$A$ wave travelling along a string is described by,
$y(x, t) = 0.005 \sin (80.0 x - 3.0 t)$
In which the numerical constants are in $SI$ units ($0.005 \, m, 80.0 \, rad \, m^{-1},$ and $3.0 \, rad \, s^{-1}$). Calculate
$(a)$ the amplitude,
$(b)$ the wavelength,and
$(c)$ the period and frequency of the wave.
Also,calculate the displacement $y$ of the wave at a distance $x = 30.0 \, cm$ and time $t = 20 \, s$.

(N/A) The standard wave equation is $y(x, t) = a \sin (kx - \omega t)$.
$(a)$ Comparing the given equation with the standard form,the amplitude $a = 0.005 \, m = 5 \, mm$.
$(b)$ The angular wave number $k = 80.0 \, m^{-1}$. The wavelength $\lambda$ is given by $\lambda = 2\pi / k = 2\pi / 80.0 \approx 0.0785 \, m = 7.85 \, cm$.
$(c)$ The angular frequency $\omega = 3.0 \, rad \, s^{-1}$. The period $T = 2\pi / \omega = 2\pi / 3.0 \approx 2.09 \, s$. The frequency $\nu = 1 / T = 3.0 / 2\pi \approx 0.48 \, Hz$.
For displacement at $x = 30.0 \, cm = 0.3 \, m$ and $t = 20 \, s$:
$y = 0.005 \sin (80.0 \times 0.3 - 3.0 \times 20) = 0.005 \sin (24 - 60) = 0.005 \sin (-36 \, rad)$.
Since $\sin(-36 \, rad) \approx \sin(-36 + 12\pi) \approx \sin(1.699 \, rad) \approx 0.992$,
$y \approx 0.005 \times 0.992 \approx 0.00496 \, m \approx 5 \, mm$.