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प्रश्न
For the travelling harmonic wave
y (x, t) = 2.0 cos 2π (10t – 0.0080x + 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 4 m.
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उत्तर १
Equation for a travelling harmonic wave is given as:
y (x, t) = 2.0 cos 2π (10t – 0.0080x + 0.35)
= 2.0 cos (20 π t – 0.016 π 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 = "k"x = (2pi)/lambda`
For x = 4 m = 400 cm
Φ = 0.016 π × 400
= 6.4 π rad
उत्तर २
The given equation can be drawn be rewritten as under
`"y"(x, "t") = 2.0 cos [2pi (10"t" - 0.0080 x) + 2pi xx 0.35]`
or `"y"(x, "t") = 2.0 cos [2pi xx 0.0080((10"t")/0.0080 - x) + 0.7 pi]`
Comparing this equation with the standard equation of a travelling harmonic wave.
`(2pi)/lambda = 2pi xx 0.0080` or `lambda = 1/0.0080 " cm" = 125` cm
The phase difference between oscillatory motion of two points seperated by a distance `trianglex` is given by
`trianglephi = (2pi)/lambda trianglex`
When `triangle z = 4 m = 400` cm then
`trianglephi = (2pi)/125 xx 400`
`= 6.4 pi " rad"`
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