Respuesta :
Answer:
(a) 12.9m/s
(b) 193Hz
Explanation:
The principle and concept of this question is based on Doppler's effect.
Doppler's effect is a change in the observed frequency of a wave when the source or the observer move relative to the transmitting medium. Doppler's effect can be expressed mathematically as;
[tex]f = (\frac{U ± U_{0} }{U ± U_{s} })f_{0}[/tex]
Where,
f is given in terms of the sound frequency f₀
U is the speed of sound in the medium
U₀ is the speed of the observer
[tex]U_{s}[/tex] is the speed of the source
When applying Doppler's equation, the signs U₀ and [tex]U_{s}[/tex] depend on the direction and velocity of of the observer and the source.
+ Positive sign, for U₀ and [tex]U_{s}[/tex] if the velocity one is moving towards the other.
- Negative sign, for U₀ and [tex]U_{s}[/tex] if the velocity one is moving away from the other.
Now, let's applying this information to the question given.
Given:
Frequency of the train's horn, f₀ = 200Hz
Speed of sound , U = 335m/s
Frequency of the approaching train's horn the observer received, f = 208Hz
Since the observer is waiting at the crossing, he/she is at a stationary position.
So, the speed of the observer, U₀ = 0
(a) The train approaches the observer. Following the sign rule, the speed of the source [tex]U_{s}[/tex] is + positive. Applying this to the equation,
[tex]f = (\frac{U - 0}{U - (+U_{s}) })f_{0}[/tex]
[tex]f = (\frac{U}{U - U_{s} })f_{0}[/tex]
Solving for [tex]U_{s}[/tex],
[tex]\frac{f}{f_{0}}[/tex]= (\frac{U}{U - U_{s}})[/tex]
[tex]{f_{0}U=f(U - U_{s})[/tex]
[tex]{f_{0}U=fU - fU_{s}[/tex]
[tex]fU_{s} =fU - f_{0}U[/tex]
[tex]fU_{s} = U(f - f_{0})[/tex]
[tex]U_{s} =\frac{U(f - f_{0})}{f}[/tex]
[tex]U_{s} =\frac{335m/s(208Hz - 200Hz)}{208Hz}[/tex]
[tex]U_{s} =\frac{335m/s × 8Hz}{208Hz}[/tex]
[tex]U_{s} =\frac{2680}{208}[/tex]
[tex]U_{s} =12.8846m/s[/tex]
[tex]U_{s} =12.9m/s[/tex]
(b) The speed train approaches the observer. Following the sign rule, the speed of the source [tex]U_{s}[/tex] is - negative. Applying this to the equation,
[tex]f = (\frac{U}{U - (-U_{s}) })f_{0}[/tex]
[tex]f = (\frac{U}{U + U_{s}})f_{0}[/tex]
[tex]f = (\frac{335m/s}{335m/s + 12.9m/s})200Hz[/tex]
[tex]f = (\frac{335m/s}{335m/s + 12.9m/s})200Hz[/tex]
[tex]f = (\frac{335m/s}{347m/s})200Hz[/tex]
f = 193Hz
