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# AS Physics new spec HELP!

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Mark scheme:
Attachment 530855530859

The part I don't understand is: path difference = 2 * distance moved
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2. Bump!

Mark scheme:
Attachment 530855530859

The part I don't understand is: path difference = 2 * distance moved
Can you please include a picture of Figure 1? as I believe this question is connected with harmonic/fundamental frequencies and I need to see the diagram to be sure
4. (Original post by sue99)
Can you please include a picture of Figure 1? as I believe this question is connected with harmonic/fundamental frequencies and I need to see the diagram to be sure

Here's a link to the actual paper: http://filestore.aqa.org.uk/resource...-74071-SQP.PDF
Its question 3.
The markscheme can be found here: http://filestore.aqa.org.uk/resource...-74071-SMS.PDF

Here's a link to the actual paper: http://filestore.aqa.org.uk/resource...-74071-SQP.PDF
Its question 3.
The markscheme can be found here: http://filestore.aqa.org.uk/resource...-74071-SMS.PDF
The distance moved by the tube is d2-d1. The path difference is 2 times that because the wave travels that distance moved at the top of the tube and at the bottom of the tube
6. (Original post by veejn)
The distance moved by the tube is d2-d1. The path difference is 2 times that because the wave travels that distance moved at the top of the tube and at the bottom of the tube
Well you destroyed my confusion in one sentence....my teacher couldn't even tell me that! Bloody hell that was simple. Thanks very much!

Here's a link to the actual paper: http://filestore.aqa.org.uk/resource...-74071-SQP.PDF
Its question 3.
The markscheme can be found here: http://filestore.aqa.org.uk/resource...-74071-SMS.PDF
Okay, so this is how I understand it:

for sound to be at a minimum at any point, the waves must have interfered destructively, ie. the path difference is a half number of wavelengths - lamda/2

so wavelength = v / f
= ....

then because the path difference is lamda/2 , you just divide the value of the wavelength by 2
8. (Original post by sue99)
Okay, so this is how I understand it:

for sound to be at a minimum at any point, the waves must have interfered destructively, ie. the path difference is a half number of wavelengths - lamda/2

so wavelength = v / f
= ....

then because the path difference is lamda/2 , you just divide the value of the wavelength by 2
Yep so path difference is 0.2125m. As veejn mentioned above; the path difference is divided by two and this was what I didn't quite understand. It's because there's the top part of the tube and bottom part of the tube as seen in figure 1. That's why we divide by two since both lengths are equal and when we divide by two, we find the length of just one of the top/bottom part of the tube which is the length that the tubes been extended by.
Yep so path difference is 0.2125m. As veejn mentioned above; the path difference is divided by two and this was what I didn't quite understand. It's because there's the top part of the tube and bottom part of the tube as seen in figure 1. That's why we divide by two since both lengths are equal and when we divide by two, we find the length of just one of the top/bottom part of the tube which is the length that the tubes been extended by.
Haha you're too nice; I didn't really help at all! But I'm glad you understand it now!
Well you destroyed my confusion in one sentence....my teacher couldn't even tell me that! Bloody hell that was simple. Thanks very much!
No worries

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