Wave on a string problem


by Sdawg1969
Tags: string, wave
Sdawg1969
Sdawg1969 is offline
#1
Feb5-08, 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
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Sdawg1969
Sdawg1969 is offline
#2
Feb5-08, 11:21 AM
P: 27
anyone?
Sdawg1969
Sdawg1969 is offline
#3
Feb5-08, 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 T1-T2?

does anyone have any input here?

Sdawg1969
Sdawg1969 is offline
#4
Feb5-08, 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)]=600t1-600t2

finally,

([sin[tex]^{}-1[/tex](1/3)]-[sin[tex]^{}-1[/tex](-1/3)])/600=t1-t2

do the math and [tex]\Delta[/tex]t is .00113s

I think this is solved
Puchinita5
Puchinita5 is offline
#5
Jul27-09, 08:43 PM
P: 162
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???
Puchinita5
Puchinita5 is offline
#6
Jul27-09, 08:49 PM
P: 162
oops nevermind...my calculator was in degree mode instead of radian mode


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