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For the travelling harmonic wave y (x, t) = 2.0 cos 2π (10t – 0.0080x + 0.35) Calculate the phase difference between oscillatory motion of two points separated by a distance of 4 m.

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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

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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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अध्याय 14: Waves - Exercises [पृष्ठ ३८७]

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एनसीईआरटी Physics Part 1 and 2 [English] Class 11
अध्याय 14 Waves
Exercises | Q 10.1 | पृष्ठ ३८७

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