(a) 2250 Hz, 0.152 m
In this situation, both the ambulance and observer are stationary.
This means that there is no shift in frequency/wavelength due to the Doppler effect. So, the frequency heard by the observer is exactly identical to the frequency emitted by the ambulance:
f = 2250 Hz
While the wavelength is given by the formula:
[tex]\lambda=\frac{v}{f}[/tex]
where
v = 343 m/s is the speed of sound
f = 2250 Hz is the frequency of the sound
Substituting, we find
[tex]\lambda=\frac{343 m/s}{2250 Hz}=0.152 m[/tex]
(b) 2439.2 Hz, 0.141 m
The Doppler effect formula for a moving source is
[tex]f'=(\frac{v}{v+v_s})f[/tex]
where
f' is the apparent frequency
f is the original frequency
v is the speed of sound
[tex]v_s[/tex] is the velocity of the source (the ambulance), which is positive if the source is moving away from the observer, negative otherwise
Here the ambulance is moving toward the observer, so
[tex]v_s = -26.6 m/s[/tex]
Substituting into the formula, we find the frequency heard by the observer:
[tex]f'=(\frac{343 m/s}{343 m/s-26.6 m/s})(2250 Hz)=2439.2 Hz[/tex]
while the wavelength seen by the observer will be:
[tex]\lambda' = \frac{v}{f'}=\frac{343 m/s}{2439.2 Hz}=0.141 m[/tex]
(c) 2517.4 Hz, 0.136 m
In this situation, we must use the most general formula for the Doppler effect, which is
[tex]f'=(\frac{v+v_r}{v+v_s})f[/tex]
where
[tex]v_r[/tex] is the velocity of the observer, which is positive if the observer is moving toward the source, negative otherwise
[tex]v_s[/tex] is the velocity of the source (the ambulance), which is positive if the source is moving away from the observer, negative otherwise
In this situation,
[tex]v_s = -26.6 m/s[/tex]
[tex]v_r = +11.0 m/s[/tex]
Therefore, the frequency heard by the observer is
[tex]f'=(\frac{343 m/s+11.0 m/s}{343 m/s-26.6 m/s})(2250 Hz)=2517.4 Hz[/tex]
while the wavelength seen by the observer will be:
[tex]\lambda' = \frac{v}{f'}=\frac{343 m/s}{2517.4 Hz}=0.136 m[/tex]
