Doppler Effect Question

[Strawberries]

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Can someone explain the following? (ie the math set up). Thanks!

 

[ilovemcat]

Can someone explain the following? (ie the math set up). Thanks!

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.

 

[Strawberries]

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.

 

[dmplz707]

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!

 

[Arctangent]

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.

 

[dmplz707]

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.

Thank you very much for the response! Definitely helps out visualizing it and trying to understand it conceptually.

 

[becomingadoctor]

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!

There is an error in the math. the numerator should be 340 + 5 because the cyclist is moving TOWARDS the bus so it is trying to achieve an increase in frequency and an increase in frequency would be achieved by increasing the numerator. On the contrary, the bus is moving AWAY from the cyclist so it is trying to achieve a decrease in frequency and a decrease in frequency would be achieved by increasing the denominator. The net effect is movement away in terms of the bus. It is best to look at the numerator and denominator separately to make the judgment call. Look at it in terms of what effect each of the relative movements are trying to achieve in terms of frequency. If something moves towards another thing, the perceived frequency will be higher. If something moves away from the source, the perceived frequency will be lower.

Doing the math gives you the source frequency to be about 1057 Hz which rounds to 1060 Hz. The equation where 340-5 is in the numerator gives you about 1089 Hz.