A fire truck emitting a 450 Hz signal passes by a stationary detector. The difference in frequency measured by the detector is 58 Hz. If the speed of sound is 345 m/s, how fast is the fire truck moving?
The answer should be 22 m/s but I don't even know where to start. Could there not be two answers?
To solve this problem, we can use the concept of the Doppler effect, which describes the change in frequency of a wave as it is emitted or received by a moving object.
The formula we use to calculate the observed frequency is:
f' = f * (v + v_obs)/(v + v_source)
Where:
- f' is the observed frequency
- f is the emitted frequency
- v is the speed of sound
- v_obs is the velocity of the observer
- v_source is the velocity of the source
In this case, the detector is stationary (so v_obs = 0), and the fire truck is moving towards the detector. This implies that the frequency observed (f') will be higher than the emitted frequency (f).
We are given:
- f = 450 Hz
- f' - f = 58 Hz
- v = 345 m/s
We need to find v_source, the velocity of the fire truck.
Substituting the given values into the Doppler effect formula, we have:
f' = f * (v + v_obs)/(v + v_source)
450 Hz + 58 Hz = 450 Hz * (345 m/s + 0)/(345 m/s + v_source)
Simplifying the equation:
508 Hz = 450 Hz * (345 m/s)/(345 m/s + v_source)
508 Hz * (345 m/s + v_source) = 450 Hz * 345 m/s
(508 Hz * 345 m/s)/(450 Hz) = 345 m/s + v_source
v_source = (508 Hz * 345 m/s)/(450 Hz) - 345 m/s
v_source ≈ 26.067 m/s - 345 m/s
v_source ≈ -318.933 m/s
Since the velocity of the fire truck is positive (it is moving towards the detector), we discard the negative value.
Therefore, the velocity of the fire truck is approximately 318.933 m/s.
It's important to note that the fire truck's speed is the magnitude of the velocity, so the speed is approximately 318.933 m/s.
To solve this problem, we can use the Doppler effect equation:
Δf = (vD / vS) * fS
Where:
- Δf is the difference in frequency measured by the detector
- vD is the velocity of the detector (in this case, the stationary detector)
- vS is the velocity of the source (in this case, the fire truck)
- fS is the emitted frequency
In this case, Δf = 58 Hz, fS = 450 Hz, and vD is assumed to be 0 since the detector is stationary.
Plugging in the known values into the formula:
58 Hz = (0 / vS) * 450 Hz
Since the denominator is zero, there are two possibilities:
1. When the fire truck is approaching the detector (vS > 0)
2. When the fire truck is moving away from the detector (vS < 0)
Since we're interested in the speed of the fire truck, we'll consider the case where the fire truck is approaching the detector (vS > 0). In this case, the emitted frequency will be higher, resulting in a positive Δf.
Now, let's solve for vS:
58 Hz = (0 / vS) * 450 Hz
58 Hz * vS = 0 * 450 Hz
58 Hz * vS = 0
To avoid dividing by zero, we assume that vS approaches 0, but is not actually zero. Thus, we have:
58 Hz * vS = 0
vS = 0
Since the fire truck is moving towards the stationary detector (vS > 0), this value is not valid for our case. Therefore, the speed of the fire truck cannot be determined based on the given information.
Hence, we cannot conclude that the speed of the fire truck is 22 m/s based on the provided information.
If the listener does not move
Doppler:
Fl = { v/(v+vs) ] Fs
Fl is frequency heard
Fs is source frequency
vs = speed of source AWAY from listener
v = speed of sound
here:
when it is coming at you
FL = [345/(345-vs) ] 450
going awy:
Fl = [ 345/(345+vs) ] 450
so
22= 345*450 / [1/(345-vs) -1/(345+vs)]
1.417*10^-4 = [345+vs -345 +vs]/(345^2-vs^2)
1.417*10^-4 = [2 vs]/(345^2-vs^2)
I am going to say 345^2 >> vs^2
then
vs = 8.43 m/s