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Karnataka Board PUCPUC Science Class 11

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

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Question

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 `λ/2`.

Numerical
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Solution 1

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 = "k"x = (2pi)/lambda`

For  `x = lambda/2`

`phi = (2phi)/lambda xx lambda/2`

`= pi " rad"`

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Solution 2

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  = lambda/2 = 125/2` cm , then

`triangle phi  = (2phi)/125 xx 125/2`

`= pi " rad"`

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Chapter 14: Waves - Exercises [Page 387]

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NCERT Physics Part 1 and 2 [English] Class 11
Chapter 14 Waves
Exercises | Q 10.3 | Page 387

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