Question

Question:Equation 9.36 describes the most general linearly polarized wave on a string. Linear (or "plane") polarization (so called because the displacement is parallel to a fixed vector n) results from the combination of horizontally and vertically polarized waves of the same phase (Eq. 9.39). If the two components are of equal amplitude, but out of phase by (say,${\delta }_{\nu }=0,{\delta }_{\mathrm{h}}=90°$,), the result is a circularly polarized wave. In that case:

(a) At a fixed point, show that the string moves in a circle about the axis. Does it go clockwise or counter clockwise, as you look down the axis toward the origin? How would you construct a wave circling the other way? (In optics, the clockwise case is called right circular polarization, and the counter clockwise, left circular polarization.)

(b) Sketch the string at time t =0.

(c) How would you shake the string in order to produce a circularly polarized wave?

Short answer

Answer

(a)

The wave is circles counter clockwise.

Use to make the wave circle in a different way is${\delta }_{\text{h}}=-90°$.

(b) Sketch the string at time t = 0 is obtained and is shown in Figure 3

(c) We must shake the string in a circular motion in order to create a circularly polarised wave.

Step-by-step solution

Write the given data from the question.

Consider the wave equation for vertical polarization is$f_v\left(z,t\right)=A\cos \left(kz-\omega t\right)\hat{x}$.

Consider the wave equation for horizontal polarization is,

localid="1658317731880" $$\begin{gathered} f_h\left(z,t\right)=A\cos \left(kz-\omega t+90°\right)\hat{y} \\ =-A\sin \left(kz-\omega t\right)\hat{y} \end{gathered}$$

Determine the formula of vertical and horizontal acceleration

Write the formula of vertical and horizontal acceleration.

${\mathbf{f}}_{\mathbf{v}}^{\mathbf{2}}+{\mathbf{f}}_{\mathbf{h}}^{\mathbf{2}}={\mathbf{A}}^{\mathbf{2}}$ …… (1)

Here, ${\mathbf{f}}_{\mathbf{v}}^{\mathbf{2}}+{\mathbf{f}}_{\mathbf{h}}^{\mathbf{2}}$is vector sum lies on a circle of radius A.

(a) Determine the wave is circles counter clockwise and wave to circle in another way use.

The condition between vertical and horizontal acceleration is given as follows:

At time,

Substitute f for $f_v+f_h$ and $A\cos \left(kz\right)\hat{x}-A\sin \left(kz\right)\hat{y}$ for into equation (1).

$f=A\cos \left(kz\right)\hat{x}-A\sin \left(kz\right)\hat{y}$

Now, at time.

$$\begin{gathered} f=A\cos \left(kz-90°\right)\hat{x}-A\sin \left(kz-90°\right)\hat{y} \\ =A\sin \left(kz\right)\hat{x}+A\cos \left(kz\right)\hat{y} \end{gathered}$$

Thus, the circles counter clockwise.

Use to make the wave circle in a different way is ${\delta }_{\text{h}}=-90°$.

(b) Sketch the string at time .

As reference of equation 9.36

The most widespread linearly polarised wave on a string is described by equation 9.36. It is possible to create linear (or "plane") polarisation by combining waves that are both horizontally and vertically polarised and have the same phase. This is possible because the displacement is parallel to a fixed vector, termed n.

At time t = 0,

Consider the wave equation for vertical polarization is $f_v\left(z,t\right)=A\cos \left(kz-\omega t\right)\hat{x}$

Figure 1

Consider the wave equation for horizontal polarization is,

$$\begin{gathered} f_h\left(z,t\right)=A\cos \left(kz-\omega t+90°\right)\hat{y} \\ =-A\sin \left(kz-\omega t\right)\hat{y} \end{gathered}$$

Figure 2

Now, construct the following sketch shows the string position at t = 0 by using vertical and horizontal acceleration.

Figure 3

(c) Determine shake the string in order to produce a circularly polarized wave.

This wave equation may alternatively be seen as the superposition of two parallel linearly polarised waves that have the same amplitude but are phased differently by a factor of $\frac{\pi }{2}$ .

We must shake the string in a circular motion in order to create a circularly polarised wave.