Respuesta :
Answer:
e. f2 < f < f1
Explanation:
According to Doppler's Effect:
[tex]\frac{f_o}{f_s} =\frac{S+v_o}{S-v_s}[/tex] ......................................(1)
where:
[tex]f_o\ \&\ f_s[/tex] are observed frequency and source frequency respectively.
S = velocity of sound in the air from a stationary source
[tex]v_o\ \&\ v_s[/tex] are the velocity of the observer and the velocity of sound source with respect to a stationary frame of reference.
- When the ambulance approaches a stationary observer
Here [tex]v_o=0\[/tex]
Then eq. (1) becomes:
[tex]\frac{f}{f1} =\frac{S}{S-v_s}[/tex]
Now, the value:
[tex]\frac{f1}{f} =\frac{S}{S-v_s}>1[/tex]
[tex]\therefore f<f1[/tex]
- Now according to the given condition the source is moving away from the observer i.e. the velocity of the source is opposite to the velocity of sound with respect to the stationary observer.
Now the eq. (1) becomes
[tex]\frac{f2}{f} =\frac{S}{S-(-v_s)}[/tex]
∵the direction of motion of the source is away from the observer so a negative sign has been introduced.
Now, the value:
[tex]\frac{f2}{f} =\frac{S}{S+v_s}<1[/tex]
[tex]\therefore f>f2[/tex]
