Answer:
f' = 358442.3 Hz
Explanation:
This is a problem about the Doppler's effect. To find the perceived frequency of the observers you use the following formula:
[tex]f'=f(\frac{v}{v\pm v_s})[/tex]
v: speed of sound = 342 m/s
vs : speed of the source (jet) = 1220 km/h (1h/3600s)*(1000 m/ 1km) = 338.88 m/s (it is convenient to convert the units of vs to m/s)
f: frequency emitted by the source = 3270 Hz
f': perceived frequency
Due to the jet is getting closer to the observers, the sing of the denominator in equation (1) is minus (-). Then, you replace the values of f, vs and v:
[tex]f'=(3270s^{-1})(\frac{342m/s}{342m/s-338.88m/s})=358442.3 Hz[/tex]
Hence, the perceived frequency by the observers is 358442.3 Hz
Answer:
[tex]f_o=359466.42Hz[/tex]
Explanation:
You can solve this problem using doppler effect formula. The Doppler effect is the phenomenon by which the frequency of the waves perceived by an observer varies when the emitting focus or the observer itself moves relative to each other
Doppler effect general case is given by:
[tex]f_o=f\frac{v\pm v_o}{v \mp v_s} \\\\Where:\\\\f=Actual\hspace{3}frequency\\f_o=Observed\hspace{3}frequency\\v=Speed\hspace{3}of\hspace{3}the\hspace{3}sound\hspace{3}waves\\v_o=Velocity\hspace{3}of\hspace{3}the\hspace{3}observer\\v_s=Velocity\hspace{3}of\hspace{3}the\hspace{3}source[/tex]
Now, you can use the following facts:
[tex]+v_o[/tex] Is used when the observer moves towards the source
[tex]-v_o[/tex] Is used when the observer moves away from the source
[tex]-v_s[/tex] Is used when the source moves towards the observer
[tex]+v_s[/tex] Is used when the source moves away from the observer
Since the source is alone in motion towards the observer, the formula is given by:
[tex]f_o=f\frac{v}{v-v_s} \\\\Where:\\\\v_s=1220km/h\approx 338.8888889m/s\\v=342m/s\\f=3270Hz[/tex]
Therefore:
[tex]f_o=(3270) \frac{342}{342-338.8888889} = 359466.42Hz[/tex]
