Suppose that a police car on the highway is moving to the right at 20 m/s, while a speeder is coming up from almost directly behind at a speed of 31 m/s, both speeds being with respect to the ground. The police officer aims a radar gun at the speeder. Assume that the electromagnetic wave emitted by the radar gun has a frequency of 9.0 109 Hz.
Find the magnitude of the difference between the frequency of the emitted wave and the wave that returns to the police car after reflecting from the speeder's car.
___________Hz
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Delta f = fv/c
where f=transmitted frequency
v=difference in velocity
c=velocity of light
To find the magnitude of the difference between the frequency of the emitted wave and the wave that returns to the police car after reflecting from the speeder's car, we need to consider the Doppler effect.
The Doppler effect is the change in frequency or wavelength of a wave in relation to an observer who is moving relative to the wave source. The formula for the Doppler effect is:
Δf/f = v/c
Where:
Δf = change in frequency
f = frequency of the source wave
v = velocity of the observer relative to the medium
c = speed of the wave in the medium
In this case, the observer (police officer) is in the moving police car, and the source of the wave is the speeder's car. The observer is moving towards the speeder's car, so the relative velocity (v) is the difference between the velocity of the police car (20 m/s) and the velocity of the speeder (31 m/s). The speed of the wave in the medium (c) is the speed of light, which is approximately 3.0 x 10^8 m/s.
Using the formula, we can calculate the magnitude of the difference in frequency:
Δf/f = v/c
Δf/9.0 x 10^9 Hz = (31 m/s - 20 m/s) / (3.0 x 10^8 m/s)
Simplifying the equation:
Δf / 9.0 x 10^9 Hz = 11 m/s / 3.0 x 10^8 m/s
Cross-multiplying:
Δf = (11 m/s / 3.0 x 10^8 m/s) * 9.0 x 10^9 Hz
Calculating the magnitude of the difference in frequency:
Δf = 3.3 x 10^2 Hz
Therefore, the magnitude of the difference between the frequency of the emitted wave and the wave that returns to the police car after reflecting from the speeder's car is 3.3 x 10^2 Hz.
To find the magnitude of the difference between the frequency of the emitted wave and the wave that returns to the police car after reflecting from the speeder's car, we need to consider the Doppler effect.
The Doppler effect is the change in frequency of a wave when there is relative motion between the source of the wave and the observer. In this case, the radar gun is the source and the police car is the observer.
To calculate the frequency shift, we can use the formula:
Δf/f = vc / c
Where:
Δf is the difference in frequency
f is the frequency of the emitted wave
v is the relative velocity between the source and the observer
c is the speed of light in a vacuum (approximately 3.0 x 10^8 m/s)
In this case, the relative velocity (v) is the sum of the velocities of the speeder and the police car, since they are moving in opposite directions:
v = v_speeder - v_police
Plugging in the given values:
v = 31 m/s - (-20 m/s) [since the police car is moving to the right, its velocity is positive]
v = 51 m/s
Now we can calculate the magnitude of the frequency shift:
Δf/f = (51 m/s) / (3.0 x 10^8 m/s)
Δf/f = 1.7 x 10^-7
Finally, we can find the magnitude of the difference in frequency by multiplying it by the frequency of the emitted wave:
Δf = (1.7 x 10^-7) * (9.0 x 10^9 Hz)
Δf = 1.53 Hz
Therefore, the magnitude of the difference between the frequency of the emitted wave and the wave that returns to the police car after reflecting from the speeder's car is 1.53 Hz.