Doppler Effect and Beat

Imagine a person in the middle of a plaza, in front of a church. The person starts to walk away from the Church in a straight line, when the bell inside the church begins to chime. Suppose we know the frequency of the bell ringing $f_b$ and the person's velocity $v_p$.

I'm curious about the beat that will occur. When the person is walking away, it percieves a frequency slightly smaller than $f_b$, say $f_p = f_b - \Delta f \quad (\Delta f>0)$. Now, this person will reflect the wave, and thus we will have two different waves: $y_b$, coming from the bell with a frequency $f_b$, and $y_r$, the wave that the person reflected.

This will result in a beat effect with beat frequency $f_{bt}$. My question is:

Will $f_{bt} = \Delta f$?

I think i could rephrase this as: Knowing $f_{bt}$ and $f_b$, how can we determine the person's velocity?

The relation between the two questions is that if we know the relation between $\Delta f$ and $f_{bt}$, then we can use the Doppler's Effect equations to determine the person's velocity.

• What does your analysis indicate? You can guess, but you should have a good reason. Or you can calculate what it will be. – Bill N Jul 3 '17 at 0:08
• The analysis seems very confusing to me. I saw in a website (portuguese) that we could establish a relation between a wall that reflects sounds and a mirror that reflects light, but i couldn't understand it properly. That's why i posted it in here. – Vitor C Goergen Jul 3 '17 at 0:26
• Your analysis seems sound; generally the beat frequency is the difference between two frequencies that are interfering. – probably_someone Jul 3 '17 at 8:04

The person walking is first the listener/receiver and the bell is the source. The person receives a shifted frequency $f_r=R_1f_0$. $R_1<1,$ is based on a stationary (with respect to the air) source and moving receiver. You could also calculate the frequency decrease $\Delta f_1=(1-R_1)f_0.$
That wave is reflected from the person, so the person walking is now the source and people at rest back toward the bell are the new listeners who receiver a shifted frequency $f_{Final}=R_2f_r$. $R_2<1$ is based on a moving source and stationary receiver, so it is slightly different from $R_1$. (It's similar to the difference in $\frac{A}{(A+B)}$ vs. $\frac{(A-B)}{B}$ where B is smaller than A.) We could calculate this decrease $\Delta f_2=(1-R_2)f_r$
The beat frequency will then be $f_0-f_{Final}=\Delta f_1+\Delta f_2\simeq 2\Delta f_1$.