Answer to Question #139984 in Physics for Belal

Question

A police officer with an exceptionally good ear hears an approaching motorcycle. The engine sound he hears as the bike approaches is 195 Hz.

(a) Is the actual sound the bike makes higher or lower in pitch?

(b) What causes this frequency shift? Indicate with a diagram.

c) Then after the bike passes, he hears 147 Hz as the perceived pitch. If it was a nice sunny 20 °C day, how fast was the bike moving?

Answer 1

Because of Doppler shift, the actual frequency of the bike's sound is lower than the perceived frequency (the pitch) during the approach.

$$f=f_0\frac{c}{c+v}\to v=c\left(\frac{f_0}{f}-1\right)$$

At 20 °C, the speed of sound in air is 343 m/s, so

$$v=343\left(\frac{f_0}{195}-1\right)$$

After the bike passes,

$$v=343\left(\frac{f_0}{147}+1\right)\\343\left(\frac{f_0}{147}+1\right)=343\left(\frac{f_0}{195}-1\right)\\f_0=167.6\ Hz$$

$$v=343\left(\frac{167.6}{195}-1\right)=-48.2\frac{m}{s}$$

The sign is arbitrary, depending on the assigned direction. Simply, the speed is slightly more than 48 m/s.