Why Standing-Wave Nodes Stay Still: Phase and Reflection

A guitar string can have two waves travelling through it in opposite directions, yet certain points on the string never move. The sections between those points swing up and down, flatten out, and reverse direction, but the same points remain stationary.

Why?

A standing-wave node stays still because the two waves have equal amplitudes and are always exactly out of phase at that position. Their displacements cancel at every instant, not just occasionally. The waves themselves continue travelling, but the positions where they cancel remain fixed.

The important detail is that the phase difference between the two waves stays constant at each position. Once you understand why that happens, stationary nodes become much easier to explain.

01Why do the waves keep cancelling at the same positions?

Imagine a stretched string with one end fixed. A wave travels along the string, reflects from the fixed end, and travels back in the opposite direction.

The two waves overlap. According to the principle of superposition, their displacements add together at every position and instant.

Before going further, consider a prediction.

Suppose the first wave displaces a particular point upwards by 2 mm, while the reflected wave displaces it downwards by 2 mm. What happens?

The resultant displacement is zero.

Now suppose the first wave changes to an upward displacement of 1 mm. If the reflected wave simultaneously changes to a downward displacement of 1 mm, what happens?

The resultant displacement is still zero.

This is exactly what happens at a node.

The two waves do not need to have zero displacement individually. They need to have equal and opposite displacements.

Superposition creates the node

The principle of superposition states that the resultant displacement is the algebraic sum of the individual wave displacements:

\[ y=y_1+y_2 \]

where:

• \(y\) is the resultant displacement, in metres.
• \(y_1\) is the displacement caused by the first wave, in metres.
• \(y_2\) is the displacement caused by the second wave, in metres.

At a node:

\[ y_1=-y_2 \]

Therefore:

\[ y=0 \]

This cancellation must happen at every instant for the position to qualify as a node.

Consider two waves with an amplitude of 2.0 mm. The following is illustrative data showing their displacements at one particular node.

Here, \(T\) is the period of oscillation, measured in seconds.

Notice something important: at \(T/4\), both waves have their maximum displacement, but the resultant displacement is still zero.

The node does not stay still because the individual waves disappear. It stays still because their effects cancel.

02Phase difference explains why nodes stay stationary

To understand why cancellation keeps happening at the same location, we need to look at phase.

The phase of a wave describes its position within an oscillation cycle.

Two waves are:

• In phase when they are at the same stage of their cycles.
• Completely out of phase when their phase difference is \(180^\circ\), or \(\pi\) radians.

When two equal-amplitude waves are completely out of phase, their displacements cancel.

For a standing wave, the two component waves travel in opposite directions but have the same frequency.

This produces an important result:

At any fixed position, the phase difference between the two waves does not change with time.

Both waves oscillate at the same frequency, so their phases advance at the same rate. One wave does not gradually catch up with the other.

The phase difference varies from one position to another, but it remains constant over time at each position.

This is why some positions experience permanent destructive interference, while others experience permanent constructive interference.

What is the difference between phase difference and path difference?

These two quantities are closely related, but they describe different things.

Path difference is the difference between the distances travelled by two waves to reach the same point.

We calculate it using:

\[ \Delta r=|r_1-r_2| \]

where:

• \(\Delta r\) is the path difference, in metres.
• \(r_1\) and \(r_2\) are the distances travelled by the two waves, in metres.

Phase difference describes how far apart the two waves are in their oscillation cycles.

For waves travelling through the same medium with wavelength \(\lambda\), the phase difference caused by a path difference is:

\[ \Delta\phi=\frac{2\pi\Delta r}{\lambda} \]

Here:

• \(\Delta\phi\) is the phase difference, in radians.
• \(\Delta r\) is the path difference, in metres.
• \(\lambda\) is the wavelength, in metres.

This expression assumes there is no additional phase shift from the sources or reflection.

For two initially in-phase sources, the relationship is:

A path difference of half a wavelength means the waves arrive half a cycle apart.

However, there is an important complication when dealing with reflected waves.

