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A transverse harmonic wave on a string is described by y(x, t) = 3.0 sin (36t + 0.018x + π/4) where x and y are in cm and t is in s. The positive direction of x is from left to right. - Physics

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प्रश्न

A transverse harmonic wave on a string is described by y(x, t) = 3.0 sin (36t + 0.018x + π/4) where x and y are in cm and t is in s. The positive direction of x is from left to right.

  1. The wave is travelling from right to left.
  2. The speed of the wave is 20 m/s.
  3. Frequency of the wave is 5.7 Hz.
  4. The least distance between two successive crests in the wave is 2.5 cm.
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उत्तर

a, b and c

Explanation:

The general equation of a plane progressive wave with initial phase is

Various forms of the progressive wave function:

  • `y = a sin (ωt - kx)`
  • `y = a sin (ωt - (2π)/λ x)`
  • `y = a sin 2π [t/T - x/λ]`
  • `y = a sin  (2π)/T (t - x T/λ)`
  • `y = a sin  (2π)/λ (vt - x)`
  • `y = a sin ω(t - x/v)`

Given equation is `y(x, t) = 3.0 sin(36t + 0.018x + π/4)`

Option (a): Since there is +ve sign between wr and kx, the wave travels from right to left (the positive direction of x is from left to right). Hence it is correct.

Option (b): Speed of the wave, `v = ω/k = 36^-1/(0.018  cm)` = 2000 cm/s = 20 m/s. Hence it is correct.

Option (c): Frequency of the wave, `v = ω/(2π) = (36  s^-1)/(2π)` = 5.7 Hz. Hence it is correct.

Option (d): Least distance between two successive crests, `λ = (2π)/k = (2π)/(0.018  cm^-1)` = 349 cm. Hence it is wrong.

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अध्याय 15: Waves - Exercises [पृष्ठ १०८]

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एनसीईआरटी एक्झांप्लर Physics [English] Class 11
अध्याय 15 Waves
Exercises | Q 15.11 | पृष्ठ १०८

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