T
Thanks, I think I get it. just to be sure. Theoretically if I found the phase shift to be 3/5 pi and all the other variable were the same as in the problem. My T would be 2/5 T because one sine wave is 2/5 pi ahead of the other, right?
ok I just got stuck half way into a problem, I would like it if someone explained it!
Ok the question says, two identical sinusodial waves with a wavelength of 3.0 m and traveling in the same direction with a v of 2m/s. Starting from the same point, just the second waves starts later. and the...
sorry, I am terrible at physics. I get 20(5) -32(2) = 36, 25(5) - 40(2) = 45, 45 - 36 = 9 radians if you convert that to degrees, pi/20.
therefore, I have no idea how to get to the answer from there. any n2pi won't give me 152 radians. I get like 171 or 351.
It might be simple, but how do you...
Thanks for the help. but I still have a question, What do you mean by ci? b/c I thought I could get the answer by subtracting the difference of the two -> kix+wi , but I get totally the wrong answer. I get 9 radians and the answer is 152 degrees. So I would suppose that variable c[SUB]i is what...
ok, I would just like to know in general b/c we have to do this a lot.
The equations have the same amplitude, but different k and w
suppose you have y1 = Asin(k1x-w1t) and y2 = Asin(k2-w2t)
and only other information is they are on a string, at a point x, and a time t.
side note...
How do you find the force extended if given the amplitude? Is main questions, I also have one slight question.
Ok, doing a problem. There is no damping. A (.15kg) object is hanging from a light(6.30N/m) spring.
A sinusoidal force with an amp of 1.7 N drives the system. And the problem is...
Hey guys, I know it late its a little past one here. But I'm doing an assignment due tomorrow at I've been stuck on the last question for at least an hour.
Find the volume of the solid obtained by rotating the region bounded the curves
Y=absolute value of x. and y = square root of (...
Ok, finished it but it was ugly. And I think I messed up my algebra somewhere.
your remainder ends up being -1/4 so you get∫ 1/4 +( -1/4/ (1+4x^2))so I decided to pull out the 4 to get the form x^2+a^2 so you get 1/4x + ∫-1/4 / (4((1/4)+x^2)) --> -1/16∫ 1/((1/4)+x^2) -->-1/16(1/(1/2)) times...
One last question
to Integrate x^2/(1+4•x^2). I would assume you would do long division but 4x^2 is bigger than x^2. so would you either pull out a 1/4 and it would be 1/4 ∫ x^2/(1/4+•x^2) dx or would the first term when doing long division be 1/4? or am I just totally wrong and you...
I mean't cos(x) then du = -sintheta. But that was totally wrong! you had the right substitution. I tried u = cos(ln(x)) and dv =dx. It worked out quite nicely. I only had to integrate it twice! Thanks
Ok I have to integrate -->∫cos(lnx) dx. could I use cos =U, -sinx=du, dv=lnxdx, v = 1/x
I know the difference technically, but in this situation it is kinda weird.
because the formula f(x)g(x)= uv-∫vdu. I thinking if they were number like 9(3) it would equal 27 so f(g) = f times G? but then...