Source: High school physics (Chinese)
Problem Sets:
Problem
A long horizontal rope is fixed at one end while the other end is driven by a rod that continuously oscillates up and down in simple harmonic motion, producing a simple harmonic wave on the rope. The distance between the highest and lowest points of the rope's transverse motion is $0.50$ cm, the rod oscillates 120 times each second, and the wave speed on the rope is $19$ m/s.
- Find the amplitude and the wavelength of the wave.
- Take the rod-connected end of the rope as the origin O, the direction of wave propagation as the positive $y$-direction, and upward as the positive $x$-direction. At $t = 0$ the end at $y = 0$ is at its equilibrium position and moving upward. Write the displacement expression of the wave.
Amplitude $A = 0.25$ cm; wavelength $\lambda = \dfrac{19}{120} \approx 0.16$ m. Displacement expression: $x = 0.25\sin\left[240\pi\left(t - \dfrac{y}{19}\right)\right]$ cm ($t$ in s, $y$ in m).
Part 1. The highest-to-lowest distance equals twice the amplitude, so $A = \dfrac{0.50}{2} = 0.25$ cm. The frequency is $f = 120$ Hz, hence the wavelength is $\lambda = \dfrac{v}{f} = \dfrac{19}{120} \approx 0.16$ m.
Part 2. The angular frequency is $\omega = 2\pi f = 240\pi$ rad/s. The driven end ($y = 0$) starts at equilibrium moving in the $+x$ direction, so its motion is $x = A\sin\omega t$. Because the wave travels in $+y$ with speed $v$, a point at position $y$ lags the source by $y/v$, giving $x = A\sin\left[\omega\left(t - \dfrac{y}{v}\right)\right]$. Substituting the values, $x = 0.25\sin\left[240\pi\left(t - \dfrac{y}{19}\right)\right]$ cm, with $t$ in s and $y$ in m.