Question

At an air show a jet flies directly toward the stands at a speed of 1230 km/h, emitting a frequency of 3140 Hz, on a day when the speed of sound is 342 m/s.

a. What frequency (in Hz) is received by the observers?
b. What frequency do they receive as the plane flies directly away from them?

Answer 1

a.
`f_o=1570.75\ Hz`

b.
`f_o=3254181.82\ Hz` which won't be audible to the humans as the audible capacity of the human ear is from 20 Hz to 20,000 Hz.

Step-by-step explanation:

Given:

• speed of jet,
  `v=1230\ km.hr^(-1)=341.67\ m.s^(-1)`

• speed of sound in air,
  `s=342\ m.s^(-1)`

• frequency of sound,
  `f=3140\ Hz`

a.

Since the sound source is moving therefore we can determine the observed frequency by the Doppler's effect given as:

`(f_o)/(f) =(s-v_o)/(s+v)`

where:

`f_o=` observed frequency when the jet plane approaches the observers

`v_o=` observer's velocity = 0

`(f_o)/(3140) =(342-0)/(342+341.67)`

`f_o=1570.75\ Hz`

b.

When the plane flies away from the observers then we take its velocity as negative.

`(f_o)/(f) =(s-v_o)/(s+v)`

where:

`f_o=` observed frequency when the jet plane flies away from the observers

`v_o=` observer's velocity = 0

`(f_o)/(3140) =(342-0)/(342+(-341.67))`

`f_o=3254181.82\ Hz` which won't be audible to the humans as the audible capacity of the human ear is from 20 Hz to 20,000 Hz.

Answer 2

Step-by-step explanation:

Given that,

Speed of the jet,
`v_s=1230\ km/h=341.67\ m/s`

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 :

`f'=f((v+v_o)/(v-v_s))`

`v_o` is the speed of observer

`f'=3140* ((342)/(342-341.67))\\\\f'=3254181.8\\\\f'=3254\ kHz`

(b) Let f' is the frequency received as the plane flies directly away from them. So,

`f'=f((v+0)/(v-(-v_s)))\\\\f'=f((v+0)/(v+v_s))`

`f'=3140* ((342)/(342+341.67))\\\\f'=1570.75`

Hence, this is the required solution.