How to Sketch the Resultant of Two Opposite Wave Pulses

An upward pulse of displacement +3.0 cm travels along a stretched string while an identical downward pulse of displacement −3.0 cm approaches from the opposite direction. They meet. What should the string look like when the two pulses completely overlap?

It should be momentarily flat, sitting along its equilibrium position. But the pulses haven’t disappeared. A moment later, they emerge on opposite sides and continue travelling.

This is the principle of superposition. To sketch the resultant of two opposite wave pulses, add their signed vertical displacements at the same horizontal position and the same instant. The important part is getting the signs, positions, and timing right.

01How to sketch the resultant of two opposite pulses

Start with one rule:

\[ y_R=y_1+y_2 \]

Here:

• \(y_R\) is the resultant displacement.
• \(y_1\) is the displacement caused by the first pulse.
• \(y_2\) is the displacement caused by the second pulse.

Choose the equilibrium position as \(y=0\), and take upward displacement as positive. Downward displacement is negative.

For example, if one pulse produces a displacement of \(+4\) cm at a particular point and the other produces \(-2\) cm, the resultant is:

\[ y_R=4+(-2)=+2\text{ cm} \]

The string is still 2 cm above equilibrium at that point.

The rule is simple, but you must apply it position by position. Don’t just subtract the two maximum amplitudes and assume that gives you the entire resultant shape.

The five-step sketching method

Step 1: Draw the equilibrium line.

This is the undisturbed position of the string. Draw it horizontally and use it as your zero-displacement reference.

Step 2: Establish the sign of each pulse.

An upward pulse has positive displacement. A downward pulse has negative displacement.

Step 3: Place both pulses at the same instant.

Use the same horizontal position scale for both pulses. Pay attention to where their bases, peaks, and slopes sit.

Step 4: Add the displacements at matching positions.

At each horizontal position \(x\), calculate:

\[ y_R(x)=y_1(x)+y_2(x) \]

For example:

All three columns use the same sign convention.

Step 5: Draw the resultant profile.

Plot the calculated displacements against their horizontal positions. Connect the points using the appropriate shape of the original pulses.

Where the pulses don’t overlap, the resultant is simply the displacement of whichever pulse is present. Where they overlap, add their displacements.

02Why do opposite pulses cancel?

Imagine two people holding opposite ends of a long, stretched rope. One flicks their end upwards while the other flicks their end downwards.

The pulses travel towards each other. When they overlap, each small section of rope responds to both pulses simultaneously.

The rope doesn’t choose one pulse over the other. Its displacement is the algebraic sum of both contributions.

This is the principle of superposition:

\[ \boxed{y_R=y_1+y_2} \]

It applies to waves travelling through a linear medium, where the individual disturbances can be added without significantly changing the medium’s response.

What happens when two identical opposite pulses overlap?

Suppose both pulses have:

• The same triangular shape.
• The same maximum displacement, 3.0 cm.
• Opposite displacement directions.
• Exactly matching horizontal positions during full overlap.

At every position within their shared profile, one pulse has a displacement equal in magnitude but opposite in sign to the other.

Therefore:

\[ y_R=+3.0+(-3.0)=0 \]

The resultant is a straight line along equilibrium.

This is complete destructive interference at that instant.

However, cancellation describes the resultant displacement, not the permanent destruction of the pulses. In an ideal linear string, the pulses continue through one another and retain their original shapes and directions.

What if the pulses have different amplitudes?

They won’t completely cancel.

Suppose an upward pulse has an amplitude of 5.0 cm and a downward pulse has an amplitude of 2.0 cm. When their peaks align:

\[ y_R=+5.0+(-2.0) \]

\[ \boxed{y_R=+3.0\text{ cm}} \]

The resultant is still upward, but smaller than the original upward pulse.

The same principle applies if the downward pulse is larger. The resultant then points downwards.

Remember that this subtraction only gives the displacement where the relevant parts of the pulses actually overlap. The complete resultant profile still requires adding the displacements at every position.

03Testing superposition with a string experiment

How could you check whether the pulses really pass through each other, rather than simply disappearing when they meet?

A stretched-string experiment makes the process visible.

Setting up the experiment

Imagine a long, horizontal string stretched between two supports.

