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A bat flies toward a wall, emitting a steady sound of frequency 1.70 \(\mathrm{kHz}\) . This bat hears its own sound plus the sound reflected by the wall. How fast should the bat fly in order to hear a beat frequency of 10.0 \(\mathrm{Hz}\) ?

Short Answer

Expert verified
The bat should fly at approximately 2.91 m/s to achieve a beat frequency of 10 Hz.

Step by step solution

01

Understand the Problem

This problem involves the Doppler effect and beat frequency. The bat emits a sound and hears the reflected sound from the wall. We are tasked to find the speed of the bat to create a beat frequency of 10 Hz.
02

Identify Given Values

We are given the frequency of the bat's emitted sound, \( f_s = 1.70 \: \text{kHz} = 1700 \: \text{Hz} \), and the beat frequency the bat hears, \( f_b = 10 \: \text{Hz} \).
03

Apply Doppler Effect Formula

The frequency of sound the bat hears due to the reflection is affected by the Doppler effect. When the bat moves towards the wall, the frequency of the sound the bat hears (after reflection) is: \( f_r = \frac{f_s (v + v_b)}{v - v_b} \), where \( v = 343 \: \text{m/s} \) is the speed of sound and \( v_b \) is the bat's speed.
04

Introduce Beat Frequency Concept

Beat frequency \( f_b \) is the difference between the emitted frequency \( f_s \) and the frequency after reflection \( f_r \): \( f_b = | f_r - f_s | = 10 \: \text{Hz} \).
05

Solve for Bat's Speed

Substitute \( f_r \) into the beat frequency equation: \( | \frac{f_s (v + v_b)}{v - v_b} - f_s | = 10 \). Simplify and solve this equation for \( v_b \). We derive:\[ \frac{v_b}{v - v_b} \times 1700 = 10 \]By solving this equation, we find \( v_b \approx 2.91 \: \text{m/s} \).

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Key Concepts

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

Beat Frequency
Beat frequency occurs when two sounds of slightly different frequencies interfere with one another, creating a pulsating effect that is heard as a "beating" sound. Think of it as two musical notes played together, but with frequencies so close that instead of hearing both as distinct sounds, you hear a single tone that rises and falls in volume.

The beat frequency (\( f_b \)) is simply the absolute difference between the two frequencies involved. In this exercise, the bat emits a sound with a frequency of 1700 Hz and hears the reflected sound off the wall. When calculating to produce a beat frequency of 10 Hz, it's about determining how close the original sound and the echoed sound's frequencies need to be.
  • If two frequencies are very close, the beating effect is slow (a low beat frequency).
  • If they are more apart, the beating effect happens faster (a high beat frequency).
Understanding beat frequency helps us solve how fast the bat must fly in order to hear the desired 10 Hz beat when hearing both the original and reflected sounds.
Sound Reflection
Sound reflection is akin to echo, where sound waves bounce off a surface and return to the source or listener. When the bat's sound hits the wall, part of the wave reflects back, and this is the sound the bat hears in addition to its own emitted sound.

This reflection is governed by simple principles:
  • Reflected sound maintains the original frequency, assuming the reflecting surface is stationary.
  • The intensity of reflected sound depends on the surface type; smooth and hard surfaces reflect sound better than rough or absorbent ones.
In our exercise, the key is understanding how the Doppler effect will alter this reflected sound, changing its frequency as perceived by the bat due to its motion towards the wall. This change in frequency due to sound reflection is crucial for calculating the ultimate beat frequency heard by the bat.
Frequency Calculation
The core of tackling this problem lies in frequency calculation, specifically understanding how frequencies change due to motion, a phenomenon explained by the Doppler effect.

For sound waves, the Doppler effect affects the observed frequency based on the relative motion of the source and observer:
  • Approaching the source increases the observed frequency.
  • Moving away decreases it.
In the exercise, the bat moving towards the wall means that the frequency of the sound it hears (reflected) increases because of the Doppler effect.

The calculation involves substituting the known values into the Doppler effect formula. For this problem, use:\[ f_r = \frac{f_s (v + v_b)}{v - v_b} \]This formula computes the frequency of the reflected sound the bat perceives, influencing the beat frequency. After setting the known beat frequency into context, you solve for the bat's speed.
Speed of Sound
Speed of sound in air is typically around 343 m/s, though it can vary with atmospheric conditions. It’s a crucial parameter in calculations involving sound travel, used in this exercise while applying the Doppler effect formula.

The speed of sound is affected by:
  • Temperature: Sound travels faster in warmer air.
  • Medium: Sound speeds differently in gases, liquids, and solids.
For solving our problem, we assume a constant speed of 343 m/s for simplicity. This speed forms part of the equation calculating the perceived frequency change due to the bat's motion. The consistent use of this speed keeps our Doppler effect calculations accurate, ensuring the derived bat speed of approximately 2.91 m/s is correct for the conditions.

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Most popular questions from this chapter

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

Two organ pipes, open at one end but closed at the other, are each 1.14 m long. One is now lengthened by 2.00 \(\mathrm{cm} .\) Find the frequency of the beat they produce when playing together in their fundamental.

A long tube contains air at a pressure of 1.00 atm and a temperature of \(77.0^{\circ} \mathrm{C}\) . The tube is open at one end and closed at the other by a movable piston. A tuning fork near the open end is vibrating with a frequency of 500 \(\mathrm{Hz}\) . Resonance is produced when the piston is at distances \(18.0,55.5,\) and 93.0 \(\mathrm{cm}\) from the open end. (a) From these measurements, what is the speed of sound in air at \(77.0^{\circ} \mathrm{C} ?\) (b) From the result of part (a), what is the value of \(\gamma ?\) (c) These data show that a displacement antinode is slightly outside of the open end of the tube. How far outside is it?

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?

BIO The Human Voice. The human vocal tract is a pipe that extends about 17 \(\mathrm{cm}\) from the lips to the vocal folds (also called "vocal cords") near the middle of your throat. The vocal folds behave rather like the reed of a clarinet, and the vocal tract acts like a stopped pipe. Estimate the first three standing-wave frequencies of the vocal tract. Use \(v=344 \mathrm{m} / \mathrm{s} .\) The answers are only an estimate, since the position of lips and tongue affects the motion of air in the vocal tract.)

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