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

Answer:

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

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

$k = \frac{2\pi}{\lambda}$

$k = \frac{2\pi}{2.6}$

$k = 2.42 per m$

also we have

$T = 1.2 s$

so we have

$\omega = \frac{2\pi}{T}$

$\omega = \frac{2\pi}{1.2}$

$\omega = 5.23 rad/s$

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

so we have

$\phi_0 = \frac{\pi}{2}$

so we have

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