Why Standing Waves Carry No Net Energy Along an Ideal String

Imagine a string vibrating in a stable standing-wave pattern. Every point between the nodes is moving up and down, sometimes quite rapidly. Yet the pattern itself doesn't travel along the string.

That creates an interesting puzzle. If the string is moving and contains energy, why doesn't that energy travel from one end to the other?

**In the ideal model, a standing wave carries zero net energy along the string because it consists of two equal-amplitude travelling waves moving in opposite directions.** Each wave transports the same average amount of energy per second, but in the opposite direction. Their energy transfers cancel.

This doesn't mean the string has no energy. Its energy is continually exchanged between kinetic energy and elastic potential energy. Energy can also flow locally in either direction during a cycle, but there is no net transfer along the string when averaged over a complete cycle.

Before going further, consider one prediction: if the entire string is momentarily straight, has its energy disappeared?

No. At that instant, the string can have maximum transverse velocity and its energy is kinetic rather than elastic potential energy.

[IMAGE: A taut horizontal string displaying a standing-wave pattern between two fixed ends, with stationary nodes and moving antinodes suggested by faint successive positions. Include two equal-amplitude travelling-wave arrows pointing in opposite directions along the string. Show that the combined pattern remains stationary.]

## 1. Two travelling waves create a standing wave

Start with a familiar travelling wave.

Imagine flicking one end of a stretched string. A disturbance moves along the string, carrying energy with it. Individual points on the string move perpendicular to the direction in which the wave travels.

The distinction is important:

- The wave travels along the string.
- The string particles move up and down.
- Energy is transferred along the string, even though the material itself does not travel along with the wave.

Now imagine a second wave travelling along the same string in the opposite direction.

If the two waves have the same frequency, wavelength, and amplitude, they can combine to form a standing wave.

This happens naturally when a wave reflects from a boundary and overlaps with the returning wave.

### What happens when the waves overlap?

The principle of superposition states that the resultant displacement at any point is the sum of the individual wave displacements.

For two equal-amplitude waves travelling in opposite directions, we can write:

\[
y_1=A\sin(kx-\omega t)
\]

\[
y_2=A\sin(kx+\omega t)
\]

Here:

- \(A\) is the amplitude of each travelling wave, in metres.
- \(x\) is position along the string, in metres.
- \(t\) is time, in seconds.
- \(k=2\pi/\lambda\) is the wave number, in radians per metre.
- \(\omega=2\pi f\) is the angular frequency, in radians per second.
- \(\lambda\) is wavelength, in metres.
- \(f\) is frequency, in hertz.

Adding the two displacements gives:

\[
\begin{aligned}
y&=y_1+y_2\\
&=2A\sin(kx)\cos(\omega t)
\end{aligned}
\]

Notice something interesting about this equation.

The position-dependent term, \(\sin(kx)\), determines the amplitude at each location. The time-dependent term, \(\cos(\omega t)\), determines how the string oscillates.

The resulting pattern does not travel along the string. Its nodes and antinodes remain in fixed positions.

[IMAGE: Three consecutive snapshots of a fixed-end standing wave at t = 0, t = T/4, and t = T/2. Show the string at maximum positive displacement, momentarily straight, and maximum negative displacement respectively. Keep the nodes in exactly the same positions in all three snapshots. Include arrows at the antinodes in the straight-string snapshot to indicate transverse motion.]

For this particular arrangement, the nodes occur where:

\[
\sin(kx)=0
\]

Therefore:

\[
x=\frac{n\lambda}{2}
\]

where \(n=0,1,2,\ldots\).

The antinodes occur halfway between neighbouring nodes:

\[
x=\frac{(2n+1)\lambda}{4}
\]

These positions describe a string with a node at \(x=0\).

The important point is that the standing-wave pattern is produced by two waves that are still travelling. The individual waves haven't stopped moving just because their resultant pattern is stationary.

For a refresher on how the string's length and boundary conditions determine its allowed harmonics, see [Standing Waves on Strings: Harmonics and Wave Speed](https://tutorgum.com/learn/resources/hsc-physics-standing-waves-on-strings/).

## 2. Phase difference determines the interference pattern

To understand why the two waves cancel at some points and reinforce at others, we need to consider their phase difference.

