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Can someone explain the following? (ie the math set up). Thanks!
Can someone explain the following? (ie the math set up). Thanks!
Thanks for the help. I just realized I misread the question - I thought that the frequency of the source was 1000 Hz.This is a Doppler Effect Question. The relative velocity between the source of the frequency (bus), and the observer (bike) is 20 m/s. Since the bus is moving further and further away each second, we can make a prediction that the frequency observed by the cyclist will decrease. Since we are told that the frequency OBSERVED by the cyclist is 1000 Hz, the SOURCE frequency must be HIGHER. We can rule out choices A and B.
Frequency Source: ?
Frequency Detected: 1000 Hz
Velocity Detector: 5 m/s
Velocity Source: 25 m/s
Speed of Sound: 340 m/s
Figure it out the rest. The answer should seem fairly obvious.
PS - Where did you get this question from? If it's from AAMC, you should put that in title in the future because people don't want to preview questions before hand.
Thanks for the help. I just realized I misread the question - I thought that the frequency of the source was 1000 Hz.
No its not AAMC; its TPR.
Hey guys sorry to bring up this old thread. I searched it through the archives since I was looking for some doppler effect questions to practice on. Anyways, I wanted to know if anyone knows of a more "intuitive" way to solve this question since plugging in the numbers and solving for the frequency of the the source (fs) would take a too long on a real mcat. Here's the work that I've done;
fo = fs (v-vo / v+vs)
1000 = fs (340-5 / 340+25)
So what would be the more intuitive way to solve this question? Thanks guys!
One way I think of it when I want a general answer (greater/less than/equal to fs) is to imagine the waves being emitted with respect to if the source/observer are converging or diverging.
In this problem, the bus and the biker are diverging, or getting farther apart (25m/s > 5m/s). The waves continually need to travel a greater and greater distance to go from source to observer. Longer time between cycles/waves, or in other words, less frequent waves/cycles means that frequency decreases compared to if the distance between the two objects were constant. Or in other words, the perceived frequency should be less than the actual frequency.
In this question then, A and B would be ruled out, and from experience or by looking at the formula for a bit, you know you would need very extreme conditions for about a 2-fold change in frequency, so D is very unlikely. Thus, you would choose C.
Hey guys sorry to bring up this old thread. I searched it through the archives since I was looking for some doppler effect questions to practice on. Anyways, I wanted to know if anyone knows of a more "intuitive" way to solve this question since plugging in the numbers and solving for the frequency of the the source (fs) would take a too long on a real mcat. Here's the work that I've done;
fo = fs (v-vo / v+vs)
1000 = fs (340-5 / 340+25)
So what would be the more intuitive way to solve this question? Thanks guys!