Physics: Post your doubts here!

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hela

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dannydan56 said:
I1/I2= A1^2/A2^2

I1= Initial intensity
I2= New intensity
A1= Initial amplitude
A2 = New amplitude

I1/ 2I1 = (2.8)^2 / A2^2.. We have to find A2.. So solve this you'd get the answer.. I think its 3.959 ..
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Thank you very much ! this was very helpful
I wanted to know whether this formula can be applied to ALL the intensity questions Thank you again!
 
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selena said:
may/june 12 qp 21 question 3 (b).can some1 explain
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3bi) speed max at mean position which means any time when disp 0 hence 0.2
3bii) max potential energy must b when it is bottom most part hece t=0
3biii) 0.2
 
abruzzi

abruzzi

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Could anyone please explain how the relationship between Intensity and amplitude of a wave? I know that intensity = amplitude^2 but I always get confused when we are asked questions like.. twice the intensity, half the intensity, twice the amplitude etc. Please help me understand that.
Thanks
 
syed1995

syed1995

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abruzzi said:
Could anyone please explain how the relationship between Intensity and amplitude of a wave? I know that intensity = amplitude^2 but I always get confused when we are asked questions like.. twice the intensity, half the intensity, twice the amplitude etc. Please help me understand that.
Thanks
Click to expand...

I is directly propotional to A^2

I = kA^2
I/A^2 = k

We always find the value of k and then find the twice the amplitude etc etc..

Like if I is 6 and A is 2 .. and another wave has Double the amplitude..

6= k(2^2)
6= 4k
k= 1.5
I=kA^2
I=1.5(2*2)^2
I=24 Answer..

similarly in terms of I and A..

I = kA^2
k = I/A^2

Doubling A means ..

I = kA^2
I = k (2A)^2
I = k*4A^2
I = 4A^2 * I/A^2
I = 4I
 
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lyuchenqujoy

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I wonder if amplitude is a vector quality or scalar? The definition is max displacement. Displacement is a vector, but some books also said that amplitude is the MAGNITUDE of max displacement. I don't think that amplitude needs a direction, right? So I just got confused.........
 
asd

asd

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A star said:
no the weight is taken perp to the ground so y we taking paralell to slope this concept bugged me throughout the year
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You do realize if you release a ball from rest on a slope it starts to move down the slope. Clearly, its the weight component "DOWN THE SLOPE" (along the slope) which makes it move down. You have to find that component which does this.
In this case the car is already moving down the slope but the forward force will always be slightly larger than the driving force because of that little contribution from the weight component down the slope.
 
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yeah i get it now took a while i knew there had to be i couldnt resolve it properly
 
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syed1995

syed1995

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A star said:
no the weight is taken perp to the ground so y we taking paralell to slope this concept bugged me throughout the year
Click to expand...

Can you explain this concept? Because I don't know this one... I am weak in forces.. How do you do this? cuz with what I was doing.. I was getting a wrong answer in the component of weight in that question...


PS. Would appreciate a diagram on which you show the direction of weight.. and then resolve the component of weight..
 
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