Doppler effect: why do I find this exercise so difficult?

[mcastillo356]

Homework Statement

The frequency of a car horn is $400\,\text{Hz}$. If the horn is honked as the car moves with speed $u_s=34\,\text{m/s}$ through still air toward a stationary receiver, find $(a)$ the wavelenth of the sound passing the receiver, and $(b)$ the frequency recieved. Take the speed of sound in air as $343\,\text{m/s}$. $(c)$Find the wavelength of the sound passing the receiver and find the frequency received if the car is stationary as the horn is honked and a receiver moves with a speed $u_r=34\,\text{m/s}$ toward the car.

Relevant Equations

$\lambda=\cfrac{v-u_s}{f_s}$
$f_r=\cfrac{v\pm{u_r}}{v\pm{u_s}}\,f_s$
$\lambda=\cfrac{v\pm{u_s}}{f_s}$

PICTURE $(a)$ The waves in front of the source are compressed, so we use the minus sign in $\lambda=(v\pm{u_s})/f_s$. $(b)$ We calculate the received frequency using $f_r=[(v\pm{u_r})/(v\pm{u_s})]f_s$. $(c)$ For a moving receiver, we use the same equations as in Parts $(a)$ and $(b)$.

SOLVE
Calculate the wavelength in front of the car. In front of the source the wavelength is shorter, so choose the sign accordingly:

$$\lambda=\cfrac{v-u_s}{f_s}=\cfrac{343\,\text{m/s}-34\,\text{m/s}}{400\,\text{Hz}}=0.758\,\text{m}=\boxed{0.76\,\text{m}}$$

Solve for the received frequency with $u_r=0$

$$f_r=\cfrac{v\pm{u_r}}{v\pm{u_s}}\,f_s=\cfrac{v}{v-u_s}\,f_s=\Bigg (\cfrac{343}{343-34}\Bigg )(400\,\text{Hz})=453\,\text{Hz}=450\,\text{Hz}$$

Why the sign minus in the second equation?

$(c)$ 1. Calculate the wavelength in front of the source with $u_s=0$

$$\lambda=\cfrac{v\pm{u_s}}{f_s}=\cfrac{343\,\text{m/s}}{400\,\text{Hz}}=0858\,\text{m}=\boxed{0.86\,\text{m}}$$

2. The received frequency is given with $u_s$. The source is approaching the receiver, so the frequency is shifted upward. Choose the sign accordingly:

$$f_r=\cfrac{v\pm{u_r}}{v\pm{u_s}}\,f_s=\cfrac{v+u_r}{v}\,f_s=\Bigg (1+\cfrac{u_r}{v}\Bigg )\,f_s=\Bigg (1+\cfrac{34}{343}\Bigg )(400\,\text{Hz})=\boxed{440\,\text{Hz}}$$

In front of the source the wavelength is shorter. Why then is bigger than the wavelength in front of the car?

 

[anuttarasammyak]

Why the sign minus in the second equation?

Because by $\lambda$ you got

$$ f_r=\frac{v}{\lambda} $$

In front of the source the wavelength is shorter. Why then is bigger than the wavelength in front of the car?

0.86 m is for $u_s=0$, a parked car, which is larger than 0.76m which is for in front of the moving car.

 

[mcastillo356]

Thank you very much @anuttarasammyak, you've solved every doubt.

 

[physicsstudent21]

After reading the following article, your understanding to the Doppler effect will be much more clear. You may find this article by searching its title or simply searching "doppler effect intuitive" in Google or Bing.
H. Lei, “An intuitive model for illustrating the classical Doppler effect equation,” The Physics Teacher, vol. 63, pp. 652-654, 2025.

 

[Steve4Physics]

After reading the following article, your understanding to the Doppler effect will be much more clear. You may find this article by searching its title or simply searching "doppler effect intuitive" in Google or Bing.
H. Lei, “An intuitive model for illustrating the classical Doppler effect equation,” The Physics Teacher, vol. 63, pp. 652-654, 2025.

That’s a nice approach. A copy of the article can be found here:
https://arxiv.org/pdf/2504.18159

Also, the equation $f_r=\cfrac{v\pm{u_r}}{v\pm{u_s}}\,f_s$ is often quoted without any explanation of the sign-convention – which is risky.

A better (IMO) approach is to define the positive direction to be from the receiver to the source. Then a simple unambigous equation does the job:
$f_r=\cfrac{v + u_r}{v + u_s}\,f_s$
where $u_r$ and $u_s$ are now signed values representing the velocities of the receiver and source.

 

[neilparker62]

That’s a nice approach. A copy of the article can be found here:
https://arxiv.org/pdf/2504.18159

Thanks for the link.