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Homework Statement
Two sinusoidal waves in a string are defined by the wave functions
y1 = 2.00 sin (20.0x – 32.0t)
y2 = 2.00 sin (25.0x – 40.0t)
where x, y1, and y2 are in centimeters and t is in seconds.
(a) What is the phase differencebetween these two waves at the point x = 5.00 cm at t = 2.00 s?
(b) What is the positive x value closest to the origin for which the two phases differ by ±π at t = 2.00 s?
(At that location, the two waves add to zero.)
Homework Equations
The Attempt at a Solution
I solved (a) (I'm just typing the abridged version to save some time, I know about the units):
Δφ = (25*x - 40*t) - (20*x - 32*t) = 5*x 8*t
We put in t = 2s and x = 5 cm and that's it.
But (b)'s giving me some trouble, because I'm not sure exactly what I'm meant to use. My official solutions manual, gives this answer:
(b) The sine functions repeat whenever their arguments change by an integer number of cycles, an integer multiple of 2π radians. Then the phase shift equals ±π whenever Δφ = π + 2nπ, for all integer values of n.
Substituting this into the phase equation, we have:
π + 2nπ = −(5.00 rad/cm)x + (8.00 rad/s)t
At t = 2.00 s, π + 2nπ = −(5.00 rad/cm)x + (8.00 rad/s)(2.00 s)
or (5.00 rad/cm)x = (16.0 − π − 2nπ) rad
The smallest positive value of x is found when n = 2:
x = (16.0 5 )rad/5.00 rad/cm = 0.058 4 cm
Okay, I understand what he says about the phase increasing by 2π with every cycle. And, after he gets his result, I understand why n = 2 ( n = 0 & n = 1 give larger numbers, n = 3 gives us a negative answer, so n = 2 gives us the smallest positive value of x). What I don't get, is how he came to the conclusion of Δφ = π + 2nπ. Also, Δφ = +- π, so where is the (-)?
As I've said before, english isn't my native tongue, so I'm basically missing some "key words", and I'm moving "faster" than I'd like to (i.e. not so much time spent reading the chapters), so any help would be apreciated.
Thanks!