Why Destructive Interference Does Not Destroy Energy

Illuminate two narrow slits with a laser and look at the screen behind them. You see alternating bright and dark bands. At a dark band, almost no light arrives. If the light waves have cancelled each other, where has their energy gone?

It hasn’t been destroyed. Destructive interference reduces the energy delivered to particular locations, while constructive interference increases it elsewhere. The energy is redistributed across the interference pattern. In an ideal lossless system, the total energy is conserved.

The important distinction is that waves interfere by adding their disturbances, not by subtracting their energy. Understanding that difference also explains why reflections from boundaries can produce standing-wave nodes without absorbing the energy of the waves.

01Why cancelling waves do not mean cancelling energy

Start with two identical light waves arriving at the same point.

At one instant, suppose the electric field produced by the first wave points in the positive direction, while the second points equally strongly in the negative direction.

Their electric fields add:

\[ E_{\text{resultant}}=E_1+E_2 \]

If the fields have equal magnitudes and opposite directions:

\[ E_{\text{resultant}}=E_0-E_0=0 \]

The resultant electric field is zero. This is complete destructive interference.

But notice what has actually cancelled: the electric fields, not two quantities of energy.

For light, intensity is proportional to the square of the resultant electric-field amplitude. A zero resultant amplitude therefore produces zero intensity at that location in the ideal model.

The waves have not been absorbed or destroyed. Their combined fields produce a spatial pattern in which some locations receive much more energy than others.

This is the same underlying principle for other types of waves:

• For light, the electric fields superpose.
• For sound, pressure variations superpose.
• For waves on a string, transverse displacements superpose.

In every case, the resultant disturbance determines the local wave behaviour. You cannot calculate the energy simply by adding or subtracting the individual displacements.

The intensity equation explains where the energy goes

For two coherent waves of equal amplitude, the resultant intensity is:

\[ \boxed{I=2I_0(1+\cos\delta)} \]

Here:

• \(I\) is the resultant intensity, measured in watts per square metre (\(\text{W m}^{-2}\)).
• \(I_0\) is the intensity produced by either wave individually at that location.
• \(\delta\) is the phase difference between the waves, measured in radians.

Coherent waves have the same frequency and maintain a constant phase relationship.

Consider three possible phase differences.

The interesting comparison is between the first and third rows.

When the waves arrive in phase, their amplitudes reinforce each other. The resultant intensity is four times the intensity of either individual wave.

When they arrive completely out of phase, their amplitudes cancel, producing zero intensity.

So where is the energy?

Look at the entire interference pattern rather than one position.

For equal waves, the average intensity across a complete fringe cycle is \(2I_0\). This is exactly the sum of the intensities of the two individual waves.

The bright regions receive increased intensity, while the dark regions receive decreased intensity. The total energy is accounted for across the pattern.

This is the central reason destructive interference does not violate conservation of energy.

There is an important qualification: real equipment is not perfectly lossless. Some light can be absorbed by the slits, screen, or surrounding materials. That energy becomes internal energy or is transferred in another form. Such absorption is a separate process from destructive interference.

02Path difference and phase difference: two different quantities

To predict where destructive interference occurs, you need to understand why two waves arrive out of phase.

Imagine two slits producing waves that travel towards the same point on a screen.

One wave might travel 2.00 m, while the other travels 2.25 m. The second wave has travelled an extra 0.25 m.

This difference is called the path difference.

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

Here:

• \(\Delta r\) is the path difference in metres.
• \(r_1\) is the distance travelled by wave 1.
• \(r_2\) is the distance travelled by wave 2.

Path difference is a distance. It is not itself a phase difference.

The connection comes from wavelength. Every time a wave travels one wavelength, it completes one full cycle and returns to the same phase.

Therefore, for two sources that start in phase and travel directly through the same medium:

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

Here, \(\lambda\) is the wavelength in metres, and \(\delta\) is the magnitude of the phase difference in radians.

For example:

• A path difference of \(\lambda\) produces a phase difference of \(2\pi\), or \(360^\circ\).
• A path difference of \(\lambda/2\) produces a phase difference of \(\pi\), or \(180^\circ\).
• A path difference of \(3\lambda/2\) produces a phase difference of \(3\pi\), or \(540^\circ\).