• A student stands at the left end and creates an upward pulse.
• A second student stands at the right end and creates a downward pulse.
• Both pulses travel towards the middle of the string.
• A ruler or marked scale sits behind the string, allowing vertical displacements and horizontal positions to be estimated.
• A camera records the string from the side, with the equilibrium position clearly visible.

The string should have consistent tension throughout. Keep the pulse amplitudes relatively small so that the string behaves approximately as a linear medium.

What should the experiment show?

Before running the experiment, make a prediction.

If the pulses have identical shapes and amplitudes but opposite displacement directions, what should happen when they overlap completely?

The predicted resultant is zero displacement across their overlapping profiles.

As the pulses approach, the string should show a smaller resultant displacement in regions where their contributions oppose each other. At complete overlap, the string should momentarily lie along equilibrium.

Afterwards, the pulses should emerge and continue travelling in their original directions.

This is the important experimental distinction: the resultant displacement changes during the interaction, but the individual pulses are not permanently destroyed.

Why might the first experiment give an unexpected result?

Suppose the string doesn’t become completely flat.

It might be tempting to conclude that superposition doesn’t work. However, the two pulses may simply be different.

For example:

• The upward pulse might have an amplitude of 4 cm.
• The downward pulse might have an amplitude of only 3 cm.
• Their widths might also differ.

Even when their peaks overlap, the resultant displacement won’t be zero.

A better experimental design is to record each pulse separately before sending both together. Compare their shapes, widths, and amplitudes. Adjust the hand movements until the pulses are as closely matched as possible.

You should also observe the interaction before either pulse reaches a support. Otherwise, reflected pulses may introduce additional disturbances.

A small measurement calculation

Suppose a side-on video records an upward pulse moving along the string.

The following values are illustrative data, not measurements from an actual experiment.

First, calculate the elapsed time:

\[ \Delta t=\frac{40}{50}=0.80\text{ s} \]

The pulse travels:

\[ \Delta x=0.60-0.20=0.40\text{ m} \]

Its estimated speed is:

\[ v=\frac{\Delta x}{\Delta t} \]

\[ v=\frac{0.40}{0.80} \]

\[ \boxed{v=0.50\text{ m/s}} \]

This measurement helps predict when two approaching pulses will meet. It also allows you to check whether both pulses travel at approximately the same speed along the same string.

The experiment supports the superposition model when the combined displacement agrees with the sum of the separately measured pulse displacements. Small differences can arise from imperfect pulse matching, damping, and measurement uncertainty.

04Phase difference and path difference: are they relevant to pulses?

Here’s a common source of confusion.

You might remember that two waves can be 180° out of phase, causing destructive interference. You might also remember that a path difference of half a wavelength produces a phase difference of 180° for two initially in-phase sources.

Does that mean every upward pulse and downward pulse are automatically 180° out of phase?

Not quite.

The terms describe different things.

When should you use phase difference?

Phase difference is particularly useful when comparing continuous, repeating waves.

For two waves with the same frequency travelling through the same medium, 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.

For two sources that begin in phase, a path difference of half a wavelength gives:

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

Therefore:

\[ \Delta\phi=\pi\text{ rad}=180^\circ \]

A crest from one wave then meets a trough from the other.

Why this is different for isolated pulses

An isolated triangular pulse doesn’t repeat in cycles like a continuous sinusoidal wave.

It may have a measurable width, but that width is not automatically a wavelength.

Therefore, you shouldn’t use the half-wavelength rule to sketch two isolated opposite pulses.

Instead, use their actual displacement profiles. At each position, add the signed displacements.

This distinction is useful in HSC questions because the wording often tells you which approach to use:

• If the question gives continuous waves, a wavelength, and distances travelled, consider phase and path difference.
• If the question gives isolated pulses and their positions or shapes, use direct superposition.
• If a pulse reflects from a boundary, determine the reflected pulse’s orientation before adding the displacements.

For a more detailed treatment of continuous waves, see How to Calculate Path Difference Between Two Wave Sources.

05Boundary effects: what happens when a pulse reflects?

So far, we’ve considered two pulses travelling towards each other without interacting with the ends of the string.

But what happens when a pulse reaches a boundary?

