Problem 9
Formula and Givens
- f' = observed frequency ?
- f = actual frequency 300 Hz
- v = velocity of sound or light waves. 343 m/s
- vo = velocity of observer 25 m/s
- vs = velocity of source. 50 m/s
f' = (v - vo) * f / (v + vs)
This time it really matters how the equations is set up. In point of fact the trains are moving away from each other and the observer and source are going in opposite directions. That will lower the frequency.
So you want the numerator to decrease and the denominator to increase. This will give you a minimum reading.
f' = (343 - 25) * f / (343 + 50)
f' = 318 * 300 / 393
f' = 242
Answer C
Problem 10
This problem is both instructive and very subtle. There are two shifts that take place. The first is the stationary car (vs = 0) to the moving car (vo ≠ 0) and the second is when the moving car reflects(vs ≠0) the signal back to the observer (who is not moving vo = 0)
First shift.
(v + vo)/(v + vs) = (v + s)/v You have to add the speed of the moving car to the speed of sound to get the maximum in the numerator.
Second shift
(v + vo) / (v + vs) = v/(v - s) You need to subtract the speed of the car from the speed of sound to get the maximum. The over all maximum effect happens when the two shifts are multiplied together. When that happens the single "v"s are cancelled and you are left with
(v+s)/(v - s) Don't forget that s is the moving speed of the car. The rest of it is straight forward. You are solving for s
f' = f * (v + s)/(v - s) Substitute known values
1250 = 1200 (343 + s)/(343 - s) Divide both sides by 1200
1.04167 = (343 + s) / (343 - s) Multiply both sides by 343 - s
1.04167 * (343 - s) = 343 + s Remove the brackets.
357.29 - 1.04167*s = 343 + s Subtract 343 from both sides.
14.292 - 1.04167*s = s Add 1.04s to both sides.
14.292 = 2.04167s Divide by 2.04167
14.292/2.04167 = s
s = 6.999988
which the question rounds to 6.9. The difference in answers comes from what to use for v. SQDF thinks the V was 340 which is an acceptable approximation.
Full credit goes to him for figuring out the double shift. This is a remarkable piece of physics. Study to realize how radar works.