A phase difference of \(540^\circ\) is equivalent to \(180^\circ\), because the waves have completed one additional full cycle.

When does destructive interference actually occur?

For two sources that begin in phase, the conditions are:

The familiar destructive-interference condition is:

\[ \boxed{\Delta r=\left(m+\frac12\right)\lambda} \]

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

However, this rule assumes the sources begin in phase and that no additional phase changes occur along the paths.

If the sources begin out of phase, or a wave reflects from a boundary, the total phase difference changes.

A more general expression, using a phase that decreases as a wave travels, is:

\[ \delta=\delta_s- \frac{2\pi(r_2-r_1)}{\lambda} +\delta_b \]

Here:

• \(\delta_s\) is the initial phase of source 2 relative to source 1.
• \(r_2-r_1\) is the signed path difference.
• \(\delta_b\) is any additional phase change caused by a reflection or boundary.
• \(\lambda\) is the wavelength.

The total phase difference, rather than path difference alone, determines the interference.

This is a useful distinction when a question introduces a reflecting surface. A wave can change phase because of the distance it travels, because of how its source starts oscillating, or because of how it reflects.

03An experiment: measuring where the energy goes

A double-slit experiment provides a way to investigate the energy question rather than simply accepting that energy is conserved.

The apparatus consists of:

1. A monochromatic laser producing light of a single frequency.
2. A narrow double-slit plate placed in front of the laser.
3. A screen positioned behind the slits.
4. A light detector that can move horizontally across the screen.

The laser illuminates both slits. Each slit acts as a coherent source of light.

The waves spread out, overlap, and produce an interference pattern on the screen. The detector measures the light intensity at different positions.

What should the experiment show?

Suppose each slit produces an intensity of \(I_0\) when measured separately.

With both slits open, the detector should record alternating intensity maxima and minima.

If destructive interference genuinely destroyed energy, the total energy measured across the screen would be lower simply because the waves interfered.

But that is not the expected result.

Instead, the intensity increases at constructive-interference positions and decreases at destructive-interference positions. When the intensity is averaged over a complete fringe cycle, it returns to the sum of the individual intensities.

Consider these illustrative idealised readings:

The average is:

\[ I_{\text{average}} =\frac{4I_0+2I_0+0+2I_0}{4} \]

\[ \boxed{I_{\text{average}}=2I_0} \]

Each slit contributes \(I_0\) individually, so their combined individual intensities are also \(2I_0\).

The average intensity across the fringe is therefore unchanged, even though the intensity at particular positions varies dramatically.

These values are theoretical illustrations, not claimed experimental measurements. Real readings will be affected by detector sensitivity, diffraction, imperfect slit illumination, and other experimental limitations.

Why measuring one dark fringe is not enough

Imagine an experimenter places a detector at a dark fringe and records almost zero intensity.

They might conclude that the light energy has disappeared.

But that experiment only measures one location. It says nothing about the energy arriving at the rest of the screen.

A better design scans the detector across the entire interference pattern. The experimenter can also measure each slit separately, then compare those readings with the two-slit pattern.

There is another important experimental control: coherence.

If two independent lamps are used instead of one laser illuminating both slits, their relative phase can change unpredictably. The interference pattern may move rapidly and disappear when averaged over time.

Using one laser provides a stable phase relationship, allowing a stationary pattern to form.

The screen must also be wide enough to capture the relevant pattern, and the detector must be sufficiently narrow to resolve the fringes. Otherwise, the measured energy balance may be misleading.

04Why unequal waves do not cancel completely

So far, we’ve assumed the two waves have equal amplitudes.

That assumption matters.

Suppose one wave has twice the amplitude of the other. Even if they arrive exactly \(180^\circ\) out of phase, the smaller wave cannot completely cancel the larger one.

The result is partial destructive interference.

