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.