Question #482da

There aren't any right answers in the four choices (although I don't really understand what you mean by "sound level").

Due to the Doppler effect, the frequency that an observer perceives (it's called the apparent frequency) is different from the real frequency, the frequency that a source makes.

The Doppler effect only affects frequency and wavelength.
It doesn't affect either the speed of a sound, the speed of the observer, the speed of the source or the amplitude of a sound.

The apparent frequency is given by :

#f' = f*(v_(wave) + v_(obs))/(v_(wave)-v_s)#,

where #v_(wave)# is the speed of sound in the medium, #v_(obs)# is the speed of the observer (#v_(obs) > 0# when the source is in front of him) and #v_s# is the speed of the source (#v_s > 0# when the observer is in front of it).

In our case, we don't care about #v_(wave)#, we suppose that #v_(obs) = 0#, and there is a source approaching the observer so #v_s > 0#.

Now, we have :

#f' = f*(v_(wave))/(v_(wave)-v_s) => f' > f#.

The wavelength is given by :

#lambda = v_(wave)/f#.

Therefore :

#lambda = v_(wave)/f > lambda' = v_(wave)/(f')#.

So the frequency that the observer perceives increases and the wavelength decreases.

Thus, there aren't any right answers in your choices.