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ALL CIE PHYSICS DOUBTS HERE !!!

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TJ-56:

--- Quote from: $!$RatJumper$!$ on November 08, 2010, 11:36:49 am ---i really dont understand 3 and 5.. please could you elaborate :)
and fir 4, the mark scheme says horizontal.. doesnt that mean 90 degrees? so why is it 36?

thankx a lot

--- End quote ---
For q5 (c):
To find the resultant intensity, you have to find the resultant amplitude first, since they are 180 degrees out of phase, find the difference between the amplitudes of 2 peaks (or troughs) which is (-) 1, so I(resultant)/I=1(sqr)/3(sqr)  simplified gives the resultant intensity of 1/9.
(you could also use the intensity 4/9 * I but with amplitude of 2 squared)
Hope that was helpful.

$!$RatJumper$!$:
ok thankx that makes sense now :) and jus a query, when finding resultant amplitudes, dont we ADD the amplitudes. like in this question, i think they added them in order to get their resultant amplitude as 0.0001

TJ-56:

--- Quote from: $!$RatJumper$!$ on November 08, 2010, 12:06:32 pm ---ok thankx that makes sense now :) and jus a query, when finding resultant amplitudes, dont we ADD the amplitudes. like in this question, i think they added them in order to get their resultant amplitude as 0.0001

--- End quote ---
That is correct, we add them.
Take for example the 1st peak on wave A, wave A's amp is 3, and wave B's amp is -2 at that time, so 3 + (-2) is 1 x 10^-4

$!$RatJumper$!$:
ok :) so is it a rule that we should ALWAYS ADD amplitudes to find the resultant?

TJ-56:

--- Quote from: $!$RatJumper$!$ on November 08, 2010, 12:11:15 pm ---ok :) so is it a rule that we should ALWAYS ADD amplitudes to find the resultant?

--- End quote ---
I think so yeah, theres another question regarding this topic on w02 qp2 q5 b iii , so, positive

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