Source: High school physics (Chinese)
Problem Sets:
Problem
A transverse wave travels along a string in the $+y$ direction. There are 6000 wavelengths in each meter of the string, the period is $0.20$ s, and the amplitude is $3.0$ cm.
- Write the wave function of this transverse wave.
- Find the maximum speed and the maximum acceleration of any point on the string.
Wave function: $x = 3.0 \times 10^{-2}\cos(10\pi t - 12000\pi y)$ m. Maximum speed $v_{\max} = 0.3\pi \approx 0.94$ m/s; maximum acceleration $a_{\max} = 3.0\pi^2 \approx 29.6$ m/s$^2$.
With 6000 wavelengths per meter, $\lambda = \dfrac{1}{6000}$ m, so the wave number is $k = \dfrac{2\pi}{\lambda} = 12000\pi$ rad/m. The period gives $\omega = \dfrac{2\pi}{T} = \dfrac{2\pi}{0.20} = 10\pi$ rad/s, and $A = 3.0$ cm $= 3.0 \times 10^{-2}$ m. For a wave moving in the $+y$ direction, $x = A\cos(\omega t - k y) = 3.0 \times 10^{-2}\cos(10\pi t - 12000\pi y)$ m. The maximum particle speed is $v_{\max} = \omega A = 10\pi \times 3.0 \times 10^{-2} = 0.3\pi \approx 0.94$ m/s, and the maximum acceleration is $a_{\max} = \omega^2 A = (10\pi)^2 \times 3.0 \times 10^{-2} = 3.0\pi^2 \approx 29.6$ m/s$^2$.