Suppose that a police car on the highway is moving to the right at 26 m/s, while a speeder is coming up from almost directly behind at a speed of 37 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 6.00 x 10^9 Hz. Find the the difference between the frequency of the wave that returns to the police car after reflecting from the speeder's car, and the original frequency emitted by the police car.
F = ((V+Vr) / (V+Vs))*Fs.
F = ((343+37) / (343+26))*6*10^9 = 6.18*10^9 Hz.
F - Fs = 6.18*10^9 - 6.00*10^9 = 1.79*10^8 Hz.
To find the difference in frequency between the wave reflected from the speeder's car and the original frequency emitted by the radar gun, we need to use the Doppler effect.
The Doppler effect describes how the frequency of a wave changes when there is relative motion between the source of the wave and the observer. In this case, the source of the wave is the radar gun (police car) and the observer is the police officer.
The formula for the observed frequency f' is given by:
f' = f * (c + vo) / (c + vs)
Where:
f' = observed frequency (frequency of the wave that returns to the police car)
f = original frequency emitted by the radar gun (given as 6.00 x 10^9 Hz)
c = speed of light (approximately 3.00 x 10^8 m/s)
vo = velocity of the observer (velocity of the police car)
vs = velocity of the source (velocity of the speeder's car)
First, let's find the velocity of the observer (police car). Since it is moving to the right, its velocity is positive 26 m/s.
vo = 26 m/s
Next, let's find the velocity of the source (speeder's car). Since it is coming up from almost directly behind, its velocity is negative 37 m/s.
vs = -37 m/s
Now we can substitute the values into the formula for the observed frequency:
f' = (6.00 x 10^9 Hz) * (3.00 x 10^8 m/s + 26 m/s) / (3.00 x 10^8 m/s - 37 m/s)
Simplifying the equation:
f' = (6.00 x 10^9 Hz) * (3.26 x 10^8 m/s) / (2.63 x 10^8 m/s)
f' = 7.45 x 10^9 Hz
Therefore, the difference between the frequency of the wave that returns to the police car after reflecting from the speeder's car and the original frequency emitted by the police car is 7.45 x 10^9 Hz.
To find the difference in frequency between the wave reflected from the speeder's car and the original frequency emitted by the police car, we need to consider the Doppler effect.
The Doppler effect describes the change in frequency of a wave (in this case, the electromagnetic wave emitted by the radar gun) due to relative motion between the source of the wave (the police car) and the observer (the police officer).
In this scenario, the police car is moving towards the speeder's car. This means that the observer (police officer) is stationary and the source (police car) is moving.
Let's start by finding the speed of the radar wave relative to the speeder's car. We can use the formula:
v_wave = v_source - v_observer
where,
v_wave is the speed of the wave relative to the speeder's car,
v_source is the velocity of the source (police car),
and v_observer is the velocity of the observer (police officer).
Given:
v_source = 26 m/s (positive because it's moving to the right)
v_observer = 0 m/s (police officer is stationary)
Now, substituting the given values into the formula:
v_wave = 26 m/s - 0 m/s
v_wave = 26 m/s
So, the speed of the radar wave relative to the speeder's car is 26 m/s.
Next, we'll apply the Doppler effect formula to find the difference in frequency.
Δf/f = ((v_wave ± v_observer) / v_wave)
where,
Δf/f is the relative change in frequency,
v_wave is the speed of the radar wave relative to the speeder's car (26 m/s),
and v_observer is the velocity of the observer (police officer, 0 m/s).
Substituting the values into the formula:
Δf/f = ((26 m/s - 0 m/s) / 3 x 10^8 m/s)
Δf/f = (26 m/s) / (3 x 10^8 m/s)
Calculating this, we get:
Δf/f ≈ 8.67 x 10^-8
This means that the difference in frequency between the wave that returns to the police car after reflecting from the speeder's car and the original frequency emitted by the police car is approximately 8.67 x 10^-8.