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 6.00 109 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.
deltaf = 2*fo*(deltaV)/c
= 2*6*10^9*11/3*10^8 = 440 Hz
The shift is toward higher frequency
The factor of 2 takes into account frequency shifts of the incident and reflected wave.
There is a more accurate formula for the Doppler shift of electromagnetic waves, but this is close enough for motion velocities much less than c.
To find 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, you can use the Doppler Effect equation.
The Doppler effect equation in this case can be written as:
Δf/f = (v_r - v_o) / v_o
where Δf is the change in frequency, f is the original frequency emitted by the police car, v_r is the velocity of the reflector (speeder's car) relative to the radar gun, and v_o is the velocity of the observer (police car) relative to the radar gun.
In this case, the radar gun emits a wave with a frequency of 6.00 × 10^9 Hz.
Given that the police car is moving to the right at 26 m/s and the speeder is coming up from almost directly behind at a speed of 37 m/s, we need to calculate the relative velocity of the reflector and the observer.
The relative velocity of the reflector relative to the radar gun (speeder's car relative to the police car) can be calculated by subtracting the velocity of the observer (police car) from the velocity of the reflector (speeder's car):
v_r = v_s - v_p
where v_s is the velocity of the speeder's car (37 m/s) and v_p is the velocity of the police car (26 m/s).
Substituting the values:
v_r = 37 m/s - 26 m/s
v_r = 11 m/s
Now, we can calculate the relative velocity of the observer (police car) with respect to the radar gun by subtracting the velocity of the radar gun (0 m/s) from the velocity of the observer (police car):
v_o = v_p - v_g
where v_p is the velocity of the police car (26 m/s) and v_g is the velocity of the radar gun (0 m/s).
Substituting the values:
v_o = 26 m/s - 0 m/s
v_o = 26 m/s
Now, we can substitute the values of v_r and v_o into the Doppler effect equation to calculate the change in frequency:
Δf/f = (v_r - v_o) / v_o
Δf/f = (11 m/s - 26 m/s) / 26 m/s
Δf/f = -15 m/s / 26 m/s
Δf/f ≈ -0.577
The negative sign indicates that the frequency observed by the police car after reflecting from the speeder's car will be lower than the original frequency emitted.
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 approximately -0.577.
To find the difference in frequency, also known as the Doppler shift, between the wave that returns to the police car and the original frequency emitted by the radar gun, we can use the Doppler effect equation:
Δf = (f_r - f_e) / f_e
Where:
Δf is the difference in frequency,
f_r is the frequency of the wave that returns to the police car, and
f_e is the original frequency emitted by the radar gun.
In this case, the police car is the source of the wave, and the speeder's car is the observer.
First, let's calculate the frequency of the wave that returns to the police car, f_r. The Doppler effect equation for the frequency observed by the observer, in this case, the police car, is given by:
f_r = f_s * (v + v_o) / (v - v_s)
Where:
f_s is the original frequency emitted by the source, which is the speed of the radar gun, 6.00 * 10^9 Hz.
v is the speed of the wave, which is the speed of light, 3.00 * 10^8 m/s.
v_o is the speed of the observer, which is the speed of the police car, 26 m/s.
v_s is the speed of the source, which is the speed of the speeder's car, -37 m/s (negative since it's moving towards the observer).
Now we can calculate f_r:
f_r = (6.00 * 10^9 Hz) * (3.00 * 10^8 m/s + 26 m/s) / (3.00 * 10^8 m/s - (-37 m/s))
= (6.00 * 10^9 Hz) * (3.00 * 10^8 m/s + 26 m/s) / (3.00 * 10^8 m/s + 37 m/s)
≈ 6.02 * 10^9 Hz
Next, we substitute the values into the Doppler shift equation to find Δf:
Δf = (6.02 * 10^9 Hz - 6.00 * 10^9 Hz) / (6.00 * 10^9 Hz)
= 0.02 / (6.00 * 10^9 Hz)
≈ 3.33 * 10^-12 Hz
Therefore, the difference in frequency is approximately 3.33 * 10^-12 Hz.