Phase describes the position of a point within a wave cycle. Two waves arriving at a point in phase have matching displacements and oscillate together. Waves arriving in antiphase have opposite displacements and cancel if their amplitudes are equal.

The phase difference depends on two things:

1. The path difference between the waves.
2. Any phase change caused by reflection at a boundary.

The relationship between path difference and phase difference is:

\[
\Delta\phi=\frac{2\pi\Delta r}{\lambda}
+\phi_{\mathrm{reflection}}
\]

Here, \(\Delta r\) is the path difference in metres, and \(\phi_{\mathrm{reflection}}\) is the phase change introduced by reflection.

### What does a fixed boundary do?

Consider a wave travelling towards a rigidly fixed end of a string.

The endpoint cannot move. When the wave reflects, the reflected wave is inverted.

This is a phase change of:

\[
\phi_{\mathrm{reflection}}=\pi
\]

or \(180^\circ\).

[IMAGE: An annotated horizontal string approaching a rigid fixed boundary at the right. Show an incident upward pulse and its inverted reflected pulse travelling in the opposite direction. Label the fixed endpoint, incident wave, reflected wave, and 180-degree phase reversal.]

Now consider a point \(P\) located a distance \(x\) from that fixed boundary.

The incident and reflected waves have a path difference of \(2x\). This is because one wave has travelled an additional distance equal to the round trip between the boundary and point \(P\).

Therefore, the relative phase difference is:

\[
\Delta\phi=\frac{2\pi(2x)}{\lambda}+\pi
\]

The phase difference determines whether the waves interfere constructively or destructively.

**At the fixed boundary**, \(x=0\).

There is no additional path difference, so the phase difference is simply \(\pi\). The waves cancel completely.

This produces a node.

**One-quarter of a wavelength from the boundary**, \(x=\lambda/4\).

The path difference is:

\[
\Delta r=2x=\frac{\lambda}{2}
\]

This creates an additional phase difference of \(\pi\). Combined with the \(\pi\) phase change from reflection, the total phase difference is \(2\pi\).

The waves are now in phase and reinforce each other.

This produces an antinode.

The pattern continues, with nodes separated by half a wavelength.

### Don't confuse phase difference with energy transfer

These ideas answer different questions.

| Concept | What it tells you |
|---|---|
| Path difference | How far apart the travelling waves' routes are |
| Phase difference | Whether the waves arrive in phase or out of phase |
| Superposition | What the resultant displacement will be |
| Boundary conditions | How waves reflect and where nodes and antinodes form |
| Energy transfer | How much energy travels in each direction |

A phase difference tells you how the waves combine at a particular point. It does not, by itself, tell you whether there is net energy transfer.

That requires us to consider the energy carried by each travelling wave.

### Worked example: Find the interference at a point on the string

A sinusoidal wave reflects from a fixed end of a string. Its wavelength is \(0.80\text{ m}\).

Point \(P\) is located \(0.20\text{ m}\) from the fixed end. Each travelling wave has an amplitude of \(1.5\text{ mm}\).

Determine the phase difference between the waves at \(P\), and calculate the resultant amplitude.

**Step 1**: Calculate the path difference.

The path difference is twice the distance from the reflecting boundary:

\[
\begin{aligned}
\Delta r&=2x\\
&=2(0.20)\\
&=0.40\text{ m}
\end{aligned}
\]

**Step 2**: Calculate the phase difference due to the path difference.

\[
\begin{aligned}
\Delta\phi_{\mathrm{path}}
&=\frac{2\pi\Delta r}{\lambda}\\
&=\frac{2\pi(0.40)}{0.80}\\
&=\pi
\end{aligned}
\]

**Step 3**: Include the phase change from reflection.

Because the end is fixed, reflection introduces an additional phase change of \(\pi\).

\[
\Delta\phi=\pi+\pi=2\pi
\]

The waves are therefore in phase at \(P\).

**Step 4**: Calculate the resultant amplitude.

The two waves reinforce each other completely:

\[
\begin{aligned}
A_{\mathrm{resultant}}
&=1.5+1.5\\
&=3.0\text{ mm}
\end{aligned}
\]

The resultant amplitude is \(3.0\text{ mm}\), so \(P\) is an antinode.

Notice that the reflection phase change was essential. Ignoring it would give the wrong interference prediction.

## 3. Where does the energy actually go?

The key is to remember that a wave carries energy through the motion of the medium, not through the permanent movement of the medium itself.

