Question

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 displacement of 0.19 m at t = 0 and x = 0.

Answer 1

`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))`