What phase difference between two identical traveling waves, moving in the same direction along a stretched string, results in the combined wave having an amplitude1.50 times that of the common amplitude of the two combining waves? (a)Express your answer in degrees, (b) Express your answer in radians, and (c) Express your answer in wavelengths.

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, $y\left(x,t\right)=y_m\sin \left(kx-\omega t\pm \varphi \right)$ (i)

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

$y=y_m\sin \left(kx-\omega t\right)$

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

$y=y_m\sin \left(kx-\omega t+\varphi \right)$

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

$y=y_m\sin \left(kx-\omega t\right)+y_m\sin \left(kx-\omega t+\varphi \right)$

By using trigonometric relation

$$\begin{gathered} y=y_m\sin \left(kx-\omega t\right)+y_m\left[\sin \left(kx-\omega t\right)\cos \left(\varphi \right)+\cos \left(kx-\omega t\right)\sin \left(\varphi \right)\right] \\ =y_m\sin \left(kx-\omega t\right)+y_m\left[1+\cos \left(\varphi \right)\right]+y_m\cos \left(kx-\omega t\right)\sin \left(\varphi \right) \\ =2y_m\sin \left(kx-\omega t\right)\left[\frac{1+\cos \left(\varphi \right)}{2}\right]+2y_m\cos \left(kx-\omega t\right)\sin \left(\frac{\varphi }{2}\right)\cos \left(\frac{\varphi }{2}\right) \\ =2y_m\sin \left(kx-\omega t\right)\left(\frac{\varphi }{2}\right)+2y_m\cos \left(kx-\omega t\right)\sin \left(\frac{\varphi }{2}\right)\cos \left(\frac{\varphi }{2}\right) \\ =2y_m\cos \left(\frac{\varphi }{2}\right)\sin \left(kx-\omega t+\varphi \right) \end{gathered}$$

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

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

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

Phase in radian is given as:

$$\begin{gathered} \varphi =\frac{\mathrm{\pi }}{180}\times 82.8°(\because 1Radian=\frac{\mathrm{\pi }}{180}\times 1Degree) \\ =1.45\mathrm{rad} \end{gathered}$$

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 $2\mathrm{\pi }$radian. So $\varphi$in terms of wavelength is given as:

$$\begin{gathered} \varphi =\frac{1.45\mathrm{rad}}{2\mathrm{\pi }} \\ =0.230\mathrm{wavelength} \end{gathered}$$

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