Reflection of waves on string. Impedance of medium.

Problem 2. A string of impedance $Z1 = 2 kg/s$ is joined to a second string of impedance Z2. The strings are along the z-direction and an incident wave $y1 = (0.1 m) cos ((2\pi(100)/s)*t – k1*z)$ is launched in string one moving towards positive z and towards the junction of the two strings.

A) Find Z2 if 50% of the incident power is reflected and the reflected wave has the same orientation as the incident wave.

B) Find Z2 if 50% of the incident power is reflected and the reflected wave is inverted relative to the incident wave.

Answer

The reflection coefficient for wave amplitudes is found here

(http://farside.ph.utexas.edu/teaching/315/Waves/node38.html)

$\Gamma = Ar/Ai = \frac {Z(S)-Z(L)} {Z(S) +Z(L)}$

 (Ar and Ai are the reflected and incident amplitudes)

where Z(L) is load impedance and Z(S) is source impedance

Since the energy density is proportional to A^2 then the reflection coefficient for powers is

$R = \gamma^2 = Pr/Pi = [\frac { (Z(S)-Z(L)}{ Z(S) +Z(L) } ]^2$

In case R=0.5 and amplitudes having the same orientation we have

$[Z(S) -Z(L)] / [Z(S) +Z(L)] = \sqrt{0.5} =0.707$

$0.293*Z(S) =1.707*Z(L)$

$Z(L) =0.172*Z(S) =0.172*2 =0.343 kg/s$

In case R =0.5 and amplitudes are inverted we have

$[Z(S) -Z(L)] / [Z(S) +Z(L)] = -\sqrt{0.5} =-0.707$

$1.707*Z(S) =0.293*Z(L)$

$Z(L) =8.845*Z(S) =8.845*2 =17.689 kg/s$

Observation: The thumb rule is that half of wavelength is lost when the wave is reflecting on a higher impedance medium.