Respuesta :
Explanation:
Given that,
Speed of the jet, [tex]v_s=1230\ km/h=341.67\ m/s[/tex]
Frequency of the sound produced by jet, f = 3140 Hz
Speed of sound in air, v = 342 m/s
(a) The formula of the observed frequency is given in terms of Doppler's effect as :
[tex]f'=f(\dfrac{v+v_o}{v-v_s})[/tex]
[tex]v_o[/tex] is the speed of observer
[tex]f'=3140\times (\dfrac{342}{342-341.67})\\\\f'=3254181.8\\\\f'=3254\ kHz[/tex]
(b) Let f' is the frequency received as the plane flies directly away from them. So,
[tex]f'=f(\dfrac{v+0}{v-(-v_s)})\\\\f'=f(\dfrac{v+0}{v+v_s})[/tex]
[tex]f'=3140\times (\dfrac{342}{342+341.67})\\\\f'=1570.75[/tex]
Hence, this is the required solution.
Answer:
a. [tex]f_o=1570.75\ Hz[/tex]
b. [tex]f_o=3254181.82\ Hz[/tex] which won't be audible to the humans as the audible capacity of the human ear is from 20 Hz to 20,000 Hz.
Explanation:
Given:
- speed of jet, [tex]v=1230\ km.hr^{-1}=341.67\ m.s^{-1}[/tex]
- speed of sound in air, [tex]s=342\ m.s^{-1}[/tex]
- frequency of sound, [tex]f=3140\ Hz[/tex]
a.
Since the sound source is moving therefore we can determine the observed frequency by the Doppler's effect given as:
[tex]\frac{f_o}{f} =\frac{s-v_o}{s+v}[/tex]
where:
[tex]f_o=[/tex] observed frequency when the jet plane approaches the observers
[tex]v_o=[/tex] observer's velocity = 0
[tex]\frac{f_o}{3140} =\frac{342-0}{342+341.67}[/tex]
[tex]f_o=1570.75\ Hz[/tex]
b.
When the plane flies away from the observers then we take its velocity as negative.
[tex]\frac{f_o}{f} =\frac{s-v_o}{s+v}[/tex]
where:
[tex]f_o=[/tex] observed frequency when the jet plane flies away from the observers
[tex]v_o=[/tex] observer's velocity = 0
[tex]\frac{f_o}{3140} =\frac{342-0}{342+(-341.67)}[/tex]
[tex]f_o=3254181.82\ Hz[/tex] which won't be audible to the humans as the audible capacity of the human ear is from 20 Hz to 20,000 Hz.
