# Doppler Effect?

## You are standing on a train station platform as a train goes by close to you. As the train approaches, you hear the whistle sound at a frequency of f1 = 92 Hz. As the train recedes, you hear the whistle sound at a frequency of f2 = 79 Hz. Take the speed of sound in air to be v = 340 m/s. Find the numeric value, in hertz, for the frequency of the train whistle fs that you would hear if the train were not moving.

May 19, 2017

$\textsf{85 \textcolor{w h i t e}{x} H z}$

#### Explanation:

As an estimate I would expect the frequency to be about halfway between the two values i.e $\cong$ 85.5 Hz.

We can show this as follows:

When the train is approaching we get:

$\textsf{{f}_{o b s} = \left[\frac{v}{v - {v}_{s o u r c e}}\right] {f}_{s o u r c e}}$

Where v is the speed of sound.

We can find the speed of the train sf(v_(source), from which we can get $\textsf{{f}_{s o u r c e}}$.

When the train is receding we get:

$\textsf{{f}_{o b s} = \left[\frac{v}{v + {v}_{s o u r c e}}\right] {f}_{s o u r c e}}$

Putting in the numbers we get:

$\textsf{92 = \left[\frac{340}{340 - {v}_{s o u r c e}}\right] {f}_{s o u r c e} \text{ } \textcolor{red}{\left(1\right)}}$

$\textsf{79 = \left[\frac{340}{340 + {v}_{s o u r c e}}\right] {f}_{s o u r c e} \text{ } \textcolor{red}{\left(2\right)}}$

Now divide sf(color(red)((1)) by $\textsf{\textcolor{red}{\left(2\right)} \Rightarrow}$

sf(92/79=([(340)/(340-v_(source))])/([(340)/(340+v_(source))])xxcancel(f_(source))/(cancel(f_(source)))

$\textsf{\frac{92}{79} = \left[\frac{\cancel{340}}{340 - {v}_{s o u r c e}}\right] \times \left[\frac{340 + {v}_{s o u r c e}}{\cancel{340}}\right]}$

$\textsf{1.1645 = \frac{340 + {v}_{s o u r c e}}{340 - {v}_{s o u r c e}}}$

sf(1.1645(340-v_(source))=(340+v_(source))

sf(395.95-1.1645v_(source)=(340+v_(source))

$\textsf{2.1645 {v}_{s o u r c e} = 395.95 - 340 = 55.949}$

$\therefore$$\textsf{{v}_{s o u r c e} = \frac{55.949}{2.1645} = 25.85 \textcolor{w h i t e}{x} \text{m/s}}$

This is the speed of the train.

Now we can substitute this value back into $\textsf{\textcolor{red}{\left(1\right)} \Rightarrow}$

$\textsf{92 = \left[\frac{340}{340 - 25.85}\right] {f}_{s o u r c e}}$

$\textsf{92 = \left[1.0823\right] {f}_{s o u r c e}}$

$\textsf{{f}_{s o u r c e} = \frac{92}{1.0823} = 85 \textcolor{w h i t e}{x} H z}$

From the prediction this seems reasonable.