For the travelling harmonic wave
`y(x,t) = 2.0 cos 2 pi (10 t - 0.0080 x + 0.35)`
where x and y in cm and t in s. Calculate the phase difference between oscillatory motion of two points seperated by a distance of
(a) 4 m,
(b) 0.5 m,
(c ) `lamda//2`,
(d) `3lamda//4`

Explanation

Comparing give wave equation,
`y = 2 cos (20pi t - 0.016pi x + 7pi)` with `y =a cos (omega t - kx + phi)` we get,
`a = 2cm, omega = 20 pi rad//s, k = 0.016pi rad//cm`
Now, according to formula of phase difference between two particles taking part in the propagation of wave separted by diatance `Delta x` along the direction of propagation of wave, at any given instant is,
`Delta phi =(2pi)/(lamda) (Delta x) =k (Delta x) (because k = (2pi)/(lamda))`
(a) Here `Delta x = 400 cm`
`therefore Delta phi =0.16pi xx 400 = 6.4 pi ` rad
(b) Here `Delta x = 50 cm`
`therefore Delta phi = 9.916 pi xx 50 = 0.8 pi ` rad
(c) Here `Delta x = (lamda)/(2)`
`therefore Delta phi = (2pi )/(lamda) xx (lamda)/(2) = pi rad`
(d) Here Delta `x = (3 lamda)/( 4)`
`therefore Delta phi = (2pi)/(lamda) xx (3 lamda)/(4) = (3pi )/(2) rad`
(e) Here `T = (2pi)/(omega ) = (2pi )/( 20 pi)=0.1 s`
Phase of a particle at distance `x = 100 cm ` at time `t _(1) = T is theta_(1) = omega t _(1) - kx + phi`
`therefore theta _(1) = ((2pi)/( T) xx T) - (0.016 pi xx 100) + (7pi)`
`= 7.4 pi ` rad
At time `t _(2) = 5s`
`theta _(2) = (20 pi xx 5)- (0.16pi xx 100) + (7pi)= 105.4 pi ` rad
`therefore ` Phase difference `=theta _(2) - theta _(1)`
`= 10 5 . 4 pi - 7. 4 pi `
`= 98 pi rad`