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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 0.5 m.
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 (20πt – 0.016πx + 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`
For 0.5 m = 50 cm
Φ = 0.016 π × 50
= 0.8 π rad
उत्तर २
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((10"t")/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 = 1/0.0080 "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` x = 0.5 m = 50 cm, then
`trianglephi = (2pi)/125 xx 50`
`= 0.8 pi "rad"`
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