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 x 10^9 Hz.
(a) 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
(b) Which wave has the greater frequency?
-The emitted wave has the greater frequency.
-The wave that returns to the police car has the greater frequency.
Why?
To answer this question, we need to consider the Doppler effect, which is the change in frequency and wavelength of a wave when there is relative motion between the source of the wave and the observer.
Let's break down the problem step-by-step:
Step 1: Find the relative velocity between the police car and the speeder.
The relative velocity is the difference between the velocities of the police car and the speeder. In this case, the police car is moving to the right at 20 m/s, while the speeder is coming up from almost directly behind at 31 m/s.
Relative velocity = speeder's velocity - police car's velocity
Relative velocity = 31 m/s - 20 m/s
Relative velocity = 11 m/s
Step 2: Apply the Doppler effect equation to find the frequency shift.
The Doppler effect equation for sound waves is often used in problems involving radar waves or light waves as an approximation.
The frequency shift (Δf) is given by the equation:
Δf = [(v_receiver - v_source) / c] * f_source
where:
Δf = frequency shift
v_receiver = velocity of the receiver (police car)
v_source = velocity of the source (speeder)
c = speed of light (approximately 3.0 x 10^8 m/s)
f_source = frequency of the source wave (emitted wave)
In this case, the police car is the receiver and the speeder is the source.
Step 3: Calculate the frequency shift.
Δf = [(v_receiver - v_source) / c] * f_source
Δf = [(20 m/s - (-31 m/s)) / 3.0 x 10^8 m/s] * (9.0 x 10^9 Hz)
Simplifying:
Δf = [51 m/s / 3.0 x 10^8 m/s] * (9.0 x 10^9 Hz)
Δf = (1.7 x 10^-7) * (9.0 x 10^9 Hz)
Δf = 1.53 Hz
Step 4: Determine 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.
The magnitude of the frequency difference is the absolute value of the frequency shift:
Magnitude of the frequency difference = |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.
Now let's move on to part (b):
Step 5: Determine which wave has the greater frequency.
To determine which wave has the greater frequency, we need to compare the frequency of the emitted wave with the frequency of the wave that returns to the police car.
The frequency of the emitted wave is given as 9.0 x 10^9 Hz.
The frequency of the wave that returns to the police car is the sum of the frequency of the emitted wave and the frequency difference calculated in step 4:
Frequency of the returning wave = 9.0 x 10^9 Hz + 1.53 Hz
Since 1.53 Hz is a very small value compared to 9.0 x 10^9 Hz, we can neglect the frequency difference when compared to the emitted wave frequency.
Therefore, the emitted wave has the greater frequency.
In summary:
(a) 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.
(b) The emitted wave has the greater frequency.
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 take into account the relative motion between the police car, the speeder's car, and the radar gun.
(a) The Doppler effect is the phenomenon that describes the change in frequency of a wave due to the motion of the source or the observer. In this case, the radar gun is the source of the wave and the speeder's car is the observer. Since the speeder's car is moving towards the radar gun, the frequency of the wave that returns to the police car will be higher than the emitted frequency.
The magnitude of the difference between the frequencies of the emitted wave and the returning wave can be calculated using the following equation:
Δf = f' - f
Where Δf is the frequency difference, f' is the frequency of the returning wave, and f is the frequency of the emitted wave.
Given:
- The emitted wave has a frequency of 9.0 x 10^9 Hz.
To find the frequency of the returning wave, we need to consider the relative velocities of the police car and the speeder's car. Since the speeder's car is coming up from almost directly behind the police car, their relative velocity can be calculated by subtracting the velocity of the police car from the velocity of the speeder's car.
Relative velocity = velocity of speeder's car - velocity of police car
Given:
- The velocity of the police car is 20 m/s to the right.
- The velocity of the speeder's car is 31 m/s.
Relative velocity = 31 m/s - 20 m/s
Relative velocity = 11 m/s
Now, we can calculate the frequency of the returning wave using the Doppler effect equation:
f' = (c + vr) / (c - vs) x f
Where c is the speed of the wave (which is the speed of light since it is an electromagnetic wave), vr is the relative velocity, vs is the velocity of the source (which is the velocity of the speeder's car), and f is the frequency of the emitted wave.
Given:
- The speed of light is approximately 3.0 x 10^8 m/s.
f' = (3.0 x 10^8 m/s + 11 m/s) / (3.0 x 10^8 m/s - 0 m/s) x (9.0 x 10^9 Hz)
f' = (3.0 x 10^8 m/s + 11 m/s) / (3.0 x 10^8 m/s) x (9.0 x 10^9 Hz)
f' ≈ 9.00000367 x 10^9 Hz
Now, we can calculate the magnitude of the difference between the frequencies:
Δf = f' - f
Δf = 9.00000367 x 10^9 Hz - 9.0 x 10^9 Hz
Δf ≈ 3.67 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 approximately 3.67 Hz.
(b) Since the frequency of the returning wave is higher than the emitted frequency, the wave that returns to the police car has the greater frequency.