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The Physics PHYA2 thread! 5th June 2013

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Original post by Sukura
Can you please help me on a question?

Question 6di from June 2010!

I'd truly appreciate it


need to do 1/300 to get the answer in mm, and then convert to m by multiplying by 10 to the power of -3
Reply 861
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Edorrans
Could anyone help me with the wave questions? especially the diffraction garting and the light internally reflecting! ASAP please :smile: thank you!
Original post by iamsocoollol
hi guys can some one please upload the MS AND QP jan 2013 unit 2 pleaseeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeee

my stupid teacher forgot to email :frown:

Why not m8? Here you go! :smile:
AQA-PHYA2-W-MS-JAN13.pdf186.5KB
AQA-PHYA2-QP-Jan13.pdf686.1KB
Original post by Edorrans
Could anyone help me with the wave questions? especially the diffraction garting and the light internally reflecting! ASAP please :smile: thank you!


Please give the question and the exam paper (year). Thanks.
Does anyone know how to prove the energy stored E=1/2kdeltaL formula..?
Original post by x-Sophie-x
Does anyone know how to prove the energy stored E=1/2kdeltaL formula..?


The area under the curve (is the triangle which is given by a=1/2ab). Your a is the x-axis and b is the y-axis. Or something like that. Just replace the energy stored equation with the area and replace those delta L and K with the a and b. Is that ok? :smile:
Original post by x-Sophie-x
Does anyone know how to prove the energy stored E=1/2kdeltaL formula..?


Average force exerted on spring = 0 + f divided by 2 = f/2

Work done = force x distance

Workdone/energy stored = 1/2f x deltaL
Reply 867
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Tuya
4
Original post by x-Sophie-x
Does anyone know how to prove the energy stored E=1/2kdeltaL formula..?


The energy is area under the graph which is the triangle area if f vs delta k is a straight line.


Posted from the TSR iPad app
Reply 868
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Tuya
4
ImageUploadedByStudent Room1370364640.823812.jpg


Posted from the TSR iPad app
The area under the graph is a triangle for (straight line) graphs with extension and force on the axes, which act as base and height in area of traingle =1/2 x base x height
Original post by Tuya

Which question?
Reply 871
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Edorrans
Original post by StalkeR47
Please give the question and the exam paper (year). Thanks.


Jan 12 - Question 5 please
Reply 872
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henpen
Original post by Sukura
Can you please help me on a question?

Question 6di from June 2010!

I'd truly appreciate it


There are 300300 slits per 10−3m10^{-3}m, so there is 11 slit every 10−3300m\frac{10^{-3}}{300}m
Original post by BenChard
image.jpg
image.jpg

the ms says 1.4 my calc gives 0.6 -.-


Draw a tangent and find the gradient ?
Original post by Edorrans
Jan 12 - Question 5 please


This?
Image000006.jpg32.4KB
Done all the past papers, this question was the only one really that made no sense at all.



The question asked to state the phase relationship between x and y, and x and z, I calculate that X and Y are 135 degrees out of phase, and x and z are 315 degrees out of phase. However the mark scheme for the June 2012 paper states that the phase relationship between x and y is completely out of phase/180 degrees, and between x and z is 360/0/2pi radians completely in phase.


It makes no sense at all? Please help.
Reply 876
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Sukura
Original post by BigBadJFly
need to do 1/300 to get the answer in mm, and then convert to m by multiplying by 10 to the power of -3

Oh!! I forgot to convert... Thank you oh so much! Was wondering why I was 10x-3 out
Original post by StalkeR47
Which exam paper and which q?


jun10 q3d
Reply 878
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Edorrans
Original post by StalkeR47
This?


Yeah thats great thank you! Could you explain what the difference would be between the white light and the red laser in that drawing?
Original post by yorkshire.lad
Done all the past papers, this question was the only one really that made no sense at all.



The question asked to state the phase relationship between x and y, and x and z, I calculate that X and Y are 135 degrees out of phase, and x and z are 315 degrees out of phase. However the mark scheme for the June 2012 paper states that the phase relationship between x and y is completely out of phase/180 degrees, and between x and z is 360/0/2pi radians completely in phase.


It makes no sense at all? Please help.


I did the same thing as you :L

X and Z are in phase because they are 'doing the same thing at the same time'
X and Y are antiphase because they are doing different things.
This applies because the wave is a stationary wave :')

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