A reflection can introduce an additional phase change. We must account for that before deciding whether the waves reinforce or cancel.

03How boundary conditions affect node positions

A boundary condition describes what a wave is allowed to do at the end of a medium.

For a stretched string, two important cases are a fixed end and a free end.

1. Reflection at a fixed end

A fixed end cannot move.

When a wave reflects from a fixed end, it undergoes a phase change of \(180^\circ\), or \(\pi\) radians.

In other words, an incoming crest reflects as a trough.

The incident and reflected waves must cancel at the fixed boundary because the string cannot move there.

Therefore, the fixed end is always a displacement node.

However, the fixed boundary does not explain every node along the string. The internal nodes form because of the continuing destructive interference between the incident and reflected waves.

2. Reflection at a free end

A free end is allowed to move vertically.

For an ideal free end, the reflected wave is not inverted. There is no \(180^\circ\) phase change.

The incident and reflected waves reinforce at the boundary, producing an antinode.

This gives us an important comparison:

The boundary determines the phase relationship at the end of the string. This affects where the rest of the standing-wave pattern forms.

For sound waves in a pipe, the same reasoning applies to boundary conditions, but be careful about what is oscillating. A closed end is a displacement node for the air particles, while an open end is approximately a displacement antinode. The corresponding pressure pattern is reversed.

You can explore that application in Standing Waves in Closed Pipes: Nodes and Odd Harmonics.

04The mathematical reason nodes cannot move

We can now express the physical explanation mathematically.

Consider two waves of equal amplitude travelling in opposite directions along a string.

Let the fixed end be at \(x=0\), with positive \(x\) measured away from that end.

The incident wave travels towards the fixed end:

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

The reflected wave travels away from the fixed end and is inverted:

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

Here:

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

Using superposition:

\[ y=y_1+y_2 \]

Substituting the two wave equations:

\[ \begin{aligned} y={}&A\sin(\omega t+kx)\\ &-A\sin(\omega t-kx) \end{aligned} \]

Using a trigonometric identity:

\[ \boxed{y=2A\sin(kx)\cos(\omega t)} \]

This is the equation of the standing wave.

Look carefully at its two factors.

\[ \underbrace{2A\sin(kx)}_{\text{Position}} \quad \underbrace{\cos(\omega t)}_{\text{Time}} \]

The position-dependent factor determines the amplitude at each location.

The time-dependent factor describes how the string oscillates.

For a node, the resultant displacement must always be zero. Therefore:

\[ \sin(kx)=0 \]

This occurs when:

\[ kx=n\pi \]

where \(n\) is an integer.

Substituting \(k=2\pi/\lambda\):

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

Rearranging:

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

This tells us that the nodes occur at fixed positions, separated by half a wavelength.

The crucial point is that the node condition depends on position, not time.

The temporal factor \(\cos(\omega t)\) can change from positive to negative, but it cannot change the positions where \(\sin(kx)=0\).

That is the mathematical reason standing-wave nodes stay still.

What happens at the antinodes?

At an antinode, the standing wave has maximum amplitude.

This occurs when:

\[ |\sin(kx)|=1 \]

Therefore, the antinode positions are:

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

The amplitude at an antinode is:

\[ A_{\text{standing}}=2A \]

The maximum amplitude is twice the amplitude of either individual travelling wave.

The important spatial relationships are:

One final detail is worth understanding.

At \(t=T/4\), the entire string momentarily passes through its equilibrium position because \(\cos(\omega t)=0\).

Does this mean every point has become a node?

No. A node must remain at zero displacement for the entire cycle. The other points are simply passing through equilibrium at that instant.

05A contrasting case: When zero displacement is not a node

Imagine watching a single progressive wave travel along a rope.

At one instant, a particular point on the rope might have zero displacement because the wave crosses the equilibrium line there.

A moment later, that same point may be above or below equilibrium.

The zero-displacement position moves along the rope as the wave travels.

This is not a standing-wave node.

A standing-wave node is a permanent position of zero displacement, whereas a zero crossing in a progressive wave is simply a temporary feature of the travelling wave.

