Two sinusoidal waves with the same amplitude of 9.00 mmand the same wavelength travel together along a string that is stretched along anaxis. Their resultant wave is shown twice in Figure, as valleyAtravels in the negative direction of the xaxis by distance d=56.0 cmin 8.0 ms. The tick marks along the axis are separated by 10cm, and heightHis 8.0 mm. Let the equation for one wave be of the for$\mathbf{y}\mathbf{(}\mathbf{x}\mathbf{,}\mathbf{t}\mathbf{)}\mathbf{=}{\mathbf{y}}_{\mathbf{m}}\mathbf{}\mathbf{sin}\mathbf{(}\mathbf{kx}\mathbf{\pm }\mathbf{\omega t}\mathbf{+}{\mathbf{\phi }}_{\mathbf{1}}\mathbf{)}$, where${\mathit{\phi }}_{\mathbf{1}}\mathbf{=}\mathbf{0}$and you must choose the correct sign in front of$\mathbf{\omega }$. For the equation for the other wave, what are (a)What is${\mathit{y}}_{\mathbf{m}}$, (b)What isk, (c)What is$\mathit{\omega }$, (d)What is${\mathit{\phi }}_{\mathbf{2}}$, and (e)What is the sign in front of$\mathit{\omega }$?

i) Amplitude of first wave,$y_m=9\hspace{0.33em}\mathrm{mm}\mathrm{or}0.009\hspace{0.33em}\mathrm{m}$

ii) Distance along negative x direction,$d=56\hspace{0.33em}\mathrm{cm}\mathrm{or}0.56\mathrm{m}$

iii) Time period off the wave,$t=8.00\hspace{0.33em}\mathrm{ms}\mathrm{or}8.00\times {10}^{-3}s$

iv) Phase one,${\varphi }_1=0$

v) From fig wavelength,$\lambda =40\mathrm{cm}$

vi) Height of wave,$H=8.00\hspace{0.33em}\mathrm{mm}\mathrm{or}8.00\times {10}^{-3}m$

We use the basic formula for the solution of the wave equation, and from that, we can find the phase, amplitude, angular wave number, angular frequency, and phase of the wave.

Formula:

The general expression of the wave, $y=y_{m}sin(kx-\omega t)$(i)

The wave number of a wave, $k=\frac{2\mathrm{\pi }}{\lambda }$(ii)

The velocity of a wave,$v=dx/dt$ (iii)

The angular frequency of a wave,$\omega =vf$ (iv)

The amplitude of a wave, $A=2y_m\cos \left(\frac{\mathrm{\varphi }}{2}\right)$ (v)

Given thatthetwo waves have the same amplitude. So, the second wave also has the amplitude equal to$y_m=9\mathrm{mm}$

Hence, the value of the amplitude of the other wave is 9 mm

Using equation (ii) and the given value of wavelength, the wavenumber is given as:

$$\begin{gathered} k=\frac{2\mathrm{\pi }}{0.4} \\ =15.7\mathrm{rad}/\mathrm{m} \end{gathered}$$

Hence, the value of the wave number of the other wave is 15.7 rad/m

As, dx is the distance covered in timedt

Using equation (iii), the velocity of wave is given as:

$$\begin{gathered} v=\frac{0.56}{8.00\times {10}^{-3}} \\ =70\mathrm{m}/\mathrm{s} \end{gathered}$$

Using equation (iv) and the given values, we get the angular frequency of the other wave as:

$$\begin{gathered} \omega =70\times 15.7 \\ =1100\mathrm{rad}/\mathrm{s} \end{gathered}$$

Hence, the value of the angular frequency is 1100 rad/s

From the given data, amplitude is given as:

$$\begin{gathered} A=4.00\mathrm{mm} \\ =4.00\times {10}^{-3}\mathrm{m} \end{gathered}$$

Hence, using equation (v), we get the phase angle as:

$$\begin{gathered} \cos \left(\frac{\varphi }{2}\right)=\frac{A}{2y_m} \\ \varphi =2{\cos }^{-1}\left(\frac{A}{2y_m}\right) \\ =2{\cos }^{-1}\left(\frac{4.00\times {10}^{-3}}{2\times 9\times {10}^{-3}}\right) \\ =2{\cos }^{-1}\left(0.23\right) \\ =154.3° \\ =\frac{\mathrm{\pi }}{180}\times 154.3 \\ =2.69\mathrm{rad} \end{gathered}$$

Hence, the value of another phase is 2.69 rad

We know that the solution of the wave equation is $y(x,t)=y_{m}sin(kx\pm \omega t)$

The sign, in this equation, depends on the direction in which the wave travels. If the wave travels in the positive x direction then the sign of angular frequency is negative. If the wave travels in the negative x direction, then the sign of the angular velocity is positive. In the problem, the wave travels in negative x direction so, the sign in the wave equation is positive. Therefore, we can write,

$y(x,t)=y_{m}sin(kx+\omega t+\varphi )$Where$\varphi$ is phase difference of the two waves.

Hence, the sign in front of angular frequency is positive.