For two coherent waves with different intensities, the resultant intensity is:

\[ \boxed{ I=I_1+I_2+2\sqrt{I_1I_2}\cos\delta } \]

Here:

• \(I_1\) is the intensity of wave 1 acting alone.
• \(I_2\) is the intensity of wave 2 acting alone.
• \(\delta\) is their phase difference.
• \(I\) is their resultant intensity.

Notice that we add the intensities and an additional interference term. We do not simply subtract the two intensities.

At maximum destructive interference, \(\delta=\pi\), giving:

\[ I_{\min}= \left(\sqrt{I_1}-\sqrt{I_2}\right)^2 \]

Complete cancellation requires \(I_1=I_2\).

This is important when investigating real interference patterns. A dark fringe might not be completely black because the two waves have different amplitudes.

05What happens when waves reflect from boundaries?

There is another situation where students sometimes confuse destructive interference with energy loss: standing waves.

Consider a wave travelling along a stretched string towards a fixed end.

When the wave reaches the fixed end, it reflects and becomes inverted.

This means the reflected wave has a phase change of \(180^\circ\), or \(\pi\) radians.

The incident and reflected waves then overlap, producing a standing-wave pattern.

Fixed ends and free ends produce different phase changes

The reflection behaviour depends on the boundary.

At a fixed end, the string cannot move. The incident and reflected waves must therefore combine to give zero displacement at the boundary.

At a free end, the string can move freely. The incident and reflected waves reinforce the displacement at the boundary.

This is a boundary condition: a restriction on how the medium can move at a particular location.

Why does the reflection phase matter?

Suppose a wave reflects from a fixed end.

At the boundary itself, the reflected wave is \(180^\circ\) out of phase with the incident wave. They cancel, creating a node.

However, consider a point some distance away from the boundary.

The reflected wave has travelled further than the incident wave by twice the distance from the point to the boundary.

If the point is a distance \(x\) from the boundary, the additional path length is:

\[ \Delta r=2x \]

This creates another phase difference.

The total phase relationship therefore depends on both the reflection and the additional distance travelled.

For example, take a string with wavelength \(0.80\text{ m}\), and consider a point \(0.20\text{ m}\) from a reflecting boundary.

The additional path is:

\[ \Delta r=2(0.20)=0.40\text{ m} \]

Since:

\[ 0.40=\frac{\lambda}{2} \]

the additional travel produces a phase difference of \(180^\circ\).

At a fixed end, reflection adds another \(180^\circ\).

The total phase difference is therefore:

\[ 180^\circ+180^\circ=360^\circ \]

The waves arrive in phase at this point, producing constructive interference. It is an antinode.

If the boundary were free instead, there would be no phase inversion on reflection. The waves would remain \(180^\circ\) out of phase at this point, producing destructive interference.

This is why you cannot always apply the ordinary two-source path-difference rules without considering boundary conditions.

Does a standing-wave node absorb energy?

No.

An ideal fixed end reflects the incoming wave. It does not absorb its energy simply because the displacement at the boundary is zero.

The incident and reflected waves travel in opposite directions. They combine to produce a stationary pattern, with nodes and antinodes.

At a node, the string has zero displacement at all times. However, this does not mean the surrounding string has no energy.

The string can store elastic potential energy through its deformation, while other regions have kinetic energy as they move.

For an ideal standing wave, the energy alternates between kinetic and elastic potential forms. There is no net average energy transfer along the string.

A real boundary can absorb some energy through friction, damping, or other interactions. That energy is transferred elsewhere, rather than being destroyed by interference.

The distinction is important:

• Destructive interference changes the resultant wave.
• Reflection changes the direction and potentially the phase of a wave.
• Absorption transfers wave energy into another form.

These are related processes, but they are not interchangeable.

06A reliable way to think about energy and interference

When answering an HSC question about destructive interference, work through the following decisions.

For a more detailed explanation of how displacements add, see Wave Superposition and Interference for HSC Physics.

If you need help calculating the distances travelled by two waves, see How to Calculate Path Difference Between Two Wave Sources.

The next step is to apply these ideas to standing waves. Once you understand how reflection changes phase, you can explain why nodes and antinodes form, calculate their spacing, and determine which harmonics a system can support.

