91Ó°ÊÓ

Amplitude of the combined wave is $A=1.50y_m$where$y_m$ the amplitude of one wave.

We have to use the basic formula for the solution of the wave equation and from that; we can find the phase difference between two waves.

Formula:

The general expression of the wave, $y\left(x,t\right)=y_m\sin \left(kx-\mathrm{Ó\neg }tÂ\pm \mathrm{Ï•}\right)$ (i)

Writing the equation for the first wave, using equation (i), we get

$y=y_m\sin \left(kx-\mathrm{Ó\neg }t\right)$

For wave second, using equation (i), we get

$y=y_m\sin \left(kx-\mathrm{Ó\neg }t+\mathrm{Ï•}\right)$

The resultant equation, using the superposition principle is given as:

$y=y_m\sin \left(kx-\mathrm{Ó\neg }t\right)+y_m\sin \left(kx-\mathrm{Ó\neg }t+\mathrm{Ï•}\right)$

By using trigonometric relation

$$\begin{gathered} y=y_m\sin \left(kx-\mathrm{Ó\neg }t\right)+y_m\left[\sin \left(kx-\mathrm{Ó\neg }t\right)\cos \left(\mathrm{Ï•}\right)+\cos \left(kx-\mathrm{Ó\neg }t\right)\sin \left(\mathrm{Ï•}\right)\right] \\ =y_m\sin \left(kx-\mathrm{Ó\neg }t\right)+y_m\left[1+\cos \left(\mathrm{Ï•}\right)\right]+y_m\cos \left(kx-\mathrm{Ó\neg }t\right)\sin \left(\mathrm{Ï•}\right) \\ =2y_m\sin \left(kx-\mathrm{Ó\neg }t\right)\left[\frac{1+\cos \left(\mathrm{Ï•}\right)}{2}\right]+2y_m\cos \left(kx-\mathrm{Ó\neg }t\right)\sin \left(\frac{\mathrm{Ï•}}{2}\right)\cos \left(\frac{\mathrm{Ï•}}{2}\right) \\ =2y_m\sin \left(kx-\mathrm{Ó\neg }t\right)\left(\frac{\mathrm{Ï•}}{2}\right)+2y_m\cos \left(kx-\mathrm{Ó\neg }t\right)\sin \left(\frac{\mathrm{Ï•}}{2}\right)\cos \left(\frac{\mathrm{Ï•}}{2}\right) \\ =2y_m\cos \left(\frac{\mathrm{Ï•}}{2}\right)\sin \left(kx-\mathrm{Ó\neg }t+\mathrm{Ï•}\right) \end{gathered}$$

By comparing the equation, we can write the new amplitude as:

$$\begin{gathered} A=2y_m\cos \left(\frac{\mathrm{ϕ}}{2}\right) \\ \cos \left(\frac{\mathrm{ϕ}}{2}\right)=\frac{A}{2y_m} \\ \frac{\mathrm{ϕ}}{2}=\left(\frac{A}{2y_m}\right)(if\frac{\mathrm{ϕ}}{2}isveryverysmall) \\ \mathrm{ϕ}=2\left(\frac{1.50y_m}{2y_m}\right)\left(\mathrm{substituting}\mathrm{the}\mathrm{given}\mathrm{values}\right) \\ \mathrm{ϕ}=2\left(\frac{1.50}{2}\right) \\ =\left(0.75\right) \\ =82.8° \end{gathered}$$

Hence, the value of phase in degrees is$82.8°$

Phase in radian is given as:

$$\begin{gathered} \mathrm{Ï•}=\frac{\mathrm{Ï€}}{180}×82.8°(âˆ\mu 1Radian=\frac{\mathrm{Ï€}}{180}×1Degree) \\ =1.45\mathrm{rad} \end{gathered}$$

Hence, the value of phase in radians is$1.45rad$

To write the phase in terms of wavelength, we can use the fact that each wavelength corresponds to $2\mathrm{Ï€}$radian. So $\mathrm{Ï•}$in terms of wavelength is given as:

$$\begin{gathered} \mathrm{Ï•}=\frac{1.45\mathrm{rad}}{2\mathrm{Ï€}} \\ =0.230\mathrm{wavelength} \end{gathered}$$

Hence, the value of phase in wavelengths is$0.230\mathrm{wavelength}$