The equation of a transverse wave traveling along a string is $\mathbf{y}\mathbf{}\mathbf{=}\mathbf{}\mathbf{0}\mathbf{.}\mathbf{15}\mathbf{}\mathbf{sin}\mathbf{(}\mathbf{0}\mathbf{.}\mathbf{79}\mathbf{x}\mathbf{-}\mathbf{13}\mathbf{t}\mathbf{)}$in which xand yis in meters and tis in seconds. (a) What is the displacement yat x=2.3 m,t=0.16 s? A second wave is to be added to the first wave to produce standing waves on the string. If the second wave is of the form$\mathit{y}\mathbf{(}\mathit{x}\mathbf{,}\mathit{t}\mathbf{)}\mathbf{=}{\mathit{y}}_{\mathbf{m}}\mathit{s}\mathit{i}\mathit{n}\mathbf{(}\mathit{k}\mathit{x}\mathbf{-}\mathit{v}\mathit{t}\mathbf{)}$, what are (b)${\mathit{y}}_{\mathbf{m}}$, (c)k, (d)$\mathit{\omega }$, and (e) the correct choice of sign in front ofvfor this second wave? (f) What is the displacement of the resultant standing wave at x =2.3 m,t =0.16 s?

The given wave equation,$y=0.15\sin (0.79x-13t)$

We can find the displacementy at$\mathit{x}\mathbf{=}\mathbf{2}\mathbf{.}\mathbf{3}\mathbf{\hspace{0.33em}}\mathit{m}\mathbf{,}\mathbf{}\mathit{t}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{16}\mathbf{\hspace{0.33em}}\mathit{s}$by inserting these values in the given wave equation.

Then we can find the values of${\mathit{y}}_{\mathbf{m}}\mathbf{,}\mathit{k}\mathbf{,}\mathit{\omega }$by comparing the given equation with the general equation of the wave.

The displacement of the resultant standing wavecan be calculated using the principle of superposition.

Formula:

The wavenumber of the wave,

$k=\frac{2\mathrm{\pi }}{\lambda }$ (i)

The angular frequency of the wave,

$\omega =2\pi f$ (ii)

The transverse speed of the wave,

$v_m=\omega y_m$ (iii)

Here Here f is the frequency of the wave $\lambda$is the wavelength of the wave and$y_m$ is the amplitude of the wave.

The general equation of a translational wave on a string is given as:

$y=y_{m}sin(kx-\omega t)$ (1)

The equation of the wave is given as,

$y=0.15\sin (0.79x-13t)$ (2)

Putting values of x =2.3 m and t =0.16 s in equation (2), we can calculate the displacement y as:

$$\begin{gathered} y=0.15\sin \left(0.79\left(2.3\right)-13\left(0.16\right)\right) \\ =-0.0389 \\ ~-0.039\mathrm{m} \end{gathered}$$

Therefore, the displacement y at $x=2.3\hspace{0.33em}m,t=0.16\hspace{0.33em}s$is -0.039 m.

Comparing equations (1) and (2), we can find the amplitude of the wave as:

$y_m=0.15\mathrm{m}$

Therefore, the value of $y_m$is 0.15 m.

Comparing equations (1) and (2), we can find the value of wavenumber as:

$k=0.79\hspace{0.33em}{m}^{-1}$

Therefore, the value of k is$0.79{\mathrm{m}}^{-1}$

Comparing equations (1) and (2), we can find the value of the angular frequency as:

$\omega =13\frac{\mathrm{rad}}{\mathrm{s}}$

Therefore, the value of$\omega$ is$13\frac{\mathrm{rad}}{\mathrm{s}}$

From the given wave equation, we can conclude that the wave is traveling in the –x direction, so the sign of angular frequency ($\omega$)must be positive.

Thus, the sign of$\omega$ is positive.

The displacement y at$x=2.3\hspace{0.33em}m,t=0.16\hspace{0.33em}s$is -0.039 m.

Let's call it $y_1$.

The wave equation for the second wave can be written as,

$y_2=y_{m}sin(kx+\omega t)$

Substitute the values in the above expression, and we get,

$$\begin{gathered} y_2=0.15\sin \left(0.79\left(2.3\right)+13\left(0.16\right)\right) \\ =-0.101 \end{gathered}$$

Hence, displacement of the resulting standing wave can be calculated as,

$y(x,t)=y_1+y_2$

Substitute the values in the above expression, and we get,

$$\begin{gathered} y\left(x,t\right)=0.039-0.101 \\ =0.14\mathrm{m} \end{gathered}$$.

Hence, the value of the displacement of the resulting wave is -0.14 m.