Question

At the Indianapolis 500, you can measure the speed of cars just by listening to the difference in pitch of the engine noise between approaching and receding cars. Suppose the sound of a certain car drops by a factor of 1.80 as it goes by on the straightaway. How fast is it going?

Answer 1

Answer:

$v_s = 97.14 m/s$

Explanation:

As per Doppler's effect of sound the frequency of the sound when source is approaching the observer is given as

$f_1 = f_o\frac{v}{v- v_s}$

similarly when source is moving away from the observer then its frequency is given as

$f_2 = f_o\frac{v}{v + v_s}$

now we know that the ratio of two frequency is

$\frac{f_1}{f_2} = 1.80$

$\frac{v + v_s}{v - v_s} = 1.80$

$v + v_s = 1.80 v - 1.80 v_s$

$0.80 v = 2.80 v_s$

$v_s = \frac{0.80}{2.80}v$

$v_s = \frac{0.80 \times 340}{2.80}$

$v_s = 97.14 m/s$