Question

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 is the amplitude of the wave? (b)what is the frequency of the wave? (c) what is the wave speed? (d)find the acceleration of the particle on string at x=22 cm and t=4.3 s. (e)how much time does any given points on the string take to move between displacements y = +2.0 mm to y=-2.0 mm? homework #1

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) $\approx5.44\times10^{-3}\mathrm{s}$