Question

Slick Willy is in traffic court (again) contesting a $50.00 ticket for running a red light. "You see, your Honor, as I was approaching the light, it appeared yellow to me because of the Doppler effect. The red light from the traffic signal was shifted up in frequency because I was traveling towards it, just like the pitch of an approaching car rises as it approaches you." a. Calculate how fast Slick Willy must have been driving, in meters per second, to observe the red light (wavelength of 687 nm) as yellow (wavelength of 570 nm). (Treat the traffic light as stationary, and assume the Doppler shift formula for sound works for light as well.)

Answer 1

Answer:

61578948 m/s

Explanation:

λ$_{actual}$ = λ$_{observed}$ $\frac{c+v_{o}}{c}$

687 = 570 $(\frac{3 * 10^{8} +v_{o} }{3 * 10^{8}} )$

$v_{o}$ = 61578948 m/s

So Slick Willy was travelling at a speed of 61578948 m/s to observe this.

Answer 2

Answer:

$6.11*10^7m/s$

Explanation:

The Doppler effect formula for an observer approaching a source is given by equation (1);

$f_o=\frac{f(v+v_o)}{v-v_s}...................(1)$

where $f_o$ is the frequency perceived by the observer, v is the actual velocity of the wave in air, $v_o$ is the velocity of the observer, $v_s$ is the velocity of the source and $f$ is the actual frequency of the wave.

The actual velocity v of light in air is $3*10^8m/s$. The relationship between velocity, frequency and wavelength $\lambda$ is given by equation (2);

$v=\lambda f...........(2)$

therefore;

$f=\frac{v}{\lambda}...............(3)$

We therefore use equation (3) to find the actual frequency of light emitted and the frequency perceived by Slick Willy.

Actual wavelength $\lambda$ of light emitted is 678nm, hence actual frequency is

given by;

$f=\frac{3*10^8}{687*10^{-9}}\\f=4.37*10^{14}Hz$

Also, the frequency perceived by Slick Willy is given thus;

$f_o=\frac{3*10^8}{570*10^{-9}}\\f=5,26*10^{14}Hz$

The velocity $v_s$ of the source light is zero since the traffic light was stationary. Substituting all parameters into equation (1), we obtain the following;

$5.26*10^{14}=\frac{4.37*10^{14}(3*10^{8}+v_o)}{3*10^8-0}$

We then simplify further to get $v_o$

$10^{14}$  cancels out from both sides, so we obtain the following;

$5.26*3*10^8=4.37(3*10^8+v_o)$

$15.78*10^8=13.11*10^8+4.37v_o\\4.37v_o=15.78*10^8-13.11*10^8\\4.37v_o=2.67*10^8$

Hence;

$v_o=\frac{2.67*10^8}{4.37}\\v_o=6.11*10^7m/s$