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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 `(3λ)/4`.
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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 (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 `x = (3lambda)/4`
`phi = (2pi)/lambda xx (3lambda)/4`
`= 1.5 pi` rad
उत्तर २
The given equation can be drawn be rewritten as under
y(x, t) `= 2.0 cos [2pi (10"t" - 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 `trianglex = (3lambda)/4 = (3xx125)/4` cm, then
`triangle phi = (2phi)/125 xx (3xx125)/4`
`= (3pi)/2 "rad"`
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