07Questions and solutions

Question 1

A laser with a wavelength of \(500\text{ nm}\) illuminates two identical slits. The slits produce coherent waves that begin in phase.

At a point \(P\) on the screen, the path difference between the waves is \(0.75\ \mu\text{m}\).

Each slit, acting individually, would produce an intensity of \(0.20\text{ W m}^{-2}\) at \(P\).

Determine:

a) The phase difference between the waves in degrees.

b) Whether constructive or destructive interference occurs.

c) The resultant intensity at \(P\).

d) Whether this result means that the energy carried by the waves has been destroyed.

Solution 1

The waves arrive \(540^\circ\) out of phase, producing complete destructive interference and an ideal resultant intensity of \(0\text{ W m}^{-2}\). Energy has not been destroyed.

a) Calculate the phase difference.

Use:

\[ \delta=\frac{360^\circ\Delta r}{\lambda} \]

Convert both quantities into the same units:

\[ \Delta r=0.75\ \mu\text{m} \]

\[ \lambda=0.50\ \mu\text{m} \]

Substitute:

\[ \delta=\frac{360^\circ\times0.75}{0.50} \]

\[ \boxed{\delta=540^\circ} \]

This is equivalent to \(180^\circ\), because \(540^\circ-360^\circ=180^\circ\).

b) Identify the type of interference.

The waves are \(180^\circ\) out of phase.

A crest from one wave arrives with a trough from the other.

This produces complete destructive interference because the waves have equal amplitudes.

c) Calculate the resultant intensity.

Use:

\[ I=2I_0(1+\cos\delta) \]

Substitute:

\[ I=2(0.20)(1+\cos180^\circ) \]

Since \(\cos180^\circ=-1\):

\[ I=0.40(1-1) \]

\[ \boxed{I=0\text{ W m}^{-2}} \]

The ideal model predicts a completely dark fringe at \(P\).

d) Explain what happens to the energy.

The zero intensity only describes the energy arriving at this particular location.

Other positions on the screen experience constructive interference and receive increased intensity.

The energy is redistributed across the interference pattern. It has not been destroyed.

Question 2

Two loudspeakers produce coherent sound waves of frequency \(680\text{ Hz}\). They are connected to the same signal generator, but speaker 2 begins oscillating \(90^\circ\) ahead of speaker 1.

A listener stands at point \(P\), where:

• The distance from speaker 1 is \(4.00\text{ m}\).
• The distance from speaker 2 is \(4.25\text{ m}\).

Take the speed of sound to be \(340\text{ m s}^{-1}\).

Assume the two waves have equal amplitudes at \(P\), and each speaker acting alone produces an intensity \(I_0\).

Determine:

a) The wavelength.

b) The phase difference caused by the different travel distances.

c) The total phase difference when the waves reach \(P\).

d) The resultant intensity in terms of \(I_0\), and explain why the result is neither a maximum nor a minimum.

Solution 2

The waves arrive \(90^\circ\) out of phase (equivalently, \(270^\circ\) using the positive-angle convention), producing a resultant intensity of \(2I_0\).

a) Calculate the wavelength.

Use the wave equation:

\[ v=f\lambda \]

Rearrange:

\[ \lambda=\frac{v}{f} \]

Substitute:

\[ \lambda=\frac{340}{680} \]

\[ \boxed{\lambda=0.50\text{ m}} \]

b) Calculate the phase difference caused by the path difference.

First calculate the path difference:

\[ \Delta r=|4.25-4.00| \]

\[ \Delta r=0.25\text{ m} \]

Now calculate the phase difference:

\[ \delta_{\text{path}} =\frac{360^\circ\times0.25}{0.50} \]

\[ \boxed{\delta_{\text{path}}=180^\circ} \]

The second wave travels an additional half-wavelength, so it arrives \(180^\circ\) behind where it would otherwise have been.

c) Calculate the total phase difference.

Speaker 2 starts \(90^\circ\) ahead, but its additional travel introduces a \(180^\circ\) lag.

Using phase of wave 2 minus phase of wave 1:

\[ \delta_{\text{total}}=90^\circ-180^\circ \]

\[ \boxed{\delta_{\text{total}}=-90^\circ} \]

This is equivalent to \(270^\circ\).

d) Calculate the resultant intensity.

