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HOW WAS PHYSICS PAPER 2 AS!!!!!!!!!!!

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axes was definitely force on y axis and extension on x axis, still confused about the end point.
force cant be on the y axis

in that experiment force is the independent variable which means we change force and extension is dependent variable which means extension varies with force

So it should be Force on x axis and Extension on y axis.

If this is so the second graph should have twice the gradient of the graph already drawn
 
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force cant be on the y axis

in that experiment force is the independent variable which means we change force and extension is dependent variable which means extension varies with force

So it should be Force on x axis and Extension on y axis.

If this is so the second graph should have twice the gradient of the graph already drawn

hey, i get your point..but the question tht day was force on the y-axis n extension on x-axis..the original graph, the end point was at 3N (y.axis) and 0.25mm (x-axis)..remember?? so, how the graph should be??
 
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force cant be on the y axis

in that experiment force is the independent variable which means we change force and extension is dependent variable which means extension varies with force

So it should be Force on x axis and Extension on y axis.

If this is so the second graph should have twice the gradient of the graph already drawn
I am 100% sure that force was on y-axis!!!! And the gradient would have been halved.
 
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hey, i get your point..but the question tht day was force on the y-axis n extension on x-axis..the original graph, the end point was at 3N (y.axis) and 0.25mm (x-axis)..remember?? so, how the graph should be??
if force was on y axis then gradient of the graph for copper is half the gradient of that for steel.

But the question is wrong :mad: They contradict themselves. They teach us the independant is on x axis and they put it on y axis
 
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read the previous post. i apologized. my concepts r clear as its obvious by my answer of 500hz. it was a simple typing mistake

500 Hz was NOT the answer, I repeat. The idiots who got 500 probably made the mistake of thinking it was a uniform stationary wave, taking the distance from the start to the end of the wave to be 0.68m and the distance between three nodes. Lol, I say. It was a PART of the wave, and so the distance between two successive nodes/antinodes somewhere in the middle of the wave had to be taken.
 
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500 Hz was NOT the answer, I repeat. The idiots who got 500 probably made the mistake of thinking it was a uniform stationary wave, taking the distance from the start to the end of the wave to be 0.68m and the distance between three nodes. Lol, I say. It was a PART of the wave, and so the distance between two successive nodes/antinodes somewhere in the middle of the wave had to be taken.
the guys who got 1000 are the idiots
 
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500 Hz was NOT the answer, I repeat. The idiots who got 500 probably made the mistake of thinking it was a uniform stationary wave, taking the distance from the start to the end of the wave to be 0.68m and the distance between three nodes. Lol, I say. It was a PART of the wave, and so the distance between two successive nodes/antinodes somewhere in the middle of the wave had to be taken.
in a part of the wave distance between 3 nodes is the wavelength. even if u took distance between two nodes u had to multiply it by 2
 
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and u r saying a PART of a wave cant be its wavelength even if the part shows three nodes?
 
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wat exactly do u think da threshold will be?? i had checked out may/ june 2010 p22 it had similar questions like this yr and also num of questions were same 7 in both and threshold was 35 and it was a bit easier dan this yr
 
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in a part of the wave distance between 3 nodes is the wavelength. even if u took distance between two nodes u had to multiply it by 2

Yes, exactly. And if I remember correctly, the distance between two successive antinodes was 17cm=0.17m. So, 0.17x2=0.34m. And hence, 340/0.34=1000Hz.
 
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Yes, exactly. And if I remember correctly, the distance between two successive antinodes was 17cm=0.17m. So, 0.17x2=0.34m. And hence, 340/0.34=1000Hz.
Okay i got this question wrong but i solved it afterwards and the ans was 500Hz how? let me explain read this properly and please everyone SHUT the F___ UP..
The graph had 3 nodes and 2 antinodes.
This meant that a total of 2 complete stationery waves were formed.
The distance between 2 nodes or 2 antinodes is Lambda/2
since they had 3 nodes the distance would be Lambda/2 + lambda/2 = 0.68
Which means lambda = 0.68
Then 340/0.68 and here you go 500Hz.
Other way was by using the distance between 2 A.N.
The distance was 0.34 cm
Since they were 2 A.N it was lambda/2 =0.34
so lambda=0.68
meaning again 500Hz.
There was one other way too using distance between node and A.N tht was 0.17cm
Since distance between 1 Node and A.N is lambda/4
lamda/4 = 0.17cm
agains u'll get 0.68 and then 500hz frequency..
Now people please stop suggesting lame tricks etc if u know the concept of what a stationery wave is u'll see that 500 was the right answer :p :p :p :p :p :p
 
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Okay now ive got a question if i get wrong C value for Q1 will i get marks for part b and c..by ECF?
 
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