# Wave on a string problem

by Sdawg1969
Tags: string, wave
 P: 27 1. The problem statement, all variables and given/known data If y(x,t)=(6.0mm)sin(kx+(600rad/s)t+$$\Phi$$) 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($$\omega$$t) ? 3. The attempt at a solution well if I plug in 2.0mm for y, 6.00mm for ym and 600rad/s for $$\omega$$ 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
 P: 27 anyone?
 P: 27 I think I have it- if I then take sin$$^{}-1$$(2/6)=600*T1? sin$$^{}-1$$(-2/6)=600*T2? then $$\Delta$$T is T1-T2? does anyone have any input here?
P: 27

## Wave on a string problem

pretty sure I've got it-

y1(x,t)=ym*sin(kx+600t1+$$\Phi$$)

2.00mm=6.00mm*sin(kx+600t1+$$\Phi$$)

sin$$^{}-1$$(1/3)=kx+600t1+$$\Phi$$

y2(x,t)=ym*sin(kx+600t2+$$\Phi$$)

-2.00mm=6.00mm*sin(kx+600t2+$$\Phi$$)

sin$$^{}-1$$(-1/3)=kx+600t2+$$\Phi$$

so...
[sin$$^{}-1$$(1/3)]-[sin$$^{}-1$$(-1/3)]=(kx+600t1+$$\Phi$$)-(kx+600t2+$$\Phi$$)

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

finally,

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

do the math and $$\Delta$$t is .00113s

I think this is solved
 P: 160 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???
 P: 160 oops nevermind...my calculator was in degree mode instead of radian mode

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