Use the equal-amplitude intensity equation:

\[ I=2I_0(1+\cos\delta) \]

Substitute:

\[ I=2I_0(1+\cos(-90^\circ)) \]

Since \(\cos(-90^\circ)=0\):

\[ \boxed{I=2I_0} \]

This is equal to the sum of the two individual intensities.

The waves are neither in phase nor completely out of phase. The interference term is zero at this particular phase difference.

The tempting mistake is to consider only the \(180^\circ\) phase difference caused by the path difference. That would incorrectly predict complete destructive interference.

The initial source phase must also be included.

Question 3

A sinusoidal wave travels along a string and reflects from a boundary.

The wavelength is \(0.80\text{ m}\). Point \(P\) is located \(0.20\text{ m}\) from the boundary.

Consider two separate situations:

• Situation A: The string reflects from a fixed end.
• Situation B: The string reflects from a free end.

For each situation:

a) Calculate the additional distance travelled by the reflected wave compared with the incident wave when they reach \(P\).

b) Determine the phase difference caused by this additional travel.

c) Include the reflection phase change and determine whether \(P\) is a node or an antinode.

d) Explain whether the ideal fixed boundary absorbs the energy of the incident wave.

Solution 3

At point \(P\), the fixed-end reflection produces constructive interference and an antinode. The free-end reflection produces destructive interference and a node. Neither ideal boundary destroys the wave energy.

a) Calculate the additional distance travelled.

The reflected wave travels from \(P\) to the boundary and back to \(P\).

Therefore:

\[ \Delta r=2x \]

Substitute:

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

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

b) Calculate the phase difference from the additional travel.

Use:

\[ \delta_{\text{path}} =\frac{360^\circ\Delta r}{\lambda} \]

Substitute:

\[ \delta_{\text{path}} =\frac{360^\circ\times0.40}{0.80} \]

\[ \boxed{\delta_{\text{path}}=180^\circ} \]

The additional half-wavelength produces a \(180^\circ\) phase difference.

c) Consider each boundary separately.

Situation A: Fixed end

A fixed end produces a \(180^\circ\) phase change on reflection.

The total phase difference at \(P\) is:

\[ \delta_{\text{total}}=180^\circ+180^\circ \]

\[ \delta_{\text{total}}=360^\circ \]

The waves arrive in phase.

Therefore, constructive interference occurs at \(P\), producing an antinode.

Situation B: Free end

A free end does not invert the reflected wave.

Its reflection phase change is \(0^\circ\).

Therefore:

\[ \delta_{\text{total}}=180^\circ+0^\circ \]

\[ \delta_{\text{total}}=180^\circ \]

The waves arrive out of phase.

Destructive interference occurs at \(P\), producing a node.

Notice that the boundary itself also behaves differently:

• A fixed end is a node because its displacement must remain zero.
• A free end is an antinode because it can move freely.

d) Explain what happens to the energy.

An ideal fixed end reflects the incoming wave.

The reflected wave carries energy back along the string. The zero displacement at the boundary is a consequence of the reflection phase change, not energy absorption.

A real boundary may absorb some energy through damping or friction, but that is a separate energy-transfer process.

Question 4

A student investigates energy distribution in a double-slit experiment using a monochromatic laser and a movable light detector.

The detector records readings at four equally spaced positions across one complete fringe period.

At each position, the intensity produced by either slit acting alone is approximately \(1.0\text{ mW m}^{-2}\).

The following illustrative readings are obtained:

All intensity values are in \(\text{mW m}^{-2}\).

a) Calculate the average intensity when both slits are open.

b) Compare this with the sum of the individual slit intensities.

c) Explain why position C does not demonstrate that energy has been destroyed.

d) Identify two experimental considerations that would improve confidence in the conclusion.

Solution 4

The average intensity with both slits open is \(2.0\text{ mW m}^{-2}\), equal to the sum of the two individual intensities. The readings support energy redistribution rather than energy destruction.

a) Calculate the average intensity.

