(a) Sketch graphs of \(y\) versus \(x\) for the function $$ y(x, t)=(0.80 \mathrm{mm}) \sin (k x-\omega t) $$ for the times \(t=0,0.96 \mathrm{s},\) and \(1.92 \mathrm{s} .\) Make all three graphs of the same axes, using a solid line for the first, a dashed line for the second, and a dotted line for the third. Use the values \(k=\pi /(5.0 \mathrm{cm})\) and $\omega=(\pi / 6.0) \mathrm{rad} / \mathrm{s}$ (b) Repeat part (a) for the function $$ y(x, t)=(0.50 \mathrm{mm}) \sin (k x+\omega t) $$ (c) Which function represents a wave traveling in the \(-x\) direction and which represents a wave traveling in the \(+x\) -direction?

For both parts (a) and (b), we are given a function of the form: $$y(x, t) = A \sin(kx \pm \omega t)$$ Where A is the amplitude, \(k\) is the wave number, \(\omega\) is the angular frequency, and the sign in the argument of sine function determines the direction of the wave. The wave in part (a) has a negative sign, and part (b) has a positive sign.

For part (a), let's plot the function \(y(x, t) = (0.80\,\text{mm})\,\sin(kx - \omega t)\) for three different times: \(t_1=0\,\text{s}\), \(t_2=0.96\,\text{s}\), and \(t_3=1.92\,\text{s}\). The given values of \(k=\pi/(5.0\,\text{cm})\) and \(\omega=(\pi/6.0) \mathrm{ rad}/\mathrm{s}\) will be used. 1. For \(t_1 = 0\)s, the function becomes: \(y_1(x) = (0.80\,\text{mm})\,\sin(kx)\). 2. For \(t_2 = 0.96\)s, the function becomes: \(y_2(x) = (0.80\,\text{mm})\,\sin(kx - \omega (0.96\,\text{s}))\). 3. For \(t_3 = 1.92\)s, the function becomes: \(y_3(x) = (0.80\,\text{mm})\,\sin(kx - \omega (1.92\,\text{s}))\). Now you can sketch these three functions \(y_1(x)\), \(y_2(x)\), and \(y_3(x)\) with a solid, a dashed, and a dotted line, respectively, on the same graph.

For part (b), let's plot the function \(y(x, t) = (0.50\,\text{mm})\,\sin(kx + \omega t)\) for three different times: \(t_1=0\,\text{s}\), \(t_2=0.96\,\text{s}\), and \(t_3=1.92\,\text{s}\). The given values of \(k=\pi/(5.0\,\text{cm})\) and \(\omega=(\pi/6.0) \mathrm{ rad}/\mathrm{s}\) will be used. 1. For \(t_1 = 0\)s, the function becomes: \(y_1(x) = (0.50\,\text{mm})\,\sin(kx)\). 2. For \(t_2 = 0.96\)s, the function becomes: \(y_2(x) = (0.50\,\text{mm})\,\sin(kx + \omega (0.96\,\text{s}))\). 3. For \(t_3 = 1.92\)s, the function becomes: \(y_3(x) = (0.50\,\text{mm})\,\sin(kx + \omega (1.92\,\text{s}))\). Now you can sketch these three functions \(y_1(x)\), \(y_2(x)\), and \(y_3(x)\) with a solid, a dashed, and a dotted line, respectively, on the same graph.

To identify the direction of the wave, we need to look at the sign in the argument of the sine function while considering the wave equation. If the sign in the argument of sine function is: - Negative (i.e., \(y(x, t) = A \sin(kx - \omega t)\)), the wave is traveling in the +x direction. - Positive (i.e., \(y(x, t) = A \sin(kx + \omega t)\)), the wave is traveling in the -x direction. So, for part (a) with the given wave function \(y(x, t) = (0.80\,\text{mm})\,\sin(kx-\omega t)\), the wave is traveling in the +x direction. For part (b) with the given wave function \(y(x, t) = (0.50\,\text{mm})\,\sin(kx+\omega t)\), the wave is traveling in the -x direction.