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Wave on a string problem 
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#1
Feb508, 10:17 AM

P: 27

1. The problem statement, all variables and given/known data
If y(x,t)=(6.0mm)sin(kx+(600rad/s)t+[tex]\Phi[/tex]) describes a wave travelling along a string, how much time does any given point on the string take to move between displacements y= +2.0mm and y= 2.0mm? 2. Relevant equations I think y(t)=ym sin([tex]\omega[/tex]t) ? 3. The attempt at a solution well if I plug in 2.0mm for y, 6.00mm for ym and 600rad/s for [tex]\omega[/tex] I come up with the equation 2.0mm=6.00mm sin (600rad/s * t). Where do I go from here? are my assumptions correct so far? Other things as I am thinking 600rad/s is about 95.5Hz so each complete cycle from +6mm to 6mm should take .01s or so, so my answer should be less then that. Thanks for your help 


#2
Feb508, 11:21 AM

P: 27

anyone?



#3
Feb508, 02:32 PM

P: 27

I think I have it if I then take
sin[tex]^{}1[/tex](2/6)=600*T1? sin[tex]^{}1[/tex](2/6)=600*T2? then [tex]\Delta[/tex]T is T1T2? does anyone have any input here? 


#4
Feb508, 08:12 PM

P: 27

Wave on a string problem
pretty sure I've got it
y1(x,t)=ym*sin(kx+600t1+[tex]\Phi[/tex]) 2.00mm=6.00mm*sin(kx+600t1+[tex]\Phi[/tex]) sin[tex]^{}1[/tex](1/3)=kx+600t1+[tex]\Phi[/tex] y2(x,t)=ym*sin(kx+600t2+[tex]\Phi[/tex]) 2.00mm=6.00mm*sin(kx+600t2+[tex]\Phi[/tex]) sin[tex]^{}1[/tex](1/3)=kx+600t2+[tex]\Phi[/tex] so... [sin[tex]^{}1[/tex](1/3)][sin[tex]^{}1[/tex](1/3)]=(kx+600t1+[tex]\Phi[/tex])(kx+600t2+[tex]\Phi[/tex]) and... [sin[tex]^{}1[/tex](1/3)][sin[tex]^{}1[/tex](1/3)]=600t1600t2 finally, ([sin[tex]^{}1[/tex](1/3)][sin[tex]^{}1[/tex](1/3)])/600=t1t2 do the math and [tex]\Delta[/tex]t is .00113s I think this is solved 


#5
Jul2709, 08:43 PM

P: 166

no one ever responded to this guys problem, and now i'm actually trying to solve this as well and i tried doing the method he ended up using but i am not getting a correct answer. Although his method for the most part looks right and makes sense to me, the only thing I figure would be the problem is that x isn't a constant so they should cancel i don't think....but i'm not sure what else to do with so many unknown variables...any help???



#6
Jul2709, 08:49 PM

P: 166

oops nevermind...my calculator was in degree mode instead of radian mode



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