Problem 68 - College Physics

Question

A bat flies toward a wall, emitting a steady sound of frequency 2000 \(\mathrm{Hz}\) . The 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 \(\mathrm{Hz}\) ? (Hint: Break this problem into two parts, first with the bat as the source and the wall as the listener and then with the wall as the source and the bat as the listener.)

Step-by-Step Solution

Answer

The bat needs to fly at approximately 1.71 m/s.

Step 1: Understand the Doppler Effect Formulas

The Doppler Effect describes how the frequency of sound waves changes for an observer moving relative to the sound source. When the source is moving towards a stationary observer, the observed frequency \( f' \) is given by \( f' = f \frac{v + v_0}{v} \), where \( f \) is the emitted frequency, \( v \) is the speed of sound (343 m/s), and \( v_0 \) is the speed of the source. Conversely, when the source is stationary and the observer is moving towards it, \( f' = f \frac{v}{v - v_s} \).

Step 2: Part 1 - Bat as Source, Wall as Listener

Initially, consider the case where the bat is the source of sound and the wall is the stationary observer. Here, the bat emits sound at a frequency of 2000 Hz and is moving towards the wall. The frequency heard by the wall is \( f' = 2000 \cdot \frac{v + v_0}{v} \).

Step 3: Part 2 - Wall as Source, Bat as Listener

The sound reflects off the wall and is heard by the bat. Now, the wall acts as the source for the sound that has the frequency \( f' \) from Part 1. The bat is moving towards the stationary wall, so it hears a frequency \( f'' = f' \cdot \frac{v}{v - v_s} \).

Step 4: Setting Up the Beat Frequency Equation

The bat hears its original sound at 2000 Hz and the reflected sound at \( f'' \). The beat frequency \( f_b \) is the absolute difference between these two, \( f_b = |f'' - 2000| = 10 \), according to the problem statement.

Step 5: Solving for Bat's Velocity

Set the equation for beat frequency: \( |f'' - 2000| = 10 \). Rewrite the previous steps into equations, substitute into \( f'' = 2000 \cdot \frac{v + v_0}{v} \cdot \frac{v}{v - v_s} \), and isolate the bat's speed \( v_0 \). Simplifying yields the condition \( |2000 \cdot \frac{v + v_0}{v - v_s} - 2000| = 10 \). Solving for \( v_0 \), we find it to be approximately 1.71 m/s.

Key Concepts

Beat Frequency

Beat frequency is a fascinating phenomenon that happens when two sound waves of slightly different frequencies interfere with each other. This interference creates fluctuations in loudness, perceived as beats in the sound.
For example, if you hear two tones, one at 2000 Hz and another at 2005 Hz, you will hear 5 beats per second because the beat frequency is the absolute difference between the two frequencies:

• Beat frequency formula: \( f_b = |f_1 - f_2| \)

In the context of our exercise, the bat aims to hear a beat frequency of 10 Hz. This involves hearing its sound reflected off a wall, leading to a situation where it compares the frequency of its own emitted sound to that of the returning wave. When the conditions are right, such as when the motion of the bat is correct, the beat frequency helps determine the speed required to hear this particular frequency difference.

Sound Waves

Sound waves are mechanical waves that travel through mediums such as air, water, or solids. They are produced by vibrating objects and are perceived as sound when they reach our ears.
These waves have several key properties:

• Frequency: the number of cycles of the wave per second, measured in Hertz (Hz)
• Wavelength: the distance between successive crests of the wave
• Amplitude: related to the loudness of the sound

In our exercise, the sound waves are initially emitted by a bat. These waves travel to a wall, reflect back, and the bat hears them again. This process illustrates how sound waves can be used for echolocation, where bats determine the distance and speed of objects based on how the waves reflect back to them.

Frequency

Frequency is an essential concept in understanding sound waves, dictating the pitch of the sound perceived by listeners. The frequency, measured in Hertz (Hz), represents the number of cycles a wave completes in one second.
In the problem, the bat emits a sound at a frequency of 2000 Hz towards a wall. This frequency is central to calculating the beat frequency once the sound reflects off the wall.

• This involves using the Doppler Effect to establish how the frequency changes relative to the motion of the bat and the stationary wall.

Frequency adjustments occur due to the motion of the source (the bat) and the observer (the bat hearing the reflection), demonstrating how sound frequency varies with movement and resulting in the need to calculate the beat frequency.

Source and Observer Motion

The Doppler Effect is key to understanding how motion affects frequency. It deals with changes in the observed frequency when there is relative motion between the sound source and the observer.
For a source moving towards a stationary observer, the observed frequency increases. Conversely, if the observer moves towards the stationary sound source, the observed frequency also increases.

• Equations for these cases are:\( f' = f \left( \frac{v + v_0}{v} \right) \) when the source moves and\( f' = f \left( \frac{v}{v - v_s} \right) \) when the observer moves.

In the given problem, the bat serves as both the source and the observer at different points. Initially, it's a source moving towards the wall. After reflection, it acts as an observer moving towards the sound source (the wall). This dual role highlights how source and observer motion factors into changing the frequency perceived by the bat, crucial to solving the beat frequency and velocity in our problem.