Reflection and transmission at a junction of two strings
The most general case is a string joined to another string with a different linear mass density \(\mu\), the mass per metre. The point where they meet is a junction. When an incident wave reaches the junction, part of its energy is reflected. The rest is transmitted into the second string as the transmitted wave. The two strings are pulled by one tension \(T\), since the light knot between them can have no net horizontal force on it.
A key principle holds at any boundary: the frequency of the wave stays the same across the junction,
\[ f_{\text{incident}} = f_{\text{reflected}} = f_{\text{transmitted}}. \]
The reason is that the junction is a single point shared by both strings. It moves up and down at one rate, and it is this motion that drives the reflected and the transmitted waves.
The speed is a different matter. The speed of a wave on a string, \(v = \sqrt{T/\mu}\), from the chapter Propagating Waves on a String, depends on the mass density. So the speed changes in the second string, and so does the wavelength, \(\lambda = v/f\).
A wave of frequency \(40\,\text{Hz}\) travels at \(120\,\text{m/s}\) on a string. It reaches a junction with a second string, under the same tension, whose mass density is nine times as large. What is the wavelength in the second string, in m?
From a lighter string to a heavier string
Let a wave travel from a less dense string, \(\mu_1\), into a denser one, \(\mu_2 > \mu_1\). The wave speed then decreases, \(v_1 > v_2\). The heavier string has more inertia and is harder to move, so the junction behaves like a rigid boundary.
- The reflected pulse is inverted, a phase change of \(\pi\). It undergoes a hard reflection.
- The transmitted pulse is upright, with no phase change. It moves more slowly in the heavier string, and so it has a shorter wavelength.
From a heavier string to a lighter string
Now let the wave travel from a denser string, \(\mu_1\), into a less dense one, \(\mu_2 < \mu_1\). The wave speed then increases, \(v_1 < v_2\). The lighter string is easier to move, so the junction behaves like a free boundary.
- The reflected pulse is upright, with no phase change. It undergoes a soft reflection.
- The transmitted pulse is also upright. It moves faster in the lighter string and has a longer wavelength.
In Fig 5.1 the heavy string has four times the mass density of the light one, so a wave on it is half as fast. The beads mark the heavy string. Play the first chip. The incident pulse is \(A\) high. The reflected pulse returns inverted, \(A/3\) deep, and the transmitted pulse moves on upright, \(2A/3\) high and half as wide. Play the second chip. Now the reflected pulse is upright, \(A/3\) high, and the transmitted pulse is \(4A/3\) high and twice as wide. The last topic of this lesson derives these fractions.
Why does the width change? A pulse passes the junction in some fixed time, and the whole pulse must pass in that same time. In a slower string, it covers a shorter length in that time, so it is narrower. In a faster string it is wider. For a harmonic wave this is the change of wavelength at a fixed frequency.
A wave pulse travels along a string made of two sections. The first section is thick and heavy, and the second is thin and light. What happens when the pulse reaches the boundary between the two sections?
A pulse crosses from a light string into a heavy string under the same tension. Why is the transmitted pulse narrower than the incident pulse?
How much is reflected, and how much goes on
Put the junction at \(x = 0\). String 1, on the left, has the wave speed \(v_1\), and string 2, on the right, has the speed \(v_2\). Let the incident wave alone move the junction as \(h(t)\), any shape in time. A wave travelling towards \(+x\) at speed \(v\) can be written as a function of \(t - x/v\), because \(x - vt = -v(t - x/v)\). So the three waves are
\[ y_i = A_i\,h(t - x/v_1), \qquad y_r = A_r\,h(t + x/v_1), \qquad y_t = A_t\,h(t - x/v_2). \]
Here \(A_i\), \(A_r\) and \(A_t\) are the incident, reflected and transmitted amplitudes. The reflected wave travels back in string 1; the transmitted wave travels on in string 2.
