Source: High school physics (Chinese)
Problem Sets:
Problem
A mechanical transverse wave propagating along the $y$-axis is described by $x = 6.0 \times 10^{-2}\cos(4.0\pi t - 2.0\pi y)$ m, where $x$ is the transverse displacement (in m), $y$ the position along the axis (in m), and $t$ the time (in s).
Compare with the standard form $x = A\cos(\omega t - k y)$. The amplitude is $A = 6.0 \times 10^{-2}$ m $= 6.0$ cm. The angular frequency is $\omega = 4.0\pi$ rad/s, so $f = \dfrac{\omega}{2\pi} = 2.0$ Hz. The wave number is $k = 2.0\pi$ rad/m, so $\lambda = \dfrac{2\pi}{k} = 1.0$ m. The wave speed is $v = \dfrac{\omega}{k} = \lambda f = 2.0$ m/s. The $(\omega t - k y)$ form shows the wave travels in the $+y$ direction. The maximum particle speed is $v_{\max} = \omega A = 4.0\pi \times 6.0 \times 10^{-2} = 0.24\pi \approx 0.75$ m/s.