For a vibrating string, there are two relevant forms of energy:

- **Kinetic energy:** the string's particles are moving transversely.
- **Elastic potential energy:** the string is curved and stretched slightly, with energy associated with the restoring tension.

A standing wave continually changes between these two forms.

### Follow the energy through one complete cycle

Consider a standing wave with period \(T\). We can follow the string at four different moments.

| Time | What the string looks like | Where the energy is |
|---|---|---|
| \(t=0\) | Maximum positive displacement | Maximum elastic potential energy; transverse velocity is zero |
| \(t=T/4\) | String momentarily straight | Maximum kinetic energy; points at the antinodes move fastest |
| \(t=T/2\) | Maximum negative displacement | Maximum elastic potential energy again |
| \(t=3T/4\) | String momentarily straight | Maximum kinetic energy again, with motion in the opposite direction |

[IMAGE: Four clearly aligned snapshots of the same fixed-end standing wave at 0, T/4, T/2, and 3T/4. Show maximum upward displacement, a straight moving string, maximum downward displacement, and a straight moving string in the opposite direction. Include small transverse velocity arrows at the antinodes in the two straight-string snapshots. Label the nodes consistently.]

Notice that the string is not transporting its entire energy towards one end.

Instead, its energy changes form as the string oscillates.

At maximum displacement, the string is momentarily stationary, but its curved shape means elastic potential energy is stored in it.

A quarter of a period later, the string is straight. Its displacement is zero everywhere, but the particles are moving. The energy is now kinetic.

### Does a node contain any energy?

This is another tempting misconception.

A node is a point that never moves transversely. It might seem reasonable to conclude that a node contains no energy.

However, a node can still be associated with elastic potential energy.

Think about the string immediately around a node. Although the node itself remains at zero displacement, the neighbouring sections can be sloped. The tension in the string is associated with this deformation.

The important distinction is:

- A node has zero transverse displacement.
- A node has zero transverse velocity.
- The string around it can still store elastic potential energy.

Destructive interference cancels displacement. It does not destroy energy.

### Why do the two energy transfers cancel?

Return to the two travelling waves that produce the standing wave.

The first wave carries energy in the positive \(x\)-direction. The second carries energy in the negative \(x\)-direction.

For equal-amplitude waves of the same frequency travelling along the same ideal string, their average powers are equal.

Taking rightward energy flow as positive:

\[
\begin{aligned}
P_{\mathrm{net}}
&=P_{\rightarrow}-P_{\leftarrow}\\
&=P_0-P_0\\
&=0
\end{aligned}
\]

Here, \(P_0\) is the average power carried by either travelling component, measured in watts.

The result is zero net average power through any fixed cross-section of the string.

There is an important qualification.

**The instantaneous energy flow does not have to be zero.**

Energy can briefly flow in one direction and then reverse. Over a complete cycle, the energy transferred in each direction balances.

This is why saying that a standing wave carries no energy is slightly misleading. A standing wave contains energy, and local energy transfers can occur. What it does not produce in the ideal model is a sustained, one-way transfer of energy along the string.

[IMAGE: An annotated standing-wave string showing two equal energy-flow arrows in opposite directions, labelled equal average power. Include a marked cross-section and a small inset showing the instantaneous energy-flow direction reversing at different stages of a cycle. Make clear that the resultant standing-wave pattern remains stationary.]

## 4. How could we test whether energy is being transferred?

Looking at a standing-wave pattern is useful, but it isn't enough to establish that the net energy transfer is zero.

A pattern with stationary-looking maxima and minima can also occur when the reflected wave is weaker than the incident wave.

We need a measurement that distinguishes the two travelling components.

### Setting up the experiment

Imagine an investigation using a taut, uniform string.

The apparatus consists of:

1. A signal generator connected to a mechanical vibration driver at the left end.
2. A horizontal string held under a measured, constant tension.
3. An adjustable reflecting boundary at the right end.
4. A high-speed camera recording the string near a marked cross-section.

The right-hand boundary can be changed between a rigid reflector and an energy-absorbing termination.

[IMAGE: Experimental arrangement viewed from the side. Show a signal generator connected to a vibration driver on the left, a horizontal taut string extending rightwards, a rigid or adjustable damped reflecting boundary at the right, and a high-speed camera viewing a marked section of string near the boundary. Include incident and reflected wave directions and identify the measured section.]