There is another situation worth distinguishing: interference from two separate sound sources.

Two speakers emitting coherent waves in phase can produce stationary nodal lines where destructive interference occurs.

These cancellation positions are also determined by phase difference and path difference. However, their geometry is different from the evenly spaced nodes on a one-dimensional standing-wave string.

You cannot automatically apply the \(\lambda/2\) node-spacing rule to a two-source interference pattern.

For a closer look at this distinction, see How to Calculate Path Difference Between Two Wave Sources.

What if the two waves have different amplitudes?

A perfect node requires the two waves to have equal amplitudes.

Suppose one wave has an amplitude of 4 mm and the other has an amplitude of 3 mm.

Even when they are completely out of phase, their displacements cannot cancel completely.

The remaining amplitude is:

\[ A_{\min}=|A_1-A_2| \]

Therefore:

\[ A_{\min}=|4-3|=1\text{ mm} \]

There is no true node because the resultant amplitude never reaches zero at that position throughout the cycle.

What if the two waves have different frequencies?

This creates another problem.

If one wave has a higher frequency, its phase advances faster than the other wave’s phase.

The phase difference therefore changes with time.

A position that initially experiences destructive interference will not necessarily continue experiencing destructive interference.

The waves can move between constructive and destructive interference, producing beats rather than a stable standing-wave pattern.

For a true standing wave, the two component waves must have the same frequency, travel in opposite directions, and have equal amplitudes for complete cancellation at the nodes.

06Worked examples

07Questions and solutions

Question 1

A transverse wave travels along a string and reflects from an ideal fixed end.

The wavelength is \(0.80\text{ m}\). A student identifies a point \(0.40\text{ m}\) from the fixed end and claims that it is a node.

Determine whether the student is correct. Explain your reasoning using path difference and phase difference.

Solution 1

The student is correct. The point is a node because the incident and reflected waves are completely out of phase there.

Step 1: Calculate the path difference.

\[ \Delta r=2x \]

\[ \Delta r=2(0.40)=0.80\text{ m} \]

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

\[ \Delta\phi_{\text{path}} =\frac{2\pi(0.80)}{0.80} \]

\[ \Delta\phi_{\text{path}}=2\pi \]

Step 3: Include the fixed-end reflection.

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

Therefore:

\[ \Delta\phi_{\text{total}}=2\pi+\pi \]

\[ \Delta\phi_{\text{total}}=3\pi \]

Since \(3\pi\) is equivalent to \(\pi\) modulo \(2\pi\), the waves are completely out of phase.

Their equal amplitudes cancel at every instant.

Therefore, the point is a stationary node.

Question 2

Two waves travel in opposite directions along the same stretched string.

Both have an amplitude of \(2.0\text{ mm}\), but their frequencies are different:

• Wave 1 has a frequency of \(40\text{ Hz}\).
• Wave 2 has a frequency of \(42\text{ Hz}\).

At a particular position P, the waves are initially completely out of phase.

Calculate the time taken for the waves to first become in phase at P.

Explain why P cannot be a permanent node.

Solution 2

The waves first become in phase after \(0.25\text{ s}\). Position P cannot be a permanent node because the phase difference between the waves changes with time.

Step 1: Calculate the frequency difference.

\[ \Delta f=|f_2-f_1| \]

\[ \Delta f=|42-40| \]

\[ \Delta f=2\text{ Hz} \]

Step 2: Determine how the phase difference changes.

The relative phase changes at an angular rate of:

\[ \Delta\omega=2\pi\Delta f \]

Therefore:

\[ \Delta\omega=2\pi(2)=4\pi \text{ rad s}^{-1} \]

Initially, the phase difference is:

\[ \Delta\phi_0=\pi \]

Because the second wave has the higher frequency, the phase difference increases.

We can write:

\[ \Delta\phi(t)=\pi+4\pi t \]

Step 3: Find when the waves first become in phase.

The waves are in phase when their phase difference reaches \(2\pi\).

