Question

Two sinusoidal waves, identical except for phase, travel in the same direction along a string, producing the net wave$\mathbf{y}\mathbf{\text{'}}\mathbf{(}\mathbf{x}\mathbf{,}\mathbf{t}\mathbf{)}\mathbf{=}\mathbf{(}\mathbf{3}\mathbf{.}\mathbf{0}\mathbf{\hspace{0.33em}}\mathbf{mm}\mathbf{)}\mathbf{}\mathbf{}\mathbf{sin}\mathbf{(}\mathbf{20}\mathbf{x}\mathbf{-}\mathbf{4}\mathbf{.}\mathbf{0}\mathbf{t}\mathbf{+}\mathbf{0}\mathbf{.}\mathbf{820}\mathbf{\hspace{0.33em}}\mathbf{rad}\mathbf{)}$, with x in meters and t in seconds. (a) What is the wavelength$\mathit{\lambda }$of the two waves, (b) What is the phase difference between them, and (c) What is their amplitude${\mathit{y}}_{\mathbf{m}}$?

Short answer

a) The wavelength $\lambda$of the two waves is 0.31 m.

b) The phase difference between the two waves is 1.64 rad.

c) The amplitude of the two waves is 2.2 mm.

Step-by-step solution

The given data

The wave equation is ,$\mathrm{y}\text{'}(\mathrm{x},\mathrm{t})=(3.0\mathrm{mm})\sin (20\mathrm{x}-4.0\mathrm{t}+0.820\hspace{0.33em}\mathrm{rad})$.

Understanding the concept of the wave equation

Comparing the given equation with the general wave equation for two interfered waves, we can get the values of phase difference, amplitude, and wave constant of the net wave. From the wave constant, we can easily find the wavelength of the two waves using corresponding relations.

Formulae:

The wavenumber of the wave,

$k=\frac{2\mathrm{\pi }}{\lambda }$ (i)

The angular frequency of the wave,

$\omega =2\mathrm{\pi f}$ (ii)

The velocity of the body in angular motion,

$v_m=\omega y_m$ (iii)

The velocity of the wave,

$v=f\lambda$ (iv)

Here$y_m$ is the amplitude of the wave,f is the frequency of the wave, and$\lambda$ is the wavelength of the wave.

a) Calculation of wavelength

The general equation of an interfering wave is,

$y^{\text{'}}\left(x,t\right)=2y_m\cos \left(\frac{1}{2}\varphi \right)\sin \left(kx-\omega t+\frac{1}{2}\varphi \right)$ (1)

The given equation of the wave is,

${\mathrm{y}}^{\text{'}}(\mathrm{x},\mathrm{t})=(3.0\mathrm{mm})\sin (20\mathrm{x}-4.0\mathrm{t}+0.820\mathrm{rad})$ (2)

Comparing equations (1) and (2), we get the wavenumber as,

$k=20{\mathrm{m}}^{-1}$

Using equation (i) and the given values, we get the wavelength as:

$$\begin{gathered} \lambda =\frac{2\left(3.14\right)}{20} \\ =0.31m \end{gathered}$$

Hence, the value of the wavelength is 0.31 m

b) Calculation of phase difference between two waves

Comparing equations (1) and (2), we get the phase difference,

$$\begin{gathered} \frac{1}{2}\mathrm{\varphi }=0.820\mathrm{rad} \\ \mathrm{\varphi }=1.64\mathrm{rad} \end{gathered}$$

Therefore, the phase difference between the two waves is 1.64 rad.

c) Calculation of the amplitude

Comparing equations (1) and (2), we get the amplitude of the wave as:

$$\begin{gathered} 2{\mathrm{y}}_{\mathrm{m}}\cos \left(\frac{1}{2}\mathrm{\varphi }\right)=3.0\mathrm{mm} \\ 2{\mathrm{y}}_{\mathrm{m}}\cos \left(\frac{1}{2}\left(1.64\right)\right)=3.0\mathrm{mm} \\ 1.36{\mathrm{y}}_{\mathrm{m}}=3.0\mathrm{mm} \\ {\mathrm{y}}_{\mathrm{m}}=2.198\mathrm{mm} \\ \cong 2.2\mathrm{mm} \end{gathered}$$

Therefore, the amplitude of the two waves is 2.2 mm.