GSO000 Problem 2
Problem Statement
Beats from a Chasing Sound Source and a Fleeing Observer
Problem Statement
In a windless, one-dimensional space, there is a sufficiently large stationary reflecting wall on the right. Along this line, a sound source emitting sound at frequency f0 in all directions is moving toward the wall (rightward) at speed Vs. Between the sound source and the wall, there is an observer moving in the same direction (rightward) at speed Vo.
The observer simultaneously receives the sound wave coming directly from the source and the sound wave reflected from the wall (arriving from the front). Find the number of beats per unit time (beat frequency) Δf observed by the observer. Let the speed of sound be V, and assume V>Vs>Vo>0.
Constraints
- Speed of sound V=340 m/s
- Speed of sound source Vs=20 m/s
- Speed of observer Vo=5 m/s
- Frequency of sound source f0=640 Hz
Input Format
Find the beat frequency Δf in hertz (Hz) and give the answer as a positive integer.
Solution
In this problem, we calculate the Doppler effect due to the motion of the sound source and observer. Let the rightward direction be the positive direction.
1. Frequency of Direct Sound f1
The sound source moves at velocity +Vs and the observer moves at velocity +Vo. The direct sound travels rightward (velocity +V) from the source to the observer.
By the Doppler effect formula, the frequency f1 of the direct sound heard by the observer is:
f1=V−VsV−Vof02. Frequency of Reflected Sound f2
First, find the frequency fw received by the stationary wall. The sound wave travels rightward and the wall's velocity is zero.
fw=V−VsV−0f0=V−VsVf0Next, the wall acts as a new sound source (at frequency fw), reflecting the sound wave to the left. The reflected sound velocity is −V. The observer moves rightward at velocity +Vo, moving toward the reflected source (the wall).
f2=VV+Vofw=VV+Vo(V−VsVf0)=V−VsV+Vof03. Beat Frequency Δf
The beat frequency is given by the absolute value of the difference between the two frequencies.
Δf=∣f2−f1∣=V−VsV+Vof0−V−VsV−Vof0=V−Vs2Vof0Substituting numerical values:
Δf=340−202×5×640=32010×640=321×640=20 HzAnswer: 20