Therefore:

\[ \pi+4\pi t=2\pi \]

\[ 4\pi t=\pi \]

\[ t=\frac{\pi}{4\pi} \]

\[ \boxed{t=0.25\text{ s}} \]

Step 4: Explain why P is not a permanent node.

At the beginning, the waves are completely out of phase and cancel.

After \(0.25\text{ s}\), they are in phase and reinforce each other.

The phase difference continues changing, so the cancellation does not persist.

The frequency difference produces a beat frequency of \(2\text{ Hz}\). This is inconsistent with a stable standing wave.

The key requirement is not simply that two waves travel in opposite directions. Their frequencies must also match so that their relative phase remains constant.

Question 3

Two loudspeakers produce coherent sound waves of equal amplitude and wavelength \(0.48\text{ m}\). The speakers emit their waves in phase.

A student measures the following distances:

Determine whether destructive interference occurs at P and Q.

A student then claims that P and Q must be adjacent nodes separated by half a wavelength.

Evaluate this claim.

Solution 3

Destructive interference occurs at both P and Q, but the student’s claim about their physical separation is not justified.

Step 1: Calculate the path difference at P.

\[ \Delta r_P=|3.60-3.36| \]

\[ \Delta r_P=0.24\text{ m} \]

Since:

\[ 0.24=\frac{0.48}{2} \]

the path difference is half a wavelength.

The waves arrive completely out of phase, producing destructive interference.

Step 2: Calculate the path difference at Q.

\[ \Delta r_Q=|4.08-3.36| \]

\[ \Delta r_Q=0.72\text{ m} \]

Since:

\[ 0.72=\frac{3(0.48)}{2} \]

the path difference is three half-wavelengths.

This also produces destructive interference.

Step 3: Evaluate the student’s claim.

Both points experience destructive interference, but this does not mean they must be adjacent nodes separated by \(\lambda/2\).

The half-wavelength spacing applies to adjacent nodes in a one-dimensional standing wave.

For two speakers, the nodal positions depend on the geometry of the source-to-observation paths.

The actual physical distance between P and Q cannot be calculated from the supplied path lengths alone.

This is an important distinction: path difference describes the difference in distances travelled by two waves, not necessarily the physical separation between two nodes.

Question 4

A wave with a wavelength of \(0.80\text{ m}\) reflects from two different ideal string boundaries.

In the first experiment, the boundary is fixed.

In the second experiment, the boundary is free.

Both experiments have equal-amplitude incident and reflected waves.

Consider a point \(0.20\text{ m}\) from the reflecting boundary.

Determine whether this point is a node or antinode in each experiment. Explain why the answers differ even though the wavelength and observation position are unchanged.

Solution 4

The point is an antinode for the fixed-end experiment and a node for the free-end experiment.

The difference comes from the phase change introduced by reflection.

Step 1: Calculate the path difference.

For both experiments:

\[ \Delta r=2x \]

\[ \Delta r=2(0.20) \]

\[ \Delta r=0.40\text{ m} \]

Step 2: Calculate the phase difference caused by the path difference.

\[ \Delta\phi_{\text{path}} =\frac{2\pi(0.40)}{0.80} \]

\[ \Delta\phi_{\text{path}}=\pi \]

Step 3: Consider the fixed-end experiment.

A fixed end introduces an additional phase change of \(\pi\).

Therefore:

\[ \Delta\phi_{\text{total}}=\pi+\pi \]

\[ \Delta\phi_{\text{total}}=2\pi \]

The waves are in phase and reinforce each other.

The point is an antinode.

Step 4: Consider the free-end experiment.

A free end does not introduce a phase change upon reflection.

Therefore:

\[ \Delta\phi_{\text{total}}=\pi \]

The waves are completely out of phase and cancel.

The point is a node.

The physical distance and wavelength are identical in both experiments. The different reflection conditions change the relative phase, which changes the interference result.

This demonstrates why boundary conditions must be considered before applying a path-difference rule.

Question 5

Two independent drivers produce waves along a long, uniform string from opposite ends.

Assume that:

• Both waves have equal amplitude.
• Both waves have the same frequency.
• Both waves have wavelength \(0.80\text{ m}\).
• The waves travel in opposite directions.
• Reflections from the ends are negligible in the central region.

