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A travelling wave moving from left to right has an amplitude of 0.15m, a frequen...


Question

A travelling wave moving from left to right has an amplitude of 0.15m, a frequency of 550Hz and a wavelength of 0.01m. The equation describing the wave is

Options

A) y = 0.15 sin 200π(x - 5.5t)

B) y = 0.15 sin π(0.01x - 5.5t)

C) y = 0.15 sin 5.5π(x - 200t)

D) y = 0.15 sin π(550x - 0.01t)


The correct answer is A.

Explanation:

The general wave equation is given by Y = sin A /λ [x - vt]
= Sin A [(2π X)/λ - (2π vt)/λ]
thus if A = 0.15;λ = 0.01m; f = 550Hz
then V = f x λ = 550 x 0.01 = 5.50m/s
∴ Y = 0.15 sin [(2 x λ x X)/0.01 - (2 x λ x 5.5t)/0.01]
= 0.15 sin[200λX - 200 x λ x 5.5t]
∴ Y = 0.15 sin 200λ[x - 5.5t]

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Dicussion (1)

  • The general wave equation is given by Y = sin A /λ [x - vt]
    = Sin A [(2π X)/λ - (2π vt)/λ]
    thus if A = 0.15;λ = 0.01m; f = 550Hz
    then V = f x λ = 550 x 0.01 = 5.50m/s
    ∴ Y = 0.15 sin [(2 x λ x X)/0.01 - (2 x λ x 5.5t)/0.01]
    = 0.15 sin[200λX - 200 x λ x 5.5t]
    ∴ Y = 0.15 sin 200λ[x - 5.5t]

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