Write an expression for a harmonic wave with an amplitude of 0.19 m, a wavelength of 2.6 m, and a period of 1.2 s. The wave is transverse, travels to the right, and has a displa…

Question

asked Jun 26, 2020 182k views

1 Answer

Blazs · Answer 1 · 2020-07-01T22:51:09+0000

Answer:

$y = 0.19 sin(5.23 t - 2.42x + (\pi)/(2))$

Step-by-step explanation:

As we know that the wave equation is given as

$y = A sin(\omega t - k x + \phi_0)$

now we have

A = 0.19 m

$\lambda = 2.6 m$

so we have

$k = (2\pi)/(\lambda)$

$k = (2\pi)/(2.6)$

k = 2.42 per m

also we have

T = 1.2 s

so we have

$\omega = (2\pi)/(T)$

$\omega = (2\pi)/(1.2)$

$\omega = 5.23 rad/s$

now we know that at t = 0 and x = 0 wave is at y = 0.19 m

so we have

$\phi_0 = (\pi)/(2)$

so we have

$y = 0.19 sin(5.23 t - 2.42x + (\pi)/(2))$