that the pilot hears, which involves considering the Doppler effect caused by the relative motion between the eagle and the pilot.
02

Identify Given Values

We are given the following: - Speed of the aircraft (v_observer) = 39 m/s - Speed of the eagle (v_source) = 18 m/s - Frequency of the eagle's cry (f_source) = 3400 Hz - Speed of sound (v_sound) = 330 m/s.
03

Apply the Doppler Effect Formula

The Doppler effect formula for frequency observed (f_observed) is given by \[f_{observed} = f_{source} \left( \frac{v_{sound} + v_{observer}}{v_{sound} - v_{source}} \right)\]Substitute the known values into this equation to find what frequency the pilot hears.
04

Substitute Known Values into Formula

Insert the provided values into the formula:\[f_{observed} = 3400 \left( \frac{330 + 39}{330 - 18} \right)\]
05

Solve the Equation

First, add the speeds in the numerator:\[330 + 39 = 369\]Then, subtract the speeds in the denominator:\[330 - 18 = 312\]Now calculate the observed frequency:\[f_{observed} = 3400 \times \left( \frac{369}{312} \right)\]
06

Calculate Observed Frequency

Compute the fraction first:\[\frac{369}{312} \approx 1.1827\]Then multiply by the source frequency:\[f_{observed} = 3400 \times 1.1827 \approx 4021 \text{ Hz}\]
07

Conclude the Solution

Therefore, the frequency heard by the pilot is approximately 4021 Hz.

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 91影视!

Key Concepts

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

Frequency Calculation
Understanding how to calculate frequency is crucial when dealing with the Doppler Effect. It's a phenomenon observed when the source and the observer are moving relative to each other. In this exercise, we are dealing with sound frequency that changes due to this relative motion.

The Doppler Effect provides us with a formula to calculate the observed frequency, taking into account the speeds of both the source and the observer relative to the medium through which the sound travels (often air, where sound speed is given). Here's the key formula we use:\[f_{observed} = f_{source} \left( \frac{v_{sound} + v_{observer}}{v_{sound} - v_{source}} \right)\]
  • \(f_{observed}\) is the frequency heard by the observer.
  • \(f_{source}\) is the original frequency of the sound, which in this case is the eagle's cry at 3400 Hz.
  • \(v_{sound}\) is the speed of sound in the medium, here it's 330 m/s.
  • \(v_{observer}\) and \(v_{source}\) are the speeds of the observer and the source relative to the ground.
By substituting the given values, we determine the frequency that the observer, or pilot, hears. Calculating this frequency helps to understand how sound waves behave when there is motion involved.
Relative Motion
Relative motion is the movement of one object with respect to another. It's an essential concept in physics, especially when analyzing situations involving the Doppler Effect.

In our situation, the observer (pilot in the ultralight aircraft) and the source (eagle) are moving towards each other relative to the ground. This relative motion affects how sound waves are perceived.

The speeds of both the observer and the source must be considered as they move in opposite directions towards one another. These speeds directly affect the frequency of the sound wave that the observer hears.
  • The observer's speed, in this case, is 39 m/s.
  • The source's speed is 18 m/s.
This closing speed is crucial for applying the Doppler Effect formula correctly; it accounts for the increase in perceived frequency because the two are moving closer.

Understanding this relative motion allows us to apply the correct mathematical adjustments in the formula, leading to an accurate calculation of the observed frequency.
Sound Waves
Sound waves are fundamental to understanding the Doppler Effect. They are longitudinal waves that travel through various media and are affected by factors such as speed of the medium and objects within it.

The speed of sound in air is a common reference, typically about 330 m/s under normal conditions, and it plays a key role in determining how sound is propagated when either the source or the observer is moving.
  • Sound waves consist of compressions and rarefactions.
  • They are subject to changes in frequency and wavelength based on relative motion.
When the source or observer moves, the frequency and wavelength adjust in such a way that the observer hears a different frequency from the source frequency. This change is precisely what we calculate using the Doppler Effect formula.

Sound waves are an excellent example of how wave properties like frequency and wavelength can be affected by motion. Learning how these waves behave helps in understanding and predicting phenomena like the sonic boom or the varying pitch of a passing siren.

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

A man stands at the midpoint between two speakers that are broadcasting an amplified static hiss uniformly in all directions. The speakers are 30.0 m apart and the total power of the sound coming from each speaker is 0.500 W. Find the total sound intensity that the man hears (a) when he is at his initial position halfway between the speakers, and (b) after he has walked 4.0 m directly toward one of the speakers.

A convertible moves toward you and then passes you; all the while, its loudspeakers are producing a sound. The speed of the car is a constant 9.00 m/s, and the speed of sound is 343 m/s. What is the ratio of the frequency you hear while the car is approaching to the frequency you hear while the car is moving away?

A wire is stretched between two posts. Another wire is stretched between two posts that are twice as far apart. The tension in the wires is he same, and they have the same mass. A transverse wave travels on the shorter wire with a speed of 240 m/s. What would be the speed of the wave on the longer wire?

The siren on an ambulance is emitting a sound whose frequency is 2450 Hz. The speed of sound is 343 m/s. (a) If the ambulance is stationary and you (the 鈥渙bserver鈥) are sitting in a parked car, what are the wavelength and the frequency of the sound you hear? (b) Suppose that the ambulance is moving toward you at a speed of 26.8 m/s. Determine the wavelength and the frequency of the sound you hear. (c) If the ambulance is moving toward you at a speed of 26.8 m/s and you are moving toward it at a speed of 14.0 m/s, find the wavelength and frequency of the sound you hear.

An observer stands 25 m behind a marksman practicing at a rifle range. The marksman fires the rifle horizontally, the speed of the bullets is \(840 \mathrm{m} / \mathrm{s},\) and the air temperature is \(20^{\circ} \mathrm{C}\) . How far does each bullet travel before the observer hears the report of the rifle? Assume that the bullets encounter no obstacles during this interval, and ignore both air resistance and the vertical component of the bullets鈥 motion.

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.