Question

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

Answer 1

Answer:

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Answer 2

Answer:

y = 0.15 sin 200π(x - 5.5t).

Explanation:

We know that the general form of the wave equation is given as                      Y = Asin2π/λ [x - vt].

Which can also be written as Y= Sin A [(2π X)/λ - (2π vt)/λ].

From the question we know that the values of amplitude A = 0.15, wavelength λ = 0.01m, and frequency f = 550Hz.

And we also know that the velocity V = f x λ = 550 x 0.01 which is 5.50m/s .

Therefore, substituting the values in the equation Y = 0.15 sin [(2 * λ * x)/0.01 - (2*λ*5.5t)/0.01]

Which on solving we will get the equation to be as 0.15 sin[200λx - 200*λ*5.5t]

So, the equation will be Y = 0.15 sin 200λ[x - 5.5t].