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Does anybody understand this OCR unified physics Question?

This is one of those ones where I'm no better off after looking at the mark scheme. If anybody could offer a brief explanation to each part, that would be great.
Reply 1
also waitingfor an explanation :smile:
some genius saveus please..
Reply 2
can you post the mark scheme? i might be able to explain it
Original post by Inker
can you post the mark scheme? i might be able to explain it
Here
Reply 4
Original post by MantisToboggan
This is one of those ones where I'm no better off after looking at the mark scheme. If anybody could offer a brief explanation to each part, that would be great.


Interference and Doppler shift


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Original post by Kyx
Interference and Doppler shift


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Yes, those are the two topics it is based on but I'm afraid that brings me no closer to understanding the question.
Reply 6
Original post by MantisToboggan
Yes, those are the two topics it is based on but I'm afraid that brings me no closer to understanding the question.


You know the wavelength of the radiation.

What happens when the two reflecting plates move apart? It's essentially the same as moving two sources of water waves apart :smile:

Also the reflected waves will interfere with the none reflected waves

(I assume it's the interference part that you're stuck on? :smile:)


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(edited 6 years ago)
Reply 7
Original post by Kyx
You know the wavelength of the radiation.

What happens when the two reflecting plates move apart? It's essentially the same as moving two sources of water waves apart :smile:

Also the reflected waves will interfere with the none reflected waves

(I assume it's the interference part that you're stuck on? :smile:)


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I don't know... what happens?
Reply 8
Original post by Inker
I don't know... what happens?


Imagine it as a stationary wave on a string.

Keep the frequency and wavelength the same, but make the string smaller by 1/2 a wavelength


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Reply 9
Original post by Kyx
Imagine it as a stationary wave on a string.

Keep the frequency and wavelength the same, but make the string smaller by 1/2 a wavelength


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so the stationary wave dies?
Reply 10
Original post by Inker
so the stationary wave dies?


The phase difference changes, so the resultant wave goes from maximum amplitude to minimum amplitude and back again :smile:


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Reply 11
Original post by Kyx
The phase difference changes, so the resultant wave goes from maximum amplitude to minimum amplitude and back again :smile:


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Is there a video demonstrating this anywhere? I can't imagine it.. :frown:
Reply 12
Original post by Inker
Is there a video demonstrating this anywhere? I can't imagine it.. :frown:


There might be one if you google stationary wave interference :s-smilie:


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Original post by Inker
Is there a video demonstrating this anywhere? I can't imagine it.. :frown:


I'm going to reimagine the situation for you, if you're still having issues.

The Hardboard (H) is only a partial reflector of the microwaves however the metal (M) will completely reflect the microwaves.

Hence, any microwaves passing through the hardboard will be reflected by the metal. These reflected waves will then pass through the hardboard again and interact with the waves already reflected by the hardboard.

As is mentioned in the textbook, when waves have a phase difference of pi there will be destructive interference, causing a minima. When there is a phase difference of 2pi there will be a maxima, use this information to help answer the question.

Any further queries please don't hesitate to ask.

Good Luck for Thursday
Original post by TheJonesy138
I'm going to reimagine the situation for you, if you're still having issues.

The Hardboard (H) is only a partial reflector of the microwaves however the metal (M) will completely reflect the microwaves.

Hence, any microwaves passing through the hardboard will be reflected by the metal. These reflected waves will then pass through the hardboard again and interact with the waves already reflected by the hardboard.

As is mentioned in the textbook, when waves have a phase difference of pi there will be destructive interference, causing a minima. When there is a phase difference of 2pi there will be a maxima, use this information to help answer the question.

Any further queries please don't hesitate to ask.

Good Luck for Thursday


Since the hardboard sheet is moving won't there be a Doppler shift in the frequency of the waves reflected off H? This change in frequency isn't present in the waves reflected off the metal sheet so surely the two reflected waves aren't coherent as they have different frequencies? So how can a stable interference pattern/stationary wave be set up?
Original post by TheHobbit
Since the hardboard sheet is moving won't there be a Doppler shift in the frequency of the waves reflected off H? This change in frequency isn't present in the waves reflected off the metal sheet so surely the two reflected waves aren't coherent as they have different frequencies? So how can a stable interference pattern/stationary wave be set up?


Yes, there will be a Doppler shift from the reflected waves at H, however, I also think that the waves passing through H will also experience a Doppler shift, though admittedly I'm not totally sure about that.

Bear in mind that it is more important to answer the question than to get bogged down in the physics of the question. That is to say that what the question asks and the physics of the situation are not always the same thing.

For part 1, for instance, it simply asks you to discuss the factors as to whether minima and maxima can be detected. So, if you simply discuss the fact that waves which are reflected by M -if they pass through H- will have a decreased amplitude and also that these new waves will interfere with waves reflected from H. Then you can discuss what the values of the maxima and minima are. Note how it does not ask you to say where these minima and maxima will be.

Part 1 also asks whether they can be detected, leading you to suggest that the closer H is to T, the greater its amplitude will be (this is due to the spreading out aka dissipation of the wave), Due to this increased amplitude from H it becomes more difficult to compare the maxima and minima due to amplitude from M remaining small.

Then in part 2 the Doppler shift becomes slightly more relevant and is the reason why I believe that the waves passing through H receive a Doppler shift. The path difference between a max and a min is lambda/4 which is given in the question to be 7mm. For one pulse (or cycle) the path difference is 2 times that. Then we can use the wave speed equation with lambda/2. This is another thing I'm somewhat apprehensive about but I believe it is due to the sound being produced. By which I mean that the positive amplitude sounds the same as the negative amplitude - meaning we should be able to use lamda/2. As I've said already I'm unsure on this particular part especially because it's not something which I've ever used nor seen used by anyone else.

Finally, for b, the difference in time for which the two pulses have been travelling is t0-t due to t being smaller than t0 because of the car approaching the detector. The difference in distance travelled by the pulses in this time difference is c*(t0-t) but this is two times the distance travelled by the car. This simplifies to the given expression.

All in all, this is quite a hellish question. There's quite a lot to get your head round and some funny concepts which a large percentage of candidates simply wouldn't be able to do. I would like to think that I'm a fairly strong candidate but this question is something else. I find it unlikely that we will be asked something as peculiar as this.

Sorry for the long post, all the best, and good luck :smile:

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