#### Question

For the travelling harmonic wave

*y *(*x, t*) = 2.0 cos 2π (10*t *– 0.0080*x *+ 0.35)

Where *x *and *y *are in cm and *t *in s. Calculate the phase difference between oscillatory motion of two points separated by a distance of

(a) 4 m,

(b) 0.5 m,

(c) λ/2

(d) 3λ/4.

#### Solution 1

Equation for a travelling harmonic wave is given as:

*y* (*x*, *t*) = 2.0 cos 2π (10*t* – 0.0080*x* + 0.35)

= 2.0 cos (20π*t* – 0.016π*x* + 0.70 π)

Where,

Propagation constant, *k* = 0.0160 π

Amplitude, *a *= 2 cm

Angular frequency, *ω*= 20 π rad/s

Phase difference is given by the relation:

`phi = kx = 2pi/lambda`

**(a)** For *x* = 4 m = 400 cm

Φ = 0.016 π × 400 = 6.4 π rad

**(b)** For 0.5 m = 50 cm

Φ = 0.016 π × 50 = 0.8 π rad

(c) For `x = lambda/2`

`phi = 2phi/lambda xx lambda/2 = pi "rad"`

(d) For `x = (3lambda)/4`

`phi = 2pi/lambda xx 3lambda/4 = 1.5 pi rad`

#### Solution 2

The given equation can be drawn be rewritten as under

`y(x, t) = 2.0 cos [2pi (10t - 0.0080x) + 2pi xx 0.35]`

or `y(x,t) = 2.0 cos [2pi xx 0.0080((10t)/0.0080 - x)+0.7pi]`

Comparing this equation with the standard equation of a travelling harmonic wave.

`(2pi)/lambda = 2pi xx 0.0080` or `lambda = 1/0.0080cm = 125` cm

The phase difference between oscillatory motion of two points seperated by a distance `trianglex` is given by

`trianglephi = (2pi)/lambda trianglex`

a) When `triangle z = 4 m = 400` cm then

`trianglephi = (2pi)/125 xx 400 = 6.4 pi " rad"`

b) When `triangle x = 0.5 m = 50` cm, then

`trianglephi = (2pi)/125 xx 50 = 0.8 pi "rad"`

c) When `trianglex = lambda/2 = 125/2` cm , then

`triangle phi = (2phi)/125 xx 125/2 = pi "rad"`

d) When `trianglex = (3lambda)/4 = (3xx125)/4` cm, then

`triangle phi = (2phi)/125 xx (3xx125)/4 = (3pi)/2 "rad"`