Complete Question
An ambulance is driving towards the hospital at a velocity 94.0 km/h and emitting a steady 878-Hz sound from its siren. The sound reflects off the front of the hospital and is received by the same ambulance. In addition to it's own siren, the ambulance hears a shifted tone from the reflection at what frequency? The speed of sound on this day is 343 m/s
Answer:
The frequency is [tex]f =1022.0 \ Hz[/tex]
Explanation:
From the question we are told that
The velocity of the ambulance [tex]v = 94.0km/h = 94 *\frac{1000 m}{3600s} = 26 \ m/s[/tex]
The frequency emitted is [tex]f_0 = 878 \ Hz[/tex]
The illustration of the process is shown on the first uploaded image
The frequency at the hospital wall due to Doppler's effect is mathematically repented as
[tex]f_1 = f_o [\frac{v}{v-v_o} ][/tex]
Now as seen the frequency at the ambulance is mathematically represented as
[tex]f = f_1 [\frac{v+v_o}{v} ][/tex]
Now substituting for [tex]f_w[/tex]
[tex]f = f_o [\frac{v+v_o}{v-v_o} ][/tex]
Substituting value
[tex]f = 878 * \frac{343 +26}{343 -26}[/tex]
[tex]f =1022.0 \ Hz[/tex]
Answer:
The frequency is 1022.68 Hz
Explanation:
According the Doppler`s effect, the frequency is:
[tex]f=f_{o} (\frac{v+v_{o}}{v-v_{o} } )[/tex]
Where
fo = 878 Hz
v = velocity of sound = 343 m/s
v₀ = 94 km/h = 26.11 m/s
Replacing:
[tex]f=878(\frac{343+26.11}{343-26.11} )=1022.68Hz[/tex]
