Qualification > Sciences
ALL CIE PHYSICS DOUBTS HERE !!!
Deadly_king:
--- Quote from: TJ-56 on November 09, 2010, 11:01:27 am ---W07 Q 5(a) P2
phi/360=AB/lambda
(60/360) * lambda = AB
AB = 1/6 * 6
= 1
So it leads or lags behind (question didnt state which) by 1
see attachment
S06 Q6 P2
What part didnt u understand exactly?
--- End quote ---
Good work :)
He said he need the whole question. Do you mind to take care of it?
I've to help others.
$!$RatJumper$!$:
--- Quote from: TJ-56 on November 09, 2010, 11:01:27 am ---W07 Q 5(a) P2
phi/360=AB/lambda
(60/360) * lambda = AB
AB = 1/6 * 6
= 1
So it leads or lags behind (question didnt state which) by 1
see attachment
S06 Q6 P2
What part didnt u understand exactly?
--- End quote ---
Thankx that makes sense about the phase diff. But dont we need to do anything about the amplitude since it mentions its intensity is half? even the mark scheme says something about amplitude.
And if you could do all parts on q6 it would be wonderful :)
thankx
TJ-56:
--- Quote from: Deadly_king on November 09, 2010, 11:09:48 am ---Good work :)
He said he need the whole question. Do you mind to take care of it?
I've to help others.
--- End quote ---
(b) 32.4 cm is the distance between 1 nodes, i.e. half the distance of a complete wavelength. v=f * lambda
frequencey is stated, 512 Hz
Lambda= 32.4*2 *10^-2 (im meters)
so 512*64.8 * 10^-2 = 330 m\s
(c) the exact distance of the antinode is half 32.4
So 32.4/2 = 16.2
So the antinode is 16.2 cm above the surface of water in the tube on the left.
As 15.7 cm is the length of the column of air INSIDE the tube, then 16.2-15.7 is the length of air above the where the antinode is located.
Hope that was helpful
TJ-56:
--- Quote from: $!$RatJumper$!$ on November 09, 2010, 11:28:31 am ---Thankx that makes sense about the phase diff. But dont we need to do anything about the amplitude since it mentions its intensity is half? even the mark scheme says something about amplitude.
And if you could do all parts on q6 it would be wonderful :)
thankx
--- End quote ---
oh sorry my bad, i will post the correct answer with the amplitude part in a few min
TJ-56:
--- Quote from: $!$RatJumper$!$ on November 09, 2010, 11:28:31 am ---Thankx that makes sense about the phase diff. But dont we need to do anything about the amplitude since it mentions its intensity is half? even the mark scheme says something about amplitude.
And if you could do all parts on q6 it would be wonderful :)
thankx
--- End quote ---
say the maximum amp of the given wave is 2 (taking each 5 small blocks as 1)
then we have:
I(new)/I = A(new)²/A²
0.5I/I = A(new)²/(2A)²
Making A(new) ² the subject gives us = sqr root(2A²)
= sqr root(2) * A
which is 1.4 (the new amplitude)
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