Wave on a string: time from y=2mm to -2mm. Amp. & Omega given. Best method?

[deltapapazulu]

Homework Statement

If y(x,t) = (6mm)sin(kx + (600rad/s)t + theta) describes a wave traveling along a string, how much time does any given point on the string take to move between displacements y = +2 mm and y = -2 mm?

All known data are stated in above problem.

Homework Equations

I got the answer by plotting it on a unit circle, which I describe below. But what I want to know is if there is a method particular to the use of the wave function itself to get the same answer?

The Attempt at a Solution

Because sinusoidal wave motion is exactly analogous to rotation in a unit circle, a .006m amplitude corresponds to a .006m radius(r = .006m), which together with y=.002m can be taken as arcsin(y/r) to yield the radian measure from P(x,0) to P(x, .002) on the circle. This = .34 radians. Since the motion is from .002m to -.002m you take .34*2 = .68rad. Then to find the fraction of 600rad you take .68/600 = .00113 which also represents the fraction of s=1 second in 600rad/s. Therefore it takes .00113s or 1.1ms to get from .002m to -.002m .

Is there a more appropriate method for doing this?

 

[kuruman]

Not really. The algebraic method will have you write

2 = (6mm)sin(kx + (600rad/s)t₁ + theta)

solve for t₁ in terms of the arcsine, do the same for t₂ with

-2 = (6mm)sin(kx + (600rad/s)t₂ + theta) and find the difference as

Δt = t₂-t₁=(Arcsin(-1/3)-Arcsin(1/3))/600.

Both methods involve taking the difference between two arcsines.