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

Explanation:

Step1: Identify amplitude from wave - function

The general form of a traveling 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 from angular - frequency

The angular - frequency $\omega$ and frequency $f$ are related by $\omega = 2\pi f$. Given $\omega=125\mathrm{rad/s}$, we can solve for $f$. $f=\frac{\omega}{2\pi}=\frac{125}{2\pi}\mathrm{Hz}\approx19.9\mathrm{Hz}$

Step3: Calculate wave speed from wave number and angular - frequency

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

Step4: Find acceleration of the particle

First, find the first - derivative of $y(x,t)$ with respect to $t$ to get the velocity $v_y=\frac{\partial y}{\partial t}=-(6.0\mathrm{mm})\times(125\mathrm{rad/s})\cos((8.5\mathrm{m}^{-1})x-(125\mathrm{rad/s})t)$. Then, find the second - derivative of $y(x,t)$ with respect to $t$ to get the acceleration $a_y=\frac{\partial^{2}y}{\partial t^{2}}=(6.0\mathrm{mm})\times(125\mathrm{rad/s})^{2}\sin((8.5\mathrm{m}^{-1})x-(125\mathrm{rad/s})t)$. Substitute $x = 0.22\mathrm{m}$ and $t = 4.3\mathrm{s}$ into the acceleration formula: $a_y=(6.0\times10^{-3}\mathrm{m})\times(125\mathrm{rad/s})^{2}\sin((8.5\mathrm{m}^{-1})\times0.22\mathrm{m}-(125\mathrm{rad/s})\times4.3\mathrm{s})$ $a_y=(6.0\times10^{-3})\times15625\sin(1.87 - 537.5)$ $a_y = 93.75\sin(- 535.63)$ $a_y\approx93.75\sin(14.37^{\circ})\approx23.2\mathrm{m/s}^{2}$

Step5: Find time between displacements

We know $y = A\sin(\omega t+\varphi)$. When $y = 2.0\mathrm{mm}$ and $A = 6.0\mathrm{mm}$, $\sin\theta_1=\frac{2.0}{6.0}=\frac{1}{3}$, and when $y=-2.0\mathrm{mm}$, $\sin\theta_2=-\frac{1}{3}$. The phase difference $\Delta\theta$ between these two points is $\Delta\theta = 2\sin^{-1}(\frac{1}{3})$. Since $\omega=\frac{\Delta\theta}{\Delta t}$, then $\Delta t=\frac{\Delta\theta}{\omega}=\frac{2\sin^{-1}(\frac{1}{3})}{125}\mathrm{s}\approx\frac{2\times0.3398}{125}\mathrm{s}\approx5.44\times10^{-3}\mathrm{s}$

Answer:

(a) $6.0\mathrm{mm}$ (b) $\frac{125}{2\pi}\mathrm{Hz}\approx19.9\mathrm{Hz}$ (c) $\frac{125}{8.5}\mathrm{m/s}\approx14.7\mathrm{m/s}$ (d) $\approx23.2\mathrm{m/s}^{2}$ (e) $\approx5.44\times10^{-3}\mathrm{s}$