First, the driver sends a short sinusoidal burst towards the boundary.

The camera records the incoming wave as it passes the marked section. After reflection, it records the returning wave passing the same section.

This allows us to compare the amplitudes of the two travelling components.

We can then drive the string continuously at a suitable resonant frequency to observe the resulting interference pattern.

### What should the ideal model predict?

For a perfectly reflecting fixed boundary, the reflected wave has the same amplitude as the incident wave.

Its displacement is inverted, but its average power is unchanged.

The two waves therefore carry equal average power in opposite directions.

The net energy transfer is zero.

### Why isn't just observing the pattern enough?

Suppose we only use a stroboscope to observe the continuously driven string.

We might see a stable pattern of maxima and minima. This supports the presence of interference between waves travelling in opposite directions.

However, it doesn't prove that their amplitudes are equal.

That is the limitation of the original design: it measures the resultant displacement, not the energy carried by each travelling component.

We need to measure the two amplitudes separately.

### Revising the experiment: introduce an absorbing boundary

Replace the perfectly reflecting boundary with a damper that absorbs some of the incoming wave's energy.

The reflected wave now has a smaller amplitude.

The interference pattern can still have stationary maxima and minima, but the minima are no longer perfect nodes.

More importantly, the incident wave carries more average power towards the boundary than the reflected wave carries away.

The difference is transferred to the damper and dissipated.

For sinusoidal travelling waves of the same frequency on the same string, average power is proportional to the square of amplitude:

\[
P\propto A^2
\]

This means that if the reflected amplitude decreases, its average power decreases by the square of the amplitude ratio.

Consider the following illustrative data.

<text weight="semibold">Illustrative experimental results</text>

| Measurement | Ideal rigid reflector | Damped boundary |
|---|---:|---:|
| Incident amplitude | 2.0 mm | 2.0 mm |
| Reflected amplitude | 2.0 mm | 1.2 mm |
| Reflected power / incident power | 100% | 36% |
| Net power towards boundary / incident power | 0% | 64% |

For the damped boundary:

\[
\frac{P_{\mathrm{reflected}}}{P_{\mathrm{incident}}}
=\left(\frac{1.2}{2.0}\right)^2
=0.36
\]

Therefore, 36% of the incident average power is reflected.

The remaining 64% is absorbed, assuming there are no other significant losses between the measurement section and the boundary.

This is the important experimental distinction:

- Equal incident and reflected amplitudes indicate zero net average energy transfer.
- Unequal incident and reflected amplitudes indicate a residual energy transfer towards the boundary in this arrangement.

The experiment would not establish mathematically exact equality in a real string. Measurement uncertainty, air resistance, friction, and other losses limit the precision.

However, it provides a way to test how closely the real apparatus approaches the ideal model.

For the separate practical skills of measuring wavelength and calculating wave speed from node spacing, see [Standing-String Practical: Measure Wavelength and Wave Speed](https://tutorgum.com/learn/resources/standing-string-practical-measuring-wavelength-from-node-spacing/).

## 5. Worked example: When the reflected wave is weaker

A sinusoidal wave travels along a stretched string towards a damped boundary.

The incident wave has an amplitude of \(3.0\text{ mm}\) and carries an average power of \(0.50\text{ W}\).

The reflected wave has an amplitude of \(2.4\text{ mm}\).

Both waves have the same frequency and travel along the same string.

Calculate:

a) The average power carried by the reflected wave.

b) The net average power transferred towards the boundary.

c) The percentage of incident power absorbed at the boundary, assuming no other significant losses.

[IMAGE: A horizontal string with an incoming wave travelling towards a damped termination and a smaller reflected wave travelling back. Label incident amplitude 3.0 mm and reflected amplitude 2.4 mm. Show a stationary interference envelope with minima that do not reach zero. Include an arrow indicating residual net energy transfer towards the damper.]

**Step 1**: Identify the relationship between power and amplitude.

For sinusoidal waves on the same string at the same frequency:

\[
P\propto A^2
\]

Therefore:

\[
\frac{P_r}{P_i}=\frac{A_r^2}{A_i^2}
\]

Here, \(P_r\) and \(P_i\) are the average reflected and incident powers, respectively.

