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
-
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)
Correct Answer: Option A
Explanation
The general wave equation is given by Y = sin A 2π/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]