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AS PHYSICS!!!

msk

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dude y d? see the relation is I is proportional to A square so 2I =2 A(square)...i.e square root 2 A ....so C is the answer
 

XPFMember

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Assalamoalaikum!!

well as msk said...

we have the relation I = A²
so 2I = 2A²

now amplitude in this case will be root of the whole thing which is root of (2A²) = [root of (2)] A


btw do u understand Q:22 of the same paper..i got it for my mocks and i had no idea abt it :!:
 
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msk said:
dude y d? see the relation is I is proportional to A square so 2I =2 A(square)...i.e square root 2 A ....so C is the answer
yeah i know, bt my teacher told me to find the K(constant) when dealing with these types of ques...anyways i gt it now..
 
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Math_angel said:
Assalamoalaikum!!

well as msk said...

we have the relation I = A²
so 2I = 2A²

now amplitude in this case will be root of the whole thing which is root of (2A²) = [root of (2)] A


btw do u understand Q:22 of the same paper..i got it for my mocks and i had no idea abt it :!:

the lower the spring constant the more the extension (k=F/x)
the higher the k the less the extension

for high sensitivity for low masses means we use lower k
for lower sensitivity for large masses we use higher k

we use the rigid box to prevent the sping with lower k to extent alot when there's a large mass. so the answer is A.
 
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hassam said:
@ libra why not c for that spring question...
well in C the spring with higher k is above the spring with lower k
which means that if we put a lager mass, the rigid box will prevent the higher k spring to extend
and also, the lower k spring will extend for both large and small masses..which we dont want, we want the low k spring to work for small masses only...so the answer is A.
hope u got it :cool:
 
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well u know that these particles are not travelling with wave........just vibrating up and down...and since the wave is travelling towards right, each particle on the wave will follow the movement of particle immediately to its left....since it is a pulse that is travelling forwards...means particle at C will move in the way which B previously moved i.e is up......
 
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phase is continuosly varying across progressive wave...and it varies in such a way that a particle at a particular instant in space will follow the path the previous particle took some time b4
 
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hassam said:
well u know that these particles are not travelling with wave........just vibrating up and down...and since the wave is travelling towards right, each particle on the wave will follow the movement of particle immediately to its left....since it is a pulse that is travelling forwards...means particle at C will move in the way which B previously moved i.e is up......

thanks!
 
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no u got it wrong a bit tricky try once more.....rearrange the diagram in a way in wich u myt feel it easy to deal with
 
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hassam said:
no u got it wrong a bit tricky try once more.....rearrange the diagram in a way in wich u myt feel it easy to deal with
i've already wasted 30 mins!! i'll try once more.
btw which year is this??
 
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