consider a wave y(x,t) = (6.0mm) sin ((8.5 m^(-1)) x - (125 rad/s) t), which travels along the string.\n(a)…

Explanation:

Step1: Identify amplitude

The general form of a wave is $y(x,t)=A\sin(kx - \omega t)$. Comparing with $y(x,t)=(6.0\mathrm{mm})\sin((8.5\mathrm{m}^{-1})x-(125\mathrm{rad/s})t)$, the amplitude $A$ is the coefficient of the sine - function. $A = 6.0\mathrm{mm}$

Step2: Calculate frequency

The angular frequency $\omega=125\mathrm{rad/s}$. The relationship between angular frequency $\omega$ and frequency $f$ is $\omega = 2\pi f$. So, $f=\frac{\omega}{2\pi}$. $f=\frac{125\mathrm{rad/s}}{2\pi}\approx19.9\mathrm{Hz}$

Step3: Calculate wave - speed

The wave number $k = 8.5\mathrm{m}^{-1}$ and $\omega = 125\mathrm{rad/s}$. The wave - speed $v$ is given by $v=\frac{\omega}{k}$. $v=\frac{125\mathrm{rad/s}}{8.5\mathrm{m}^{-1}}\approx14.7\mathrm{m/s}$

Step4: Find acceleration

First, find the first - derivative of $y$ with respect to $t$ to get the velocity $v_y=\frac{\partial y}{\partial t}=-A\omega\cos(kx - \omega t)$. Then, find the second - derivative of $y$ with respect to $t$ to get the acceleration $a_y=\frac{\partial^{2}y}{\partial t^{2}}=-A\omega^{2}\sin(kx - \omega t)$. Substitute $x = 0.22\mathrm{m}$, $t = 4.3\mathrm{s}$, $A = 6.0\times10^{-3}\mathrm{m}$, $k = 8.5\mathrm{m}^{-1}$, and $\omega = 125\mathrm{rad/s}$ into the acceleration formula. $a_y=-(6.0\times 10^{-3}\mathrm{m})\times(125\mathrm{rad/s})^{2}\sin((8.5\mathrm{m}^{-1})\times0.22\mathrm{m}-(125\mathrm{rad/s})\times4.3\mathrm{s})$ $a_y\approx-(6.0\times 10^{-3}\mathrm{m})\times(125\mathrm{rad/s})^{2}\sin(1.87 - 537.5)$ $a_y\approx-(6.0\times 10^{-3}\mathrm{m})\times(125\mathrm{rad/s})^{2}\sin(-535.63)$ $a_y\approx-(6.0\times 10^{-3}\mathrm{m})\times(125\mathrm{rad/s})^{2}\sin(124.37^{\circ})$ (using the property $\sin(x + 360^{\circ}n)=\sin x$) $a_y\approx-(6.0\times 10^{-3}\mathrm{m})\times15625\times0.814$ $a_y\approx - 76.6\mathrm{m/s}^{2}$

Step5: Find time between displacements

We know $y = A\sin(kx-\omega t)$. Let $y_1 = 2.0\mathrm{mm}=2.0\times10^{-3}\mathrm{m}$ and $y_2=-2.0\mathrm{mm}=-2.0\times10^{-3}\mathrm{m}$, $A = 6.0\times10^{-3}\mathrm{m}$. For $y_1$: $\sin(kx-\omega t_1)=\frac{y_1}{A}=\frac{2.0\times10^{-3}\mathrm{m}}{6.0\times10^{-3}\mathrm{m}}=\frac{1}{3}$ For $y_2$: $\sin(kx-\omega t_2)=\frac{y_2}{A}=\frac{-2.0\times10^{-3}\mathrm{m}}{6.0\times10^{-3}\mathrm{m}}=-\frac{1}{3}$ The phase difference $\Delta\phi=\omega\Delta t$. If $\sin\phi_1=\frac{1}{3}$ and $\sin\phi_2 =-\frac{1}{3}$, and assuming $\phi_1$ and $\phi_2$ are in the appropriate range, $\Delta\phi=\arcsin(\frac{1}{3})-(-\arcsin(\frac{1}{3})) = 2\arcsin(\frac{1}{3})$. Since $\Delta\phi=\omega\Delta t$, $\Delta t=\frac{2\arcsin(\frac{1}{3})}{\omega}$ $\Delta t=\frac{2\times0.3398}{125\mathrm{rad/s}}\approx5.44\times10^{-3}\mathrm{s}$

Answer:

(a) $6.0\mathrm{mm}$ (b) $19.9\mathrm{Hz}$ (c) $14.7\mathrm{m/s}$ (d) $\approx - 76.6\mathrm{m/s}^{2}$ (e) $\approx5.44\times10^{-3}\mathrm{s}$