91Ó°ÊÓ

The given wave equations are:

$$\begin{gathered} y_1=0.30\sin \left[\mathrm{Ï€}\left(5\mathrm{x}-200\mathrm{t}\right)\right] \\ {y}_2=0.30\sin \left[\mathrm{Ï€}\left(5\mathrm{x}-200\mathrm{t}\right)+\mathrm{Ï€}/3\right] \end{gathered}$$

The superposition law states that the resultant wave equation is just the algebraic sum of the two waves that are combined. The amplitude of the wave is the maximum displacement of the travelling wave. Now, the wave speed of the wave is the distance traveled by the wave in a given time. Again, the distance between any two identical points in the adjacent cycles of a waveform signal propagating in space is called the wavelength of the travelling wave.

Formulae:

If two identical waves are travelling in the same direction, with the same frequency, wavelength and amplitude but different in phase, then the resultant wave is given by,

$y=A\cos \left(kx-\mathrm{Ó\neg }t+\mathrm{Ï•}\right)..........\left(1\right)$

Where,A is the amplitude of the wave,k is the wave number,$\mathrm{Ó\neg }$ is the angular frequency and $\mathrm{Ï•}$is the phase of the travelling wave, .

The wave speed of the travelling wave, $V=\frac{\mathrm{Ó\neg }}{k}.......\left(2\right)$

The wavelength of the travelling wave,$\mathrm{λ}=\frac{2\mathrm{π}}{k}..........\left(3\right)$

Using the concept, the resultant wave due to the combination of the two given waves can be given as follows:

$$\begin{gathered} y=y_1+y_2 \\ =0.30\sin \left[\mathrm{Ï€}\left(5\mathrm{x}-200\mathrm{t}\right)\right]+0.30\sin \left[\mathrm{Ï€}\left(5\mathrm{x}-200\mathrm{t}\right)+\mathrm{Ï€}/3\right] \\ =0.30\left[2\sin \frac{\left\{\mathrm{Ï€}\left(5\mathrm{x}-200\mathrm{t}\right)+\mathrm{Ï€}\left(5\mathrm{x}-200\mathrm{t}\right)+\mathrm{Ï€}/3\right\}}{2}\cos \frac{\left\{\mathrm{Ï€}\left(5\mathrm{x}-200\mathrm{t}\right)-\mathrm{Ï€}\left(5\mathrm{x}-200\mathrm{t}\right)-\mathrm{Ï€}/3\right\}}{2}\right] \\ =0.30\left[2\sin \frac{2\left\{\mathrm{Ï€}\left(5\mathrm{x}-200\mathrm{t}\right)+\mathrm{Ï€}/3\right\}}{2}\cos \left(\frac{\mathrm{Ï€}}{6}\right)\right] \\ =\left(0.60\cos \left(\frac{\mathrm{Ï€}}{6}\right)\right)\sin \left[\mathrm{Ï€}\left(5\mathrm{x}-200\mathrm{t}\right)+\frac{\mathrm{Ï€}}{6}\right]..........\left(4\right) \end{gathered}$$

Now, comparing the above calculated equation with equation (1), the amplitude of the resultant wave can be given as follows:

$$\begin{gathered} A=\left(0.60\cos \left(\frac{\mathrm{π}}{6}\right)\right) \\ =0.60×\frac{\sqrt{3}}{2} \\ =0.52\mathrm{m} \end{gathered}$$

Hence, the value of the amplitude is 0.52 m.

Now, comparing the values of equation (4) with the equation (1), the wavenumber and angular frequency of the wave are given as:$k=5{\mathrm{Ï€³¾}}^{-1},\mathrm{Ó\neg }=200{\mathrm{Ï€²õ}}^{-1}$

Thus, the wave speed of the travelling wave can be given using equation (2) as follows:

$$\begin{gathered} v=\frac{200{\mathrm{Ï€²õ}}^{-1}}{5{\mathrm{Ï€³¾}}^{-1}} \\ =40\mathrm{m}/\mathrm{s} \end{gathered}$$

Hence, the wave speed of the wave is 40 m/s.

Now, the wavelength of the wave can be given using equation (iii) as follows:

$$\begin{gathered} \mathrm{λ}=\frac{2\mathrm{π}}{5\mathrm{π}{\mathrm{m}}^{-1}} \\ =0.4\mathrm{m} \end{gathered}$$

Hence, the value of the wavelength is 0.4 m.