The junction obeys two rules. First, the strings are tied together, so the string has one displacement at the junction. Second, the junction is a massless knot. The vertical pull on it is \(T\) times the slope on the right minus \(T\) times the slope on the left. That must be zero, or the knot would have an infinite acceleration. So the slope is the same on both sides.
\[ A_i\,h(t) + A_r\,h(t) = A_t\,h(t) \;\Rightarrow\; A_i + A_r = A_t \]
Why. the displacement is continuous: put \(x = 0\) in all three waves. The rule holds at every \(t\), so the amplitudes must match.
\[ \dfrac{\partial y_i}{\partial x} = -\dfrac{A_i}{v_1}h', \quad \dfrac{\partial y_r}{\partial x} = +\dfrac{A_r}{v_1}h', \quad \dfrac{\partial y_t}{\partial x} = -\dfrac{A_t}{v_2}h' \]
Why. the chain rule: the inner functions \(t \mp x/v\) have derivatives \(\mp 1/v\) with respect to \(x\). Here \(h'\) is the derivative of \(h\), taken at \(x = 0\).
\[ -\dfrac{A_i}{v_1} + \dfrac{A_r}{v_1} = -\dfrac{A_t}{v_2} \;\Rightarrow\; \dfrac{A_i - A_r}{v_1} = \dfrac{A_t}{v_2} \]
Why. the slope is continuous: the slope of string 1, incident plus reflected, equals the slope of string 2. Cancel \(h'\), then change every sign.
\[ (A_i - A_r)\,v_2 = (A_i + A_r)\,v_1 \]
Why. put \(A_t = A_i + A_r\) from step 1 into step 3, and multiply both sides by \(v_1 v_2\).
\[ A_r\,(v_1 + v_2) = A_i\,(v_2 - v_1) \;\Rightarrow\; \dfrac{A_r}{A_i} = \dfrac{v_2 - v_1}{v_1 + v_2} \]
Why. collect the terms in \(A_r\) on one side and those in \(A_i\) on the other.
\[ \dfrac{A_t}{A_i} = 1 + \dfrac{A_r}{A_i} = \dfrac{2v_2}{v_1 + v_2} \]
Read it. step 1 divided by \(A_i\), with step 5 put in.
These two numbers are the amplitude reflection ratio \(r = A_r/A_i\) and the amplitude transmission ratio \(t = A_t/A_i\). The sign of \(r\) carries the phase. When string 2 is heavier, \(v_2 < v_1\), so \(r\) is negative and the reflected pulse is inverted. When string 2 is lighter, \(r\) is positive and the reflected pulse is upright. The ratio \(t\) is always positive: the transmitted pulse is always upright.
The ratios contain the two earlier cases as limits. Let string 2 be infinitely heavy, so \(v_2 \to 0\). Then \(r = -1\) and \(t = 0\): all of the wave returns inverted, as at a fixed end. Let string 2 be massless, so \(v_2 \to \infty\). Then \(r = +1\) and \(t \to 2\): the wave returns upright, and the junction moves with twice the amplitude, as at a free end. For two equal strings, \(r = 0\) and \(t = 1\): the wave passes as if no junction were there.
The energy adds up
The chapter Propagating Waves on a String showed that a harmonic wave carries an average power \(P = \tfrac12\mu v\omega^2A^2\). With \(\mu = T/v^2\), this is \(P = \tfrac12(T/v)\,\omega^2A^2\). Both strings have the same tension \(T\) and the same \(\omega\), so on either string \(P\) is proportional to \(A^2/v\).
\[ \dfrac{P_r}{P_i} = \dfrac{A_r^2/v_1}{A_i^2/v_1} = r^2 \]
Why. the reflected wave travels on string 1, at the speed of the incident wave.
\[ \dfrac{P_t}{P_i} = \dfrac{A_t^2/v_2}{A_i^2/v_1} = t^2\,\dfrac{v_1}{v_2} \]
Why. the transmitted wave travels on string 2, at the speed \(v_2\).
\[ r^2 + t^2\dfrac{v_1}{v_2} = \dfrac{(v_2 - v_1)^2 + 4v_1v_2}{(v_1 + v_2)^2} \]
Why. put in the two ratios; \(t^2 v_1/v_2 = 4v_2^2 v_1/(v_2(v_1 + v_2)^2) = 4v_1v_2/(v_1 + v_2)^2\).
\[ \dfrac{(v_2 - v_1)^2 + 4v_1v_2}{(v_1 + v_2)^2} = \dfrac{(v_1 + v_2)^2}{(v_1 + v_2)^2} = 1 \]
Read it. expand: \(v_2^2 - 2v_1v_2 + v_1^2 + 4v_1v_2 = v_1^2 + 2v_1v_2 + v_2^2\).
