To determine the frequency you perceive as the train approaches, we can use the Doppler effect formula for sound. The Doppler effect describes the change in frequency or wavelength of a wave in relation to an observer moving relative to the source of the wave.
Given the stationary frequency of the train whistle [tex]\( f = 580 \text{ Hz} \)[/tex], the speed of the train [tex]\( v_{\text{train}} = 18.8 \text{ m/s} \)[/tex], and the speed of sound [tex]\( v_{\text{sound}} = 343 \text{ m/s} \)[/tex], we will use the following Doppler effect formula for a source moving towards a stationary observer:
[tex]\[
f' = f \left( \frac{v_{\text{sound}}}{v_{\text{sound}} - v_{\text{train}}} \right)
\][/tex]
Where:
- [tex]\( f'\)[/tex] is the observed frequency,
- [tex]\( f \)[/tex] is the emitted frequency of the source,
- [tex]\( v_{\text{sound}} \)[/tex] is the speed of sound in the medium,
- [tex]\( v_{\text{train}} \)[/tex] is the speed of the source moving towards the observer.
Substituting in the given values:
[tex]\[
f' = 580 \left( \frac{343}{343 - 18.8} \right)
\][/tex]
We need to solve the fraction inside the parentheses first:
[tex]\[
343 - 18.8 = 324.2
\][/tex]
Now substitute this back into the formula:
[tex]\[
f' = 580 \left( \frac{343}{324.2} \right)
\][/tex]
When we divide 343 by 324.2, we get approximately:
[tex]\[
\frac{343}{324.2} \approx 1.058
\][/tex]
Now multiply this by 580 Hz:
[tex]\[
f' = 580 \times 1.058 \approx 613.634 \text{ Hz}
\][/tex]
Thus, the frequency you hear as the train approaches is approximately:
[tex]\[
f' \approx 613.634 \text{ Hz}
\][/tex]
