Question

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.

Short answer

1. The phase difference between the two identical waves in degrees is$82.8°$
2. The phase difference between the two identical waves in radians is 1.45 rad
3. The phase difference between the two identical waves in wavelengths is 0.230 wavelength

Step-by-step solution

The given data

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

Understanding the concept of the wave equation

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)

Calculation of the phase difference in degrees

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$82.8°$

b) Calculation of phase in radians

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$1.45rad$

c) Calculation of phase in wavelengths

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$0.230\mathrm{wavelength}$