A boy on a bicycle approaches a brick wall as he sounds his horn at a frequency 400 hz. the sound he hears reflected back from the wall is at a frequency 408 hz. at what is the speed is the boy riding his bicycle toward the wall? assume the speed of sound in air is 340 m/s.

Respuesta :

As the question is about changing in frequency of a wave for an observer who is moving relative to the wave source, the concept that should come to our minds is "Doppler's effect."

Now the general formula of the Doppler's effect is:
[tex]f = (\frac{g + v_{r}}{g + v_{s}})f_{o}[/tex] -- (A)

Note: We do not need to worry about the signs, as everything is moving towards each other. If something/somebody were moving away, we would have the negative sign. However, in this problem it is not the issue.

Where,
g = Speed of sound = 340m/s.
[tex]v_{r}[/tex] = Velocity of the receiver/observer relative to the medium = ?.
[tex]v_{s}[/tex] = Velocity of the source with respect to medium = 0 m/s.
[tex]f_{o}[/tex] =  Frequency emitted from source = 400 Hz.
[tex]f[/tex] = Observed frequency = 408Hz.

Plug-in the above values in the equation (A), you would get:

[tex]408 = ( \frac{340 + v_{r}}{340 + 0})*400 [/tex]

[tex] \frac{408}{400} = \frac{340 + v_{r}}{340} [/tex]

Solving above would give you,
[tex]v_{r}[/tex] = 6.8 m/s

The correct answer = 6.8m/s



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