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Anton [14]
3 years ago
10

Slick Willy is in traffic court (again) contesting a $50.00 ticket for running a red light. "You see, your Honor, as I was appro

aching 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.)
Physics
2 answers:
Masteriza [31]3 years ago
8 0

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.

neonofarm [45]3 years ago
6 0

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

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