Answer to Question #164334 in Optics for Iron rose

Question

 An engine travelling at a constant speed towards a tunnel emits a short burst of sound of

frequency 400Hz which is reflected from the tunnel entrance. The engine driver hears an echo of

 frequency 500 Hz two seconds after the sound is emitted. Assuming the speed of sound is

         340ms^-1, calculate the speed of the engine, and its distance from the tunnel when the driver hears

the echo.

Answer 1

In this particular case the engine is the observer in motion moving towards the source where the driver hears the echo coming from the tunnel which is our given stationary source. Therefore, we apply the Doppler's effect equation:

When observer is approaching stationary source: $f_o=f_s\left(\frac{v+v_o}{v}\right)$

Where:

$f_o$ = Frequency of observer(engine)=400Hz

$f_s$ =Frequency of the source(tunnel)=500Hz

$v$ =speed of sound in air=340m/s

$v_o$ =speed of observer(engine)=?

$400=500\left(\frac{340+v_o}{340}\right)$

$\frac{400\times 340}{500}=340+v_o\:\:\:\:$

$272=340+v_o$

$v_o= 68ms^{-1}$

The negative value of speed is ignored since it has no significance. Therefore,

Speed of the engine$=$ $68ms^{-1}$

Distance of the the tunnel from where the driver heard the echo.

We take the formula:$\:Distance=speed\:\times time$

Since the engine was also moving with some speed when the sound was produced to give back an echo we take cumulative speed to be: $speed=(340+68)ms^{-1}$

$Time=2s$

$\:Distance=\left(340+68\right)\:\times 2$

$Distance=408\:\times 2$

$Distance=816m$