Use:

\[ I_{\text{average}} =\frac{\text{sum of readings}}{\text{number of readings}} \]

Substitute:

\[ I_{\text{average}} =\frac{3.8+2.1+0.2+1.9}{4} \]

\[ =\frac{8.0}{4} \]

\[ \boxed{I_{\text{average}}=2.0\text{ mW m}^{-2}} \]

b) Compare with the individual intensities.

Each slit produces an intensity of:

\[ I_1=1.0\text{ mW m}^{-2} \]

\[ I_2=1.0\text{ mW m}^{-2} \]

Their sum is:

\[ I_1+I_2=1.0+1.0 \]

\[ \boxed{I_1+I_2=2.0\text{ mW m}^{-2}} \]

This matches the average intensity measured with both slits open.

The result is consistent with the theoretical prediction that the interference term averages to zero across a complete fringe cycle.

c) Explain the low reading at position C.

Position C has an intensity of only \(0.2\text{ mW m}^{-2}\).

This is a local intensity minimum caused by destructive interference.

The intensity at other positions is greater, particularly at position A, where constructive interference produces \(3.8\text{ mW m}^{-2}\).

A low reading at one position cannot establish that energy has been destroyed across the entire system.

The experiment must consider the complete interference pattern.

d) Identify two experimental considerations.

Two important considerations are:

1. Use a coherent source, such as one laser illuminating both slits, to maintain a stable phase relationship and produce stationary fringes.

2. Scan across a sufficiently wide region with a detector small enough to resolve the fringes. A detector that is too wide may average bright and dark regions together.

It would also be useful to repeat measurements, maintain constant laser output, and account for light absorbed by the apparatus.

The data are illustrative and do not establish an exact experimental energy balance. A real investigation would need to account for measurement uncertainty and the full transmitted-light distribution.

Question 5

An ideal, long, stretched string supports two equal-amplitude sinusoidal waves travelling in opposite directions.

The waves have:

• Individual amplitude \(a=1.2\text{ mm}\).
• Frequency \(f=50\text{ Hz}\).
• Wavelength \(\lambda=0.80\text{ m}\).

They form a standing wave through reflection from a fixed end.

Taking \(x=0\) at the fixed boundary, the resultant displacement is:

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

where:

\[ k=\frac{2\pi}{\lambda}, \qquad \omega=2\pi f \]

and \(T=1/f\) is the period.

A student makes the following claim:

“When the string is straight, its displacement is zero everywhere, so all the wave energy must have disappeared. The nodes must also be places where the waves destroy their energy.”

Consider the following:

a) Calculate the position of the first antinode.

b) At \(t=T/4\), calculate the displacement and maximum transverse speed at the first antinode.

c) Explain why the straight-string position does not represent zero energy.

d) Explain what happens to energy at the nodes and why they are not energy sinks.

e) Now replace the ideal reflector with a lossy termination. It reflects a wave with amplitude \(0.60a\), with a \(180^\circ\) phase change, and absorbs all the remaining incident energy. Assume there is no transmitted wave.

Calculate:

• The resultant displacement amplitude at the boundary.
• The fraction of incident wave power reflected.
• The fraction of incident wave power absorbed.

Explain why the residual displacement amplitude cannot be used directly as the percentage of energy absorbed.

Solution 5

The first antinode is \(0.20\text{ m}\) from the fixed end, and its maximum transverse speed at \(t=T/4\) is approximately \(0.75\text{ m s}^{-1}\). The straight-string position does not mean that the wave energy has disappeared.

This question tests an important distinction between displacement, velocity, and energy.

a) Calculate the first antinode position.

For a string fixed at \(x=0\), the first antinode occurs one-quarter of a wavelength from the boundary.

Therefore:

\[ x=\frac{\lambda}{4} \]

Substitute:

\[ x=\frac{0.80}{4} \]

\[ \boxed{x=0.20\text{ m}} \]

b) Calculate displacement and transverse speed at \(t=T/4\).