**Step 2**: Calculate the reflected power.

Substitute the known values:

\[
\begin{aligned}
P_r&=P_i\left(\frac{A_r}{A_i}\right)^2\\
&=0.50\left(\frac{2.4}{3.0}\right)^2\\
&=0.50(0.64)\\
&=0.32\text{ W}
\end{aligned}
\]

The reflected wave carries an average power of \(0.32\text{ W}\).

**Step 3**: Calculate the net average power.

The incident wave carries energy towards the boundary, while the reflected wave carries energy away from it.

Taking the direction towards the boundary as positive:

\[
\begin{aligned}
P_{\mathrm{net}}&=P_i-P_r\\
&=0.50-0.32\\
&=0.18\text{ W}
\end{aligned}
\]

The net average power is \(0.18\text{ W}\) towards the boundary.

**Step 4**: Calculate the percentage of incident power absorbed.

The absorbed power is the difference between the incident and reflected powers:

\[
\begin{aligned}
P_{\mathrm{absorbed}}&=0.50-0.32\\
&=0.18\text{ W}
\end{aligned}
\]

The percentage absorbed is:

\[
\begin{aligned}
\%\,\mathrm{absorbed}
&=\frac{0.18}{0.50}\times100\\
&=36\%
\end{aligned}
\]

The boundary absorbs 36% of the incident average power.

The important conclusion is that this is not a pure standing wave. The interference pattern can have fixed maxima and minima, but the reflected wave is weaker, so the two opposing energy transfers do not balance.

## 6. A quick way to identify the difference

When analysing a string with incident and reflected waves, separate two questions.

First, what happens to the resultant displacement?

This depends on the phase difference and the principle of superposition.

Second, is there net energy transfer?

This depends on the average power carried by the waves in each direction.

The following comparison is useful.

| Situation | Resultant pattern | Net average energy transfer |
|---|---|---|
| Two equal-amplitude waves travelling in opposite directions | Pure standing wave with nodes and antinodes | Zero |
| Two unequal-amplitude waves travelling in opposite directions | Partial standing wave with non-zero minima | In the direction of the stronger wave |
| One travelling wave without a reflected component | Travelling-wave pattern | In the direction of propagation |

There is another important distinction involving boundaries.

A fixed end produces a phase reversal of \(\pi\), whereas an ideal free end produces no phase reversal.

This changes where nodes and antinodes occur. It does not change the zero-net-energy result when reflection is perfect and the two waves have equal amplitudes.

The reflection phase determines the interference pattern. The reflection amplitude determines how much energy returns.

## Questions and solutions

### Question 1

An ideal string supports a standing wave formed by two travelling waves of equal amplitude.

Each travelling wave carries an average power of \(0.075\text{ W}\). One travels towards the right, and the other travels towards the left.

Calculate the net average power transferred across a fixed cross-section of the string. Explain what the result means physically.

[IMAGE: A straight horizontal string with two equal and opposite arrows showing travelling-wave energy transfer. Label the rightward component 0.075 W and the leftward component 0.075 W. Mark a vertical cross-section through the centre.]

### Solution 1

The net average power transferred is **\(0\text{ W}\)**.

Taking rightward energy flow as positive:

\[
\begin{aligned}
P_{\mathrm{net}}
&=P_{\rightarrow}-P_{\leftarrow}\\
&=0.075-0.075\\
&=0\text{ W}
\end{aligned}
\]

The two waves carry equal average power in opposite directions, so their contributions cancel.

This does not mean the string contains no energy. The string still oscillates, and energy is continually exchanged between kinetic and elastic potential forms.

The result means that there is no net one-way energy transfer across the section when averaged over a complete cycle.

### Question 2

A sinusoidal wave reflects from a perfectly fixed end at \(x=0\).

The wavelength is \(0.72\text{ m}\), and point \(P\) is located \(0.12\text{ m}\) from the reflecting boundary.

The incident and reflected waves each have amplitude \(A\).

Calculate:

a) The path difference between the waves at \(P\).

b) The total phase difference, including the phase change on reflection.

c) The resultant amplitude at \(P\), expressed in terms of \(A\).

Determine whether \(P\) is a node, an antinode, or neither.

[IMAGE: A horizontal string with a rigid fixed boundary at x = 0 on the right and point P located 0.12 m to its left. Show the incident and reflected propagation directions, wavelength 0.72 m, and the round-trip path difference 2x. Do not label P as a node or antinode.]

