Question

1. 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 nic sunny 20 °C day, how fast was the bike moving?

2. How much faster does sound travel on a hot summer day, 40 °C, than on the coldest winter day, -40 °C?

Answer 1

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

(b) f = (c/(c+Vs))·f₀,
where f₀ is the unmodified frequency
c is the speed of the sound waves
Vs is the speed of the source of the sound.

Vs = c·f₀/f - c, or c(f₀/f - 1)

At 20 °C, the speed of sound in air is 343 m/s, so Vs during the approach is (343 m/s)((f₀/195 Hz) - 1) = ((343/195)f₀ - 343) m/s.

After the bike passes, Vs = -(343 m/s)((f₀/147 Hz) - 1) = (-(343/147)f₀ + 343) m/s. (The negative sign is used because the velocity is now directed away from the observer.)

Since the bike does not change speed, the expressions are equal:
((343/195)f₀ - 343) m/s = (-(343/147)f₀ + 343) m/s, or
(343/195)f₀ + (343/147)f₀ = 686 m/s, or
f₀ = 686/(343/195 + 343/147) = 167.6 Hz.

Use this in one of the speed equations to get Vs:
Vs = ((343/195)f₀ - 343) m/s
= ((343/195)167.6 - 343) m/s
= -48.2 m/s.

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