不同线密度弦上波的干涉及驻波问题咨询

Great question—this scenario is a perfect way to tie together wave impedance, reflection/transmission, and interference. Let’s break it down step by step, using your example of a string split into two equal lengths: first segment (0 to L) with linear density μ, second (L to 2L) with 9μ, and waves entering from the μ end.

First: Wave Speed & Impedance Basics #

Let’s start with foundational numbers to set the stage. For a string under tension T, wave speed is given by:
v = sqrt(T/μ)
So for the two segments:

• v₁ = sqrt(T/μ) (light segment, 0 to L)
• v₂ = sqrt(T/(9μ)) = v₁/3 (heavy segment, L to 2L)

Wave impedance (the property that dictates reflection/transmission behavior) is Z = μv = sqrt(Tμ). Comparing the two segments:

• Z₁ = sqrt(Tμ)
• Z₂ = sqrt(T*9μ) = 3Z₁
  Since Z₂ > Z₁, we’re moving from a low-impedance to high-impedance medium at the junction (x=L).

Reflection at the μ ↔ 9μ Junction (W₁) #

When the incident wave hits x=L, two things happen:

• Transmission: A portion of the wave continues into the 9μ segment. The transmission coefficient t (ratio of transmitted amplitude to incident amplitude) is:
  t = 2Z₁/(Z₁ + Z₂) = 2Z₁/(Z₁ + 3Z₁) = 0.5
  So the transmitted wave has half the amplitude of the incident wave, with no phase shift.
• Reflection (W₁): The reflected wave bounces back into the μ segment. Since we’re moving from low to high impedance, this reflection has a π phase shift (equivalent to flipping the wave’s sign). The reflection coefficient r is:
  r = (Z₁ - Z₂)/(Z₁ + Z₂) = (Z₁ - 3Z₁)/(4Z₁) = -0.5
  The negative sign represents that π phase shift, and the amplitude is half the incident wave’s.

Transmission & Reflection at the Far End (x=2L) #

The transmitted wave travels through the 9μ segment at speed v₁/3, taking 3L/v₁ time to reach x=2L. Assuming the far end is a fixed boundary (standard for such problems unless stated otherwise), the wave reflects here with another π phase shift.

Let’s call this reflected wave W₂. It travels back toward x=L with:

• Amplitude equal to the transmitted wave’s amplitude (fixed-end reflections preserve amplitude, only shifting phase by π).
• A total π phase shift from the fixed end reflection.

When W₂ reaches x=L again, it transmits back into the μ segment. Since we’re now moving from high to low impedance, the transmission coefficient here is t' = 2Z₂/(Z₁ + Z₂) = 1.5. This means W₂’s amplitude increases to 0.5A * 1.5 = 0.75A (where A is the incident wave’s amplitude) when entering the μ segment, with no additional phase shift.

Interference of Reflected Waves (W₁ & W₂) #

Now, W₁ (reflected directly at x=L) and W₂ (reflected at x=2L then transmitted back) will interfere in the μ segment. Here’s how their interaction plays out:

• Phase Alignment: Both waves have a π phase shift (W₁ from the junction reflection, W₂ from the fixed end). The path length difference for W₂ is 2L in the 9μ segment, which translates to 6L/λ₁ wavelengths (since λ₂ = λ₁/3). If this is an integer multiple of wavelengths, the waves will be in phase, leading to constructive interference.
• Amplitude Superposition: W₁ has amplitude 0.5A, W₂ has 0.75A. When in phase, their combined amplitude is 1.25A; when out of phase, it’s 0.25A. For steady-state continuous input, this creates a stable interference pattern that varies with position along the μ segment.

Standing Wave Formation #

For standing waves to form, the incident frequency must satisfy resonance conditions where the superposition of incident and reflected waves creates fixed nodes and antinodes. Here’s what that looks like for each segment:

• μ Segment (0 to L): Standing waves require that the boundary at x=L matches the wave behavior in the 9μ segment. Since the heavy segment’s wave speed is 1/3 of the light segment, a length L in the heavy segment acts like 3L in the light segment (in terms of wavelength count). The total effective length for resonance is L + 3L = 4L, so resonant frequencies are:
  f = n*v₁/(8L) (where n is a positive integer)
• 9μ Segment (L to 2L): The wavelength here is λ₁/3, so nodes and antinodes are 3x closer together than in the light segment. The fixed end at x=2L is always a node, and the junction at x=L will be a node or antinode depending on the resonant frequency.

内容的提问来源于stack exchange，提问作者SmarthBansal