### Solution 2

The resultant amplitude at \(P\) is **\(\sqrt{3}A\)**, so \(P\) is neither a node nor an antinode.

**Step 1**: Calculate the path difference.

For reflection at a boundary, the path difference is twice the distance from the boundary:

\[
\begin{aligned}
\Delta r&=2x\\
&=2(0.12)\\
&=0.24\text{ m}
\end{aligned}
\]

**Step 2**: Calculate the phase difference due to the path difference.

\[
\begin{aligned}
\Delta\phi_{\mathrm{path}}
&=\frac{2\pi\Delta r}{\lambda}\\
&=\frac{2\pi(0.24)}{0.72}\\
&=\frac{2\pi}{3}
\end{aligned}
\]

**Step 3**: Include the reflection phase change.

A fixed boundary introduces a phase reversal of \(\pi\).

Therefore:

\[
\begin{aligned}
\Delta\phi
&=\frac{2\pi}{3}+\pi\\
&=\frac{5\pi}{3}
\end{aligned}
\]

The waves are not in phase or in antiphase.

**Step 4**: Calculate the resultant amplitude.

For two equal-amplitude waves with phase difference \(\Delta\phi\), the resultant amplitude is:

\[
A_{\mathrm{resultant}}
=2A\left|\cos\left(\frac{\Delta\phi}{2}\right)\right|
\]

Substitute:

\[
\begin{aligned}
A_{\mathrm{resultant}}
&=2A\left|\cos\left(\frac{5\pi}{6}\right)\right|\\
&=2A\left(\frac{\sqrt{3}}{2}\right)\\
&=\sqrt{3}A
\end{aligned}
\]

The waves partially reinforce each other.

A common mistake is to assume that every point on a standing wave must be either a node or an antinode. That isn't true. Points between them have intermediate oscillation amplitudes.

### Question 3

A string is driven towards a damped boundary.

At a marked cross-section, the incident wave has an amplitude of \(2.0\text{ mm}\), while the reflected wave has an amplitude of \(1.0\text{ mm}\).

The incident wave carries an average power of \(0.18\text{ W}\).

Assume both waves have the same frequency, travel along the same string, and that there are no significant losses between the marked section and the boundary.

Calculate:

a) The average power carried by the reflected wave.

b) The net average power transferred towards the boundary.

c) The percentage of incident power absorbed by the boundary.

Explain why a stable pattern of maxima and minima may still form.

[IMAGE: A taut horizontal string approaching a damped boundary on the right. Show a stronger incident wave travelling rightwards with amplitude 2.0 mm and a weaker reflected wave travelling leftwards with amplitude 1.0 mm. Show a fixed spatial interference envelope with minima that remain above zero.]

### Solution 3

The reflected wave carries **\(0.045\text{ W}\)**, the net average power towards the boundary is **\(0.135\text{ W}\)**, and the boundary absorbs **75%** of the incident power.

**Step 1**: Calculate the reflected power.

For waves on the same string at the same frequency:

\[
P\propto A^2
\]

Therefore:

\[
\begin{aligned}
P_r&=P_i\left(\frac{A_r}{A_i}\right)^2\\
&=0.18\left(\frac{1.0}{2.0}\right)^2\\
&=0.18(0.25)\\
&=0.045\text{ W}
\end{aligned}
\]

**Step 2**: Calculate the net average power.

Taking the direction towards the boundary as positive:

\[
\begin{aligned}
P_{\mathrm{net}}&=P_i-P_r\\
&=0.18-0.045\\
&=0.135\text{ W}
\end{aligned}
\]

There is a net transfer of \(0.135\text{ W}\) towards the boundary.

**Step 3**: Calculate the percentage absorbed.

\[
\begin{aligned}
\%\,\mathrm{absorbed}
&=\frac{P_i-P_r}{P_i}\times100\\
&=\frac{0.135}{0.18}\times100\\
&=75\%
\end{aligned}
\]

The boundary absorbs 75% of the incident power.

**Step 4**: Explain the interference pattern.

The incident and reflected waves still have the same frequency, so their relative phase at each position remains fixed.

Their superposition can therefore produce a stationary pattern of maxima and minima.

