/*! 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 60 A bat, moving at \(5.00 \mathrm{... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

A bat, moving at \(5.00 \mathrm{m} / \mathrm{s},\) is chasing a flying insect (Fig. \(P 17.7) .\) If the bat emits a \(40.0 \mathrm{kHz}\) chirp and receives back an echo at \(40.4 \mathrm{kHz},\) at what speed is the insect moving toward or away from the bat? (Take the speed of sound in air to be \(v=340 \mathrm{m} / \mathrm{s} .\))

Short Answer

Expert verified
The speed at which the insect is moving towards or away from the bat is 3.4 m/s.

Step by step solution

01

Identify the given values

The frequency of the sound from the bat \(f\) is \(40.0 kHz\). The frequency of the echo \(f'\) is \(40.4 kHz\). The speed of sound in air \(v\) is \(340 m/s\). The goal is to find \(v_0\), the velocity of the insect.
02

Rearrange the equation to solve for \(v_0\)

First, rearrange Doppler Effect frequency change equation for sound towards a moving source to determine the speed of the insect. So the equation becomes \(v_0 = (v \times f' / f) - v\).
03

Insert the given values into the equation

Substitute the given values into the rearranged equation: \(v_0 = ((340 m/s \times 40.4 kHz / 40.0 kHz) - 340 m/s)\).
04

Solve the equation

Perform the above operations: \(v_0 = (340 m/s \times 1.01) - 340 m/s = 3.4 m/s\). So, the insect is moving towards the bat at a velocity of 3.4 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.

Sound Wave Frequency
When we talk about sound, frequency is one of the most important elements. It refers to how many sound waves pass a given point in one second. Frequency is measured in hertz (Hz). The higher the frequency, the higher the pitch of the sound. Bats, for example, use high-frequency sounds such as a 40.0 kHz chirp for echolocation. This means that the bat produces a sound with a frequency of 40,000 cycles per second. Understanding frequency is crucial for working with problems involving the Doppler Effect. The Doppler Effect explains how the frequency of a sound changes when the source of the sound moves relative to the observer. In our exercise, the received echo frequency of 40.4 kHz is slightly higher than the emitted 40.0 kHz, implying that the source (the insect) is moving in relation to the bat. The change in frequency helps determine the insect's speed.
Velocity Calculation
Calculating the velocity in this context involves using the Doppler Effect formula, which ties together the speed of sound, the emitted frequency, and the received frequency to find the velocity of the moving object - in this case, the insect. The formula used is:\[ v_0 = \left(\frac{v \times f'}{f}\right) - v \] Here, \( v \) is the speed of sound in air (340 m/s according to the problem), \( f \) is the frequency of the sound emitted by the bat (40.0 kHz), and \( f' \) is the frequency of the echo (40.4 kHz). By substituting these values into the equation, we can solve for \( v_0 \), the velocity of the insect. This calculation shows a fundamental aspect of the Doppler Effect: Not only does it help understand frequency changes but it also provides a means to calculate movement speeds, even when one of the objects involved is as fast and agile as a flying insect.
Sound Propagation in Air
Sound travels through air using waves that cause particles in the air to vibrate. The speed at which sound travels depends on several factors, including the medium it's traveling through and the temperature of that medium. Under normal conditions, the speed of sound in air is about 340 m/s. However, this speed can vary somewhat with changes in atmospheric conditions. Sound propagation involves the transmission of pressure waves from the source to the observer. In the case of echolocation, like with bats, sound propagates from the bat towards an object (the insect) and then back as an echo. Knowing the speed of sound is essential for solving problems related to time and distance calculations because it defines how long it takes for the sound to reach different points. By understanding how sound travels in air, we can make use of the echo's frequency shift, thanks to the Doppler Effect, to precisely determine the velocity of moving objects like the insect in our exercise.

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

This problem represents a possible (but not recommended) way to code instantaneous pressures in a sound wave into 16 -bit digital words. Example 17.2 mentions that the pressure amplitude of a \(120-\mathrm{dB}\) sound is \(28.7 \mathrm{N} / \mathrm{m}^{2}\) Let this pressure variation be represented by the digital -code \(65536 .\) Let zero pressure variation be represented on the recording by the digital word \(0 .\) Let other intermediate pressures be represented by digital words of intermediate -size, in direct proportion to the pressure. (a) What digital word would represent the maximum pressure in a \(40 \mathrm{dB}\) -sound? (b) Explain why this scheme works poorly for soft -sounds. (c) Explain how this coding scheme would clip off half of the waveform of any sound, ignoring the actual -shape of the wave and turning it into a string of zeros. By introducing sharp corners into every recorded waveform, this coding scheme would make everything sound like a buzzer or a kazoo.

A train is moving parallel to a highway with a constant speed of \(20.0 \mathrm{m} / \mathrm{s} .\) A car is traveling in the same direction as the train with a speed of \(40.0 \mathrm{m} / \mathrm{s} .\) The car horn sounds at a frequency of \(510 \mathrm{Hz},\) and the train whistle sounds at a frequency of \(320 \mathrm{Hz}\). (a) When the car is behind the train, what frequency does an occupant of the car observe for the train whistle? (b) After the car passes and is in front of the train, what frequency does a train passenger observe for the car horn?

On a Saturday morning, pickup trucks and sport utility vehicles carrying garbage to the town dump form a nearly steady procession on a country road, all traveling at \(19.7 \mathrm{m} / \mathrm{s} .\) From one direction, two trucks arrive at the dump every 3 min. A bicyclist is also traveling toward the dump, at \(4.47 \mathrm{m} / \mathrm{s}\). (a) With what frequency do the trucks pass him? (b) What If? A hill does not slow down the trucks, but makes the out-of-shape cyclist's speed drop to \(1.56 \mathrm{m} / \mathrm{s} .\) How often do noisy, smelly, inefficient, garbage-dripping, roadhogging trucks whiz past him now?

A train whistle \((f=400 \mathrm{Hz})\) sounds higher or lower in frequency depending on whether it approaches or recedes. (a) Prove that the difference in frequency between the approaching and receding train whistle is $$\Delta f=\frac{2 u / v}{1-u^{2} / v^{2}} f$$ where \(u\) is the speed of the train and \(v\) is the speed of sound. (b) Calculate this difference for a train moving at a speed of \(130 \mathrm{km} / \mathrm{h}\). Take the speed of sound in air to be \(340 \mathrm{m} / \mathrm{s}\)

The intensity of a sound wave at a fixed distance from a speaker vibrating at \(1.00 \mathrm{kHz}\) is \(0.600 \mathrm{W} / \mathrm{m}^{2} .\) (a) Determine the intensity if the frequency is increased to \(2.50 \mathrm{kHz}\) while a constant displacement amplitude is maintained. (b) Calculate the intensity if the frequency is reduced to \(0.500 \mathrm{kHz}\) and the displacement amplitude is doubled.

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.