The reflected power and the transmitted power always add to the incident power. The junction stores nothing and loses nothing; it only shares the energy out.
Move the slider in the figure above. At \(v_2/v_1 = 0.5\), the mass ratio of Fig 5.1, the bars read \(r = -0.33\) and \(t = 0.67\), and the power splits as \(0.11\) reflected and \(0.89\) transmitted. At \(1\) nothing is reflected. At \(3\) the reflection is upright, \(r = 0.50\), and a quarter of the power returns. At every setting the power bar stays full.
Two wires are joined and stretched with a tension of \(100\,\text{N}\). The first has \(\mu_1 = 0.01\,\text{kg/m}\) and the second \(\mu_2 = 0.04\,\text{kg/m}\). A harmonic wave of frequency \(50\,\text{Hz}\) and amplitude \(3\,\text{mm}\) travels along the first wire to the joint. Find the wavelengths, the reflected and transmitted amplitudes, and check that the powers add up.
\(v_1 = \sqrt{100/0.01} = 100\,\text{m/s}\) and \(v_2 = \sqrt{100/0.04} = 50\,\text{m/s}\). The frequency stays \(50\,\text{Hz}\), so \(\lambda_1 = 2\,\text{m}\) and \(\lambda_2 = 1\,\text{m}\).
\(r = (v_2 - v_1)/(v_1 + v_2)\), \(t = 2v_2/(v_1 + v_2)\), and \(P = \tfrac12\mu v\omega^2A^2\) with \(\omega = 2\pi \times 50 = 100\pi\,\text{rad/s}\).
\(r = -50/150 = -1/3\), so \(A_r = 1\,\text{mm}\), inverted. \(t = 100/150 = 2/3\), so \(A_t = 2\,\text{mm}\). \(P_i = \tfrac12 \times 0.01 \times 100 \times 98\,700 \times (0.003)^2 = 0.444\,\text{W}\), \(P_r = 0.049\,\text{W}\) and \(P_t = \tfrac12 \times 0.04 \times 50 \times 98\,700 \times (0.002)^2 = 0.395\,\text{W}\).
The wavelength halves in the heavy wire. The reflected wave is \(1\,\text{mm}\) and inverted, the transmitted wave \(2\,\text{mm}\) and upright, and \(0.049 + 0.395 = 0.444\,\text{W}\), the incident power.
A pulse of amplitude \(4\,\text{mm}\) travels from a string into a second string, under the same tension, with nine times its mass density. What is the amplitude of the transmitted pulse, in mm?
For the same junction, a second string with nine times the mass density, what fraction of the incident power is reflected?
George Green ran his father's windmill near Nottingham and taught himself mathematics. Later, as a Cambridge graduate, he wrote a paper on the reflection and refraction of sound, printed in 1838. He asked how much of a wave returns and how much goes on where two media meet. He answered it as this lesson does: the boundary must keep the medium joined and its forces balanced.
A cable-television or broadband line carries electrical waves along its length. Where two unlike cables are joined, part of each wave is reflected, exactly as at a junction of two strings. The echo can spoil a signal. So engineers join only cables that carry waves alike, the equivalent of \(v_1 = v_2\), and then \(r = 0\) and nothing comes back.
- Thinking that the frequency changes when a wave enters a new string.The junction drives both strings at its own rate. The frequency stays the same; the speed and the wavelength change.
- Expecting the transmitted pulse to be inverted when the second string is heavier.Only the reflected pulse can be inverted. The transmitted ratio \(t = 2v_2/(v_1 + v_2)\) is always positive.
- Adding the amplitudes to check energy, \(|A_r| + A_t = A_i\).Energy goes as \(A^2/v\). Check \(r^2 + t^2 v_1/v_2 = 1\).