While standing on 3rd street you hear an approaching ambulance. Its siren oscillates in frequency between about 650 Hz and 750 Hz.
Later on as you near your destination, you come upon the scene of the accident. The now-stationary ambulance runs its siren as it's about to drive off. This time, you hear its pitch oscillate between 637.65 Hz and 735.75 Hz.
Using your physics knowledge, you estimate the speed that the ambulance must have been traveling while en route to the scene of the accident. How fast was it traveling? (Assume it's an average day where the speed of sound is about 343 m/s.)
A. 25 m/s (about 55 mph)
B. 18 m/s (about 40 mph)
C. 13 m/s (about 30 mph)
D. 9 m/s (about 20 mph)
To estimate the speed of the ambulance, we can use the Doppler effect, which describes the change in frequency of a wave when the source and observer are in motion relative to each other.
First, let's calculate the change in frequency of the ambulance siren as it approaches you on 3rd street. The siren oscillates in frequency between 650 Hz and 750 Hz.
The formula for the Doppler effect is given by:
f' = f * (v + vo) / (v + vs)
Where:
f' is the observed frequency
f is the emitted frequency
v is the speed of sound
vo is the observer's velocity (in this case, 0 since you are standing still)
vs is the source's velocity (the speed of the ambulance)
Let's calculate the observed frequency (f') when the siren is at its lowest and highest frequencies:
f'1 = f1 * (v + vo) / (v + vs)
f'2 = f2 * (v + vo) / (v + vs)
Now, let's calculate the change in frequency when you reach the accident scene. The siren oscillates in frequency between 637.65 Hz and 735.75 Hz.
f'3 = f3 * (v + vo) / (v + vs)
f'4 = f4 * (v + vo) / (v + vs)
We know that the frequency at the accident scene is lower than when the ambulance was approaching you since the ambulance is now moving away from you.
f'4 < f'3 => (f4 * (v + vo) / (v + vs)) < (f3 * (v + vo) / (v + vs))
By substituting the given frequencies and the speed of sound, we have the following equation:
(735.75 Hz * (343 m/s + 0) / (343 m/s + vs)) < (637.65 Hz * (343 m/s + 0) / (343 m/s + vs))
Now, we need to find the value of vs (the speed of the ambulance).
By simplifying the equation:
(735.75 / (343 + vs)) < (637.65 / (343 + vs))
To solve this inequality, we can cross multiply:
(735.75 * (343 + vs)) < (637.65 * (343 + vs))
Simplifying further:
252337.75 + 735.75vs < 218703 + 637.65vs
Now, let's isolate vs:
735.75vs - 637.65vs < 218703 - 252337.75
98.1vs < -33634.75
vs < -343.162
Since speed cannot be negative, we discard this solution.
Therefore, it means that the speed of the ambulance is less than the speed of sound (343 m/s).
Now, let's look at the answer choices:
A. 25 m/s (about 55 mph)
B. 18 m/s (about 40 mph)
C. 13 m/s (about 30 mph)
D. 9 m/s (about 20 mph)
Since the speed of the ambulance is less than the speed of sound, we can rule out options A, B, C, and D. None of these speeds are less than 343 m/s.
Therefore, we cannot determine the exact speed of the ambulance with the given information.