Write the complex displacements as

y_{1} = (A e^{i k_{1} x} + B e^{- i k_{1} x}) e^{- iω t}

and

y_{2} = C e^{i k_{2} x - iω t}

. At a massless join, continuity of displacement and transverse force gives

A + B = C, T k_{1} (A - B) = T k_{2} C .

Eliminating

B

t = \frac{C}{A} = \frac{2 k _{1}}{k _{1} + k _{2}}, r = \frac{B}{A} = \frac{k _{1} - k _{2}}{k _{1} + k _{2}} .

The angular frequency is shared across a stationary interface, whereas the wavelength changes. The equal-tension assumption is used in the slope condition.

ii

Solution

At the same tension,

k \propto ρ

, so

k_{2} / k_{1} = 3

t = 1/2

and

r = - 1/2

t^{2}

T = \frac{ρ _{2} c _{2}}{ρ _{1} c _{1}} ∣ t ∣^{2} = \frac{ρ _{2}}{ρ _{1}} \frac{1}{4} = \frac{3}{4} .

The reflected fraction is

R = ∣ r ∣^{2} = 1/4

, so

R + T = 1

. The transmitted displacement amplitude is

2.5 mm

and its wavelength is

0.5/3 m

. The negative

r

indicates a reflected phase reversal; it does not mean negative reflected energy.

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