Two sinusoidal waves in a string are defined by the function `y_(1)=(2.00 cm) sin (20.0x-32.0t)` and `y_(2)=(2.00 cm) sin (25.0x-40.0t)` where `y_(1), y_(2)` and `x` are in centimetres and `t` is in seconds.
(a). What is the phase difference between 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 original for which the two phase differ by `+_ pi at t=2.00 s?` (That os a location where the two waves add to zero.)

Explanation

Think of the two waves as movie sinuosidal graph with different wavelength and frequencies. The phase difference between them is a function of the variables `x` and `t`. it controls how the waves add up at each point.
looking at the wave function, let us pick out the expressions for the phase as the 'angles we take the since of' then arhuments of the sine function. then algebra will give particular answers.
At any time and place `=`, the phase shift between the waves is found by subtracting the phase of the two waves, `Deltaphi=phi_(1)-Delta_(2)`.
`Deltaphi=[(20.0 rad//cm)x-(32.0 rad//s)t]-[(25.0 rad//cm)x-(40.0 rad//s)t]`
Collecting terms, `Deltaphi=-(5.00 rad//cm)x+(8.00 rad//s)t`
(a) At `x=5.00 cm` and `t=2.00 s`, the phase difference is `Deltaphi=(-5.00 rad//cm)(5.00 cm)+(8.00 rad//s)(2.00 s) |Deltaphi|=9.00 rad`
(b) The sine function repeat whenever their arguments change by an integer number of cycle, an integer multiple of `2pi rad`. then the phase shift equels `+- pi` whenever `Deltaphi=pi+2npi`, for all integer value of `n`.
subsititunig this into the phase equation, we have `pi+2npi=(5.00 rad//cm)x+(8.00 rad//s)t`
At `t=2.00 s`,
`pi+2npi=-(5.00 rad//cm)x+(8.00 rads)(2.00 s)`
or `pi+2npi=-(5.00 rad//cm)x=(16.0-pi-2npi)rad`
the smallest positive value of `x` is found when `n=2`
`x((16.0-5pi)rad)/(5.00 rad//cm)0.0584 cm`