Two ships in a heavy fog are blowing their horns, both of which produce sound with a frequency of 165.0 Hz One ship is at rest; the other moves on a straight line that passes through the one at rest.
If people on the stationary ship hear a beat frequency of 3.5 Hz, what is the speed of the approaching ship?
If people on the stationary ship hear a beat frequency of 3.5 Hz, what is the speed of the receding ship?
There are two possible received frequencies that would produce that beat frequency. One corresponds to the ships approaching each other and the other for when they are separating. They don't say clearly which case it is.
Assume they are approaching. In that case the received frequency is 168.5 Hz
Use the appropriate Doppler shift formula to compute the moving ship's velocity. Both cases should give NEARLY the same answer, about 7 m/s assuming a sound speed of 340 m/s. You will need to assume an air temperature to get an accurate sound speed.
To determine the speed of the approaching and receding ships, we can use the Doppler effect formula:
f' = f (v + vr) / (v + vs)
Where:
f' is the observed frequency
f is the actual frequency emitted by the source
v is the speed of sound in air (approximately 343 m/s)
vs is the speed of the source (ship)
vr is the speed of the receiver (stationary ship)
Since the frequency of the horn is 165.0 Hz and the observed beat frequency is 3.5 Hz, we can substitute these values into the formula.
For the approaching ship:
f' = 165.0 - 3.5 = 161.5 Hz
For the receding ship:
f' = 165.0 + 3.5 = 168.5 Hz
Using the formula, we can solve for the speed of the ships.
For the approaching ship:
161.5 = 165.0 (343 + vs) / (343)
161.5 * 343 = 165.0 * (343 + vs)
55599.5 = 165.0 * 343 + 165.0 * vs
55599.5 - 165.0 * 343 = 165.0 * vs
55599.5 - 56145 = 165.0 * vs
-545.5 = 165.0 * vs
vs = -545.5 / 165.0
vs ≈ -3.31 m/s
Therefore, the speed of the approaching ship is approximately -3.31 m/s.
For the receding ship:
168.5 = 165.0 (343 - vr) / (343)
168.5 * 343 = 165.0 * (343 - vr)
57815.5 = 165.0 * 343 - 165.0 * vr
57815.5 - 165.0 * 343 = -165.0 * vr
57815.5 - 56145 = -165.0 * vr
1670.5 = -165.0 * vr
vr = 1670.5 / -165.0
vr ≈ -10.12 m/s
Therefore, the speed of the receding ship is approximately -10.12 m/s.
To solve these problems, we need to use the formula for the Doppler effect:
f' = f((v + vr) / (v + vs)),
where f' is the observed frequency, f is the emitted frequency, vr is the speed of the receiver (stationary ship), and vs is the speed of the source (moving ship). In this case, the observed frequency is the beat frequency, which is 3.5 Hz, and the emitted frequency is 165.0 Hz.
First, let's find the speed of the approaching ship (vs). Since the ships are moving towards each other, the observed frequency will be higher than the emitted frequency. We can rearrange the formula to solve for vs:
vs = ((f' / f) - 1) * v - vr.
Plugging in the given values:
- f' = 3.5 Hz,
- f = 165.0 Hz,
- v = speed of sound (assume 343 m/s),
- vr = 0 m/s (since the ship is stationary),
we get:
vs = ((3.5 Hz / 165.0 Hz) - 1) * 343 m/s.
Now we can calculate the value of vs:
vs = ((0.0212) * 343 m/s.
vs ≈ 7.21 m/s.
Therefore, the speed of the approaching ship is approximately 7.21 m/s.
To find the speed of the receding ship, we use the formula:
vs = ((vr - f' / f) / (vr - v)).
Since the receding ship is moving away from the stationary ship, the observed frequency will be lower than the emitted frequency. Rearranging the formula to solve for vs:
vs = ((vr - f' / f) / (vr - v)).
Plugging in the given values:
- f' = 3.5 Hz,
- f = 165.0 Hz,
- v = 343 m/s (speed of sound),
- vr = 0 m/s (since the ship is stationary),
we have:
vs = (((0 m/s) - 3.5 Hz / 165.0 Hz) / ((0 m/s) - 343 m/s)).
Now we can calculate the value of vs:
vs = ((-0.0212) / (-343 m/s)).
vs ≈ 0.000062 m/s.
Therefore, the speed of the receding ship is approximately 0.000062 m/s.