How Long Does a Point on a String Take to Move Between +2.0mm and -2.0mm?

[Sdawg1969]

Homework Statement

If y(x,t)=(6.0mm)sin(kx+(600rad/s)t+\Phi) describes a wave traveling 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?

Homework Equations

I think y(t)=ym sin(\omegat) ?

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

 

[Sdawg1969]

anyone?

 

[Sdawg1969]

I think I have it- if I then take

sin^{}-1(2/6)=600*T1?

sin^{}-1(-2/6)=600*T2?

then \DeltaT is T1-T2?

does anyone have any input here?

 

[Sdawg1969]

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 \Deltat is .00113s

I think this is solved

 

[Puchinita5]

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]

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