However, because their amplitudes are unequal, destructive interference cannot produce complete cancellation.

The minima are not true nodes, and the incident wave carries more average power than the reflected wave.

This is a partial standing wave with net energy transfer towards the boundary.

### Question 4

An ideal string fixed at both ends supports a standing wave with period \(T\).

At \(t=0\), the string has maximum positive displacement. At \(t=T/4\), the string is momentarily straight.

A student makes two claims:

- The energy disappears at \(t=T/4\) because the displacement is zero everywhere.
- A node contains no energy because it never moves.

Assess both claims. Describe the energy forms present at \(t=0\) and \(t=T/4\), and explain the role of the nodes.

[IMAGE: A fixed-fixed string shown at maximum positive displacement and then momentarily straight one-quarter period later. Mark two stationary nodes and the antinodes. In the straight snapshot, show transverse velocity arrows at the antinodes.]

### Solution 4

**Both claims are incorrect.** At \(t=T/4\), the energy is kinetic rather than elastic potential energy. Nodes also remain associated with energy stored in the deformation of the surrounding string.

**Step 1**: Consider the string at \(t=0\).

The string has maximum displacement.

At this instant, its transverse velocity is zero, so its kinetic energy is zero.

However, the string is curved, and elastic potential energy is stored through the deformation associated with its tension.

The energy has not disappeared.

**Step 2**: Consider the string at \(t=T/4\).

A quarter of a period later, the string is momentarily straight.

Its displacement is zero everywhere, but the transverse velocity is maximum at the antinodes.

The energy is now kinetic.

The string is passing through its equilibrium position, not sitting motionless.

**Step 3**: Consider the nodes.

A node has zero transverse displacement and zero transverse velocity.

However, the string immediately around it can be sloped. This deformation is associated with elastic potential energy.

Therefore, a stationary point does not mean that the surrounding string has no energy.

The student's mistake is treating displacement as though it directly measures the total energy of the string.

Both the motion of the string and its elastic deformation must be considered.

### Question 5

An ideal uniform string is driven by a sinusoidal source towards an adjustable boundary.

The wavelength is \(0.72\text{ m}\), and point \(P\) is \(0.12\text{ m}\) from the boundary. The incident wave has an amplitude of \(2.0\text{ mm}\).

Consider three possible boundary conditions.

**Case A: Perfectly fixed boundary**

The reflected wave has the same amplitude as the incident wave, but undergoes a phase reversal of \(\pi\).

**Case B: Ideal free boundary**

The reflected wave has the same amplitude as the incident wave, but undergoes no phase reversal.

**Case C: Absorbing boundary**

The reflected wave has an amplitude of \(0.60A\), where \(A\) is the incident amplitude. Assume the frequency and string properties remain unchanged.

Answer the following:

a) For Cases A and B, calculate the phase difference between the incident and reflected waves at \(P\).

b) Calculate the resultant oscillation amplitude at \(P\) in each case.

c) Determine the net average energy transfer in Cases A and B. Explain why changing the reflection phase can change the positions of nodes and antinodes without creating net energy transfer.

d) For Case C, calculate the fraction of incident average power absorbed by the boundary.

e) Explain how Case C can produce a stationary spatial pattern of maxima and minima while still carrying net energy towards the boundary.

[IMAGE: Three side-by-side physical string arrangements with the same incident wave and marked point P at x = 0.12 m. First, a rigid fixed boundary showing inverted reflection; second, a frictionless free boundary showing non-inverted reflection; third, an absorbing boundary showing a smaller reflected wave. Include wavelength 0.72 m, equal component amplitudes for the first two cases, and a reflected amplitude of 0.60A for the third. Do not label resultant node or antinode positions.]

### Solution 5

The fixed and free boundaries both produce **zero net average energy transfer**, despite producing different interference patterns. The absorbing boundary transfers **64% of the incident average power** into the absorber.

This question tests an important distinction: the phase of a reflected wave determines the interference pattern, while its amplitude determines how much energy is reflected.

**Step 1**: Calculate the path difference.

The path difference at \(P\) is:

\[
\begin{aligned}
\Delta r&=2x\\
&=2(0.12)\\
&=0.24\text{ m}
\end{aligned}
\]

Since the wavelength is \(0.72\text{ m}\):

\[
\begin{aligned}
\Delta\phi_{\mathrm{path}}
&=\frac{2\pi(0.24)}{0.72}\\
&=\frac{2\pi}{3}
\end{aligned}
\]

This is the phase difference contributed by the different travel distances, before considering reflection.

