# A car sounds its horn (502 MHz) as it nears a pedestrian. The pedestrian has a perfect pitch and determines that the sound from the horn has a frequency of 520 Hz. If the speed of the sound that day was 340 m/s, how fast was the car travelling?

Dec 12, 2015

$11.8 \text{m/s}$

#### Explanation:

This is example of the Doppler Effect. You can see that more wave fronts reach the pedestrian per second as the car approaches.

This results in an increase in the frequency of the sound as detected by the pedestrian:

This is given by the expression:

${f}_{\text{obs")=[(v)/(v-v_("source")]]f_("source}}$

${f}_{\text{obs"=520"MHz}}$ . This is the frequency of the horn that the pedestrian detects.

f_("source")=502"MHz". This is the actual frequency of the horn.

We are given $v = 340 \text{m/s}$.

We need to get ${v}_{\text{source}}$.

Putting in the values $\Rightarrow$

$520 \times \cancel{{10}^{6}} = \left[\frac{340}{v - {v}_{\text{source}}}\right] \times 502 \times \cancel{{10}^{6}}$

$\therefore \frac{520}{502} = \left[\frac{340}{340 - {v}_{\text{source}}}\right] = 1.036$

$\therefore \left(340 - {v}_{\text{source}}\right) = \frac{340}{1.036} = 328.18$

:.v_("source")=340-328.18=11.8"m/s"