Respuesta :
Answer:velocity of train (source of sound) =10.68m/s
Explanation: this is a question under doppler effect.
The doppler effect formulae gives the value of sound wave detected by an observer from a source when there is a relative motion between both the observer and the source.
The formulae below
f'= [tex]\frac{v+v'}{v-vs}[/tex] * f
f'=observed frequency
f= actual frequency
v'= velocity of observer
v= velocity of sound in air=343m/s
vs= velocity of source (train)
from the question, the observer is standing thus, v'=0
The sound source (train) is moving towards the observer thus making vs positive relative to the motion of the observer(+vs), at this condition, we have that f'=447hz, v=343m/s v'=0 with the formulae
f' = [tex]\frac{v}{v-vs}[/tex] * f
putting all of these in the doppler effect formulae, we have that
447= [tex]\frac{343}{343-vs}[/tex] * f
thus we have that
447= [tex]\frac{343f}{343-vs}[/tex]
by cross mutiplying, we have that
447(343-vs)=343f
we call this equation 1
for the next condition, the source (train) is moving away from the observer thus making vs negative relative to the motion of the observer (-vs), at this condition, we have that
f'=420hz, v'=0, v=343m/s .
since vs is now negative and the formulae it was positive then the resulting sign is positive as shown below
f'=[tex]\frac{v+v'}{v-(-vs)}[/tex]
thus we have that
[tex]\frac{v+v'}{v+vs}[/tex]
but f'=420hz, v'=0, v=343m/s, by slotting in the parameters into the formulae, we have
420=[tex]\frac{343+0}{343+vs} \\[/tex]
by cross multipying we have that
420(343+vs)=343f
we call this equation 2
since both equations are equated to "343f", thus we can equate both equation 1 and 2
thus we have that
447(343-vs)=420(343+vs)
by expanding the bracket, we have
153321-447vs=144060+420vs
by collecting like terms
153321-144060= 420vs+447vs
doing the necessary algebra, we have
9261=867vs
dividing through by 867
[tex]\frac{9261}{867}[/tex] =vs
thus vs=10.86m/s