**Step 2**: Calculate the phase difference for Case A.

The fixed boundary introduces a phase reversal of \(\pi\).

Therefore:

\[
\begin{aligned}
\Delta\phi_A
&=\frac{2\pi}{3}+\pi\\
&=\frac{5\pi}{3}
\end{aligned}
\]

**Step 3**: Calculate the resultant amplitude for Case A.

Both travelling waves have amplitude:

\[
A=2.0\text{ mm}
\]

Using the superposition amplitude relationship:

\[
\begin{aligned}
A_{\mathrm{resultant},A}
&=2A\left|\cos\left(\frac{5\pi}{6}\right)\right|\\
&=\sqrt{3}A\\
&=\sqrt{3}(2.0)\\
&=3.46\text{ mm}
\end{aligned}
\]

The resultant amplitude is approximately \(3.46\text{ mm}\).

**Step 4**: Calculate the phase difference for Case B.

The ideal free boundary introduces no phase reversal.

Therefore:

\[
\Delta\phi_B=\frac{2\pi}{3}
\]

**Step 5**: Calculate the resultant amplitude for Case B.

\[
\begin{aligned}
A_{\mathrm{resultant},B}
&=2A\left|\cos\left(\frac{\pi}{3}\right)\right|\\
&=2A\left(\frac{1}{2}\right)\\
&=A\\
&=2.0\text{ mm}
\end{aligned}
\]

The resultant amplitude is \(2.0\text{ mm}\).

The two cases produce different amplitudes at the same point because their reflection phase changes are different.

The positions of nodes and antinodes also change. A fixed boundary creates a node at the boundary, while a free boundary creates an antinode.

**Step 6**: Determine the net average energy transfer in Cases A and B.

In both cases, the reflected wave has the same amplitude as the incident wave.

The two waves also have the same frequency and travel along the same string.

Therefore, they carry equal average power in opposite directions.

\[
P_{\mathrm{net}}=P_i-P_r=0
\]

Both cases have zero net average energy transfer.

This is the important conceptual hinge: a phase change of \(\pi\) at a fixed boundary and a phase change of zero at a free boundary produce different interference patterns, but neither changes the equality of the opposing energy transfers.

A student might confidently argue that the different reflection conditions must produce different energy transport. They do not. The phase relationship changes the displacement pattern, not the balance of average power when reflection is perfect.

**Step 7**: Calculate the absorbed power fraction in Case C.

The reflected amplitude is \(0.60A\).

Since average power is proportional to amplitude squared:

\[
\begin{aligned}
\frac{P_r}{P_i}
&=\left(\frac{0.60A}{A}\right)^2\\
&=0.36
\end{aligned}
\]

Therefore, 36% of the incident average power is reflected.

The absorbed fraction is:

\[
\begin{aligned}
\text{Absorbed fraction}
&=1-0.36\\
&=0.64
\end{aligned}
\]

Converting to a percentage:

\[
0.64\times100=64\%
\]

The absorbing boundary absorbs 64% of the incident average power.

**Step 8**: Explain the stationary pattern in Case C.

The incident and reflected waves still have the same frequency.

Their relative phase at each position therefore remains fixed, so their interference produces stationary spatial maxima and minima.

However, the waves have unequal amplitudes.

They cannot cancel completely, and their average powers are unequal.

Consequently, the pattern can have stationary maxima and minima while energy continues to flow towards the absorbing boundary.

This is a partial standing wave, not an ideal standing wave.

The final distinction is worth remembering:

- Reflection phase determines where constructive and destructive interference occur.
- Reflection amplitude determines how much energy returns.
- Equal opposing average powers give zero net energy transfer.
- Unequal opposing average powers produce a net energy transfer towards the weaker reflected component's boundary.

## The next useful connection

When you analyse standing waves, don't use the appearance of the pattern alone to decide whether energy is being transferred.

First, establish how the incident and reflected waves combine. Then consider whether their amplitudes are equal.

That distinction connects standing waves to resonance, reflection, and damping in real wave systems. It also explains why the ideal model is such a useful starting point, even though a real vibrating string always loses some energy through friction, air resistance, and other effects.