Answer to Question #164230 in Optics for Peter

Question

  Compare the shift in frequency if a 2000Hz source is moving towards you at 15ms-1 versus if you are moving towards it at 15ms-1. Are the two frequencies exactly the same? Are they close? Repeat

the calculation for 150ms-1 and then again for 300ms-1. What can you conclude about the asymmetry of the Doppler formulas? Show that at low speeds (relative to the speed of sound), the two formulas-source approaching and detector approaching-yield the same result.

Answer 1

$v_{snd} = 343 \;m/s$

At 15 m/s

$f’_{source} = f\frac{1}{1 -\frac{v_{stc}}{v_{snd}}} \\ = 2000 \times \frac{1}{1 -\frac{15}{343}} \\ = 2092 \;Hz \\ f’_{observer} = f(1 + \frac{v_{stc}}{v_{snd}}) \\ =2000(1 +\frac{15}{343}) \\ = 2087 \;Hz$

At 150 m/s

$f’_{source} = f\frac{1}{1 -\frac{v_{stc}}{v_{snd}}} \\ = 2000 \times \frac{1}{1 -\frac{150}{343}} \\ = 3554 \;Hz \\ f’_{observer} = f(1 + \frac{v_{stc}}{v_{snd}}) \\ =2000 (1 +\frac{150}{343}) \\ = 2874 \;Hz$

At 300 m/s

$f’_{source} = f\frac{1}{1 -\frac{v_{stc}}{v_{snd}}} \\ = 2000 \times \frac{1}{1 -\frac{300}{343}} \\ = 15,961 \;Hz \\ f’_{observer} = f\frac{1}{1 + \frac{v_{stc}}{v_{snd}}} \\ =2000 (1 +\frac{300}{343}) \\ = 3749 \;Hz$

At low speeds (relative to the speed of sound), the two formulas-source approaching and detector approaching-yield the same result.

$f’_{source} = f\frac{1}{1 -\frac{v_{stc}}{v_{snd}}} \\ f’_{source} = f(1 -\frac{v_{stc}}{v_{snd}})^{-1} \\ f’_{source} ≈ f(1 +\frac{v_{stc}}{v_{snd}}) \\ f’_{source} = f’_{observer}$