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For the harmonic travelling wave y = 2 cos 2π (10t – 0.0080x + 3.5) where x and y are in cm and t is second.

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प्रश्न

For the harmonic travelling wave y = 2 cos 2π (10t – 0.0080x + 3.5) where x and y are in cm and t is second. What is the phase difference between the oscillatory motion at two points separated by a distance of What is the phase difference between the oscillation of a particle located at x = 100 cm, at t = T s and t = 5 s?

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उत्तर

Given, wave functions are y = 2 cos 2π (10t – 0.0080x + 3.5) 

= 2 cos(20πt – 0.016πx + 7π)

Now, the standard equation of a travelling wave can be written as y = a cos(ωt – kx + `phi`)

On comparing with the above equation, we get

a = 2 cm

ω = 20π rad/s

k = 0.016π

Path difference = 4 cm

T = `(2π)/ω = (2π)/(20π) = 1/10`s

∴ At x = 100 cm, t = T

`phi`1 = 20πT – 0.016π(100) + 7π

= `20π(1/10) - 16π + 7π`

= 2π – 1.6π + 7π .......(i)

Again, at x = 100 cm, t = 5s

`phi`2 = 20π(5) – 0.016π(100) + 7π

= `100π - (0.016 xx 100)π + 7π`

= 100π – 1.6π + 7π .......(ii)

∴ From equations (i) and (ii), we get

Δ`phi` = Phase difference = `phi_1 - phi_2`

= (100π – 1.6π + 7π) – (2π – 1.6π + 7π)

= 100π – 2π 

= 98π rad

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अध्याय 15: Waves - Exercises [पृष्ठ ११२]

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एनसीईआरटी एक्झांप्लर Physics Exemplar [English] Class 11
अध्याय 15 Waves
Exercises | Q 15.36 (e) | पृष्ठ ११२

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