Initially, the two waves have the following equations:

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

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

where \(x\) increases from left to right.

A controller then delays only the wave travelling from right to left by one-quarter of a period.

After the phase delay has propagated through the central region and a new stable pattern has formed, determine:

(a) Whether a standing wave still exists.

(b) Whether the nodes remain at their original positions.

(c) The distance and direction through which the nodes shift.

Explain why the waves can still form a standing wave even though the node positions have changed.

Solution 5

A standing wave still exists, but its nodes shift \(0.10\text{ m}\) to the right.

The important distinction is that the relative phase has changed, but the waves still have the same frequency and equal amplitudes. Therefore, the new interference pattern is still stationary.

Step 1: Understand the original node positions.

Initially, the waves have zero additional phase difference.

Adding their equations:

\[ y=2A\sin(\omega t)\cos(kx) \]

For a node:

\[ \cos(kx)=0 \]

Therefore:

\[ kx=\left(n+\frac12\right)\pi \]

Since \(k=2\pi/\lambda\):

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

These are the original node positions.

Step 2: Determine the phase change caused by the delay.

The period is \(T\).

A delay of one-quarter of a period corresponds to a phase lag of:

\[ \Delta\phi=-\frac{\pi}{2} \]

The negative sign indicates that the right-to-left travelling wave is delayed.

Its new equation becomes:

\[ y_2=A\sin(\omega t+kx-\frac{\pi}{2}) \]

The resultant displacement is now:

\[ y=y_1+y_2 \]

Using a trigonometric identity:

\[ \boxed{ y=2A\sin(\omega t-\frac{\pi}{4}) \cos(kx-\frac{\pi}{4}) } \]

Step 3: Find the new node positions.

For a node, the position-dependent factor must equal zero:

\[ \cos(kx-\frac{\pi}{4})=0 \]

Therefore:

\[ kx-\frac{\pi}{4} =\left(n+\frac12\right)\pi \]

Rearranging:

\[ kx=\left(n+\frac34\right)\pi \]

Substituting \(k=2\pi/\lambda\):

\[ x'_n=\frac{(2n+\frac32)\lambda}{4} \]

The difference between the new and original positions is:

\[ \begin{aligned} \Delta x &=x'_n-x_n\\ &=\frac{\lambda}{8} \end{aligned} \]

Using the wavelength:

\[ \Delta x=\frac{0.80}{8} \]

\[ \boxed{\Delta x=0.10\text{ m}} \]

The nodes shift \(0.10\text{ m}\) to the right.

Step 4: Explain why a standing wave still exists.

Look at the new resultant equation:

\[ y=2A\sin(\omega t-\frac{\pi}{4}) \cos(kx-\frac{\pi}{4}) \]

It still contains one factor depending only on time and another depending only on position.

The positions where the spatial factor equals zero remain fixed.

The waves still have equal amplitudes, identical frequencies, and opposite directions of travel.

Therefore, they still produce a standing wave.

The tempting assumption is that changing the relative phase must destroy the standing wave. It does not. It changes the spatial position of the interference pattern.

During the transition, the phase change travels through the overlap region. Once the new phase relationship is established, the new nodes remain stationary.

This reveals the deeper idea: nodes are not permanently attached to particular points on a string. Their positions are determined by the relative phase of the component waves and the boundary conditions.

08What to understand next

The next useful step is understanding how boundary conditions determine which standing-wave patterns can exist.

For a string fixed at both ends, both endpoints must be nodes. Only certain wavelengths can fit between them, producing specific resonant frequencies.

For a closed pipe, the air-displacement pattern must satisfy a different combination of boundary conditions.

Both situations use the same underlying reasoning:

• Superposition determines the resultant displacement.
• Phase difference determines whether waves reinforce or cancel.
• Boundary conditions determine which spatial patterns are physically possible.

Once you can separate these three ideas, explaining why nodes stay still becomes much easier than memorising a diagram or a half-wavelength rule.