First calculate the period:

\[ T=\frac{1}{f} \]

\[ T=\frac{1}{50} \]

\[ T=0.020\text{ s} \]

Therefore:

\[ \frac{T}{4}=0.0050\text{ s} \]

At this time:

\[ \omega t=2\pi f\frac{T}{4} =\frac{\pi}{2} \]

Use the standing-wave equation:

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

At the first antinode, \(x=\lambda/4\), so:

\[ \sin(kx)=1 \]

Also:

\[ \cos\left(\frac{\pi}{2}\right)=0 \]

Therefore:

\[ \boxed{y=0\text{ m}} \]

The string is momentarily straight.

Now calculate the transverse velocity by differentiating displacement with respect to time:

\[ v_y=\frac{\partial y}{\partial t} \]

\[ v_y=-2a\omega\sin(kx)\sin(\omega t) \]

At the first antinode and at \(t=T/4\), both sine terms equal 1.

Therefore:

\[ |v_y|=2a\omega \]

Calculate angular frequency:

\[ \omega=2\pi f \]

\[ \omega=2\pi(50) \]

\[ \omega=314.16\text{ rad s}^{-1} \]

Substitute:

\[ |v_y|=2(1.2\times10^{-3})(314.16) \]

\[ \boxed{|v_y|=0.754\text{ m s}^{-1}} \]

The string is straight, but the points at the antinodes are moving at their maximum transverse speed.

c) Explain why the straight-string position does not mean zero energy.

The student’s mistake is treating displacement as a direct measurement of energy.

At maximum displacement, the string is curved, but its transverse velocity is zero. Energy is stored as elastic potential energy in the deformed string.

One-quarter of a period later, the string passes through equilibrium. Its displacement is zero, but the antinodes have maximum transverse velocity.

At this instant, energy is predominantly kinetic.

The energy changes form during the oscillation. It does not disappear when the string passes through equilibrium.

d) Explain what happens at the nodes.

A node is a position that remains at zero displacement throughout the standing-wave cycle.

Its transverse velocity is also zero.

However, this does not mean that a node absorbs the incident wave energy.

The string surrounding a node can still be deformed. For example, when the antinodes reach maximum displacement, the string has a non-zero slope at the nodes, storing elastic potential energy in the tensioned string.

The incident and reflected waves have equal amplitudes and travel in opposite directions. Their average energy transfers cancel, so there is no net average energy flow along the standing-wave pattern.

The energy is reflected rather than destroyed.

e) Analyse the lossy termination.

The reflected wave has amplitude:

\[ a_r=0.60a \]

It is \(180^\circ\) out of phase with the incident wave at the boundary.

Therefore, the resultant displacement amplitude at the boundary is:

\[ A_{\text{resultant}}=a-0.60a \]

\[ \boxed{A_{\text{resultant}}=0.40a} \]

There is no complete cancellation because the two wave amplitudes are unequal.

Now consider the reflected energy.

For waves travelling along the same string under the same conditions, power is proportional to the square of amplitude.

Therefore:

\[ \frac{P_r}{P_i} =\left(\frac{a_r}{a}\right)^2 \]

Substitute:

\[ \frac{P_r}{P_i}=(0.60)^2 \]

\[ \boxed{\frac{P_r}{P_i}=0.36} \]

Therefore, 36% of the incident power is reflected.

Since there is no transmitted wave, the remaining energy is absorbed:

\[ \text{Absorbed fraction}=1-0.36 \]

\[ \boxed{\text{Absorbed fraction}=0.64=64\%} \]

This is the important distinction.

The residual displacement amplitude is \(40\%\) of the incident amplitude, but that does not mean \(40\%\) of the energy remains or \(60\%\) has been absorbed.

The \(0.40a\) value describes the local resultant displacement produced by two overlapping waves. The \(64\%\) absorption is calculated from the difference between incident and reflected power.

These are different physical quantities.

The student was right that the waves can cancel their displacements. The mistake was assuming that displacement cancellation and energy destruction are the same process.

The key idea to carry into standing-wave and resonance questions is that a node describes displacement, not energy absorption. Once you combine superposition with reflection phase changes and energy transfer, you can explain why nodes form, why antinodes oscillate strongly, and how boundary conditions determine the standing-wave patterns a system can support.