The answer depends on whether the boundary is fixed or free.

Reflection at a fixed end

Imagine sending an upward pulse towards a string clamped firmly to a wall.

The end of the string cannot move vertically. When the pulse reaches the clamp, the reflected pulse is inverted.

An upward pulse reflects as a downward pulse.

Why does the inversion happen?

The fixed endpoint must remain at its equilibrium position. The incident and reflected disturbances must combine to give zero displacement at that endpoint.

The reflected pulse therefore has the opposite displacement sign.

Reflection at a free end

Now replace the clamp with an ideal free end that can move vertically without resistance.

The reflected pulse is not inverted.

An upward incident pulse reflects as an upward pulse.

The difference comes from the boundary condition, not from the direction in which the pulse travels.

How boundary effects change the resultant

Here’s the important connection.

Reflection determines the shape and orientation of the reflected pulse. Superposition then determines the resultant displacement when that reflected pulse overlaps another disturbance.

For example, imagine an upward pulse approaching a fixed end. Its reflected pulse travels back down the string with inverted displacement.

If an incoming upward pulse and its reflected downward pulse overlap, their displacements oppose each other.

At the fixed endpoint, their contributions must sum to zero.

However, you shouldn’t assume that the entire string becomes flat whenever a pulse reflects. The zero-displacement condition applies at the fixed endpoint. Away from the endpoint, you must calculate the resultant at each position.

This is why boundary effects and superposition should be treated as two separate steps.

1. Determine the reflected pulse using the boundary condition.
2. Add the incident, reflected, and any other overlapping displacements.

07Questions and solutions

Question 1

Two opposite pulses overlap at a particular point P on a stretched string.

At that instant:

• Pulse A produces a displacement of +2.8 cm.
• Pulse B produces a displacement of −1.1 cm.

Calculate the resultant displacement at P. Is the cancellation complete?

Solution 1

The resultant displacement is +1.7 cm, so point P remains 1.7 cm above equilibrium. The cancellation is only partial.

Using superposition:

\[ y_R=y_A+y_B \]

Substitute the values:

\[ y_R=2.8+(-1.1) \]

\[ \boxed{y_R=+1.7\text{ cm}} \]

The positive result means the string remains above its equilibrium position.

Complete cancellation would require the two displacements to have equal magnitudes and opposite signs at P.

Question 2

Two triangular pulses travel towards each other along a uniform string.

Their initial conditions are:

• The centres of the pulses are 1.20 m apart.
• Each pulse has a base width of 0.40 m.
• Pulse A is upward, with an amplitude of +3.0 cm.
• Pulse B is downward, with an amplitude of −2.0 cm.
• Both pulses travel at 0.50 m/s.

Assume the pulses have identical triangular shapes apart from their amplitudes.

Calculate:

(a) The time until the pulses first touch.

(b) The time until their centres coincide.

(c) The resultant shape and maximum displacement when they completely overlap.

(d) What happens after the pulses pass through each other?

Solution 2

The pulses first touch after 0.80 s. Their centres coincide after 1.20 s, when the resultant is an upward triangular pulse with a maximum displacement of +1.0 cm.

Step 1: Calculate the initial gap between the pulses.

Each pulse has a width of 0.40 m, so the distance from its centre to either edge is:

\[ \frac{0.40}{2}=0.20\text{ m} \]

The gap between the nearest edges is:

\[ d=1.20-0.20-0.20 \]

\[ \boxed{d=0.80\text{ m}} \]

Step 2: Calculate the relative speed.

The pulses travel towards each other, so their speeds add:

\[ v_{\text{relative}}=0.50+0.50 \]

\[ v_{\text{relative}}=1.00\text{ m/s} \]

The time until they first touch is:

\[ t=\frac{d}{v_{\text{relative}}} \]

\[ t=\frac{0.80}{1.00} \]

\[ \boxed{t=0.80\text{ s}} \]

Step 3: Calculate when their centres coincide.

The centres initially have a separation of 1.20 m.

Therefore:

\[ t=\frac{1.20}{1.00} \]

\[ \boxed{t=1.20\text{ s}} \]

Notice that first contact and complete alignment are different events. The pulses begin overlapping before their centres coincide.

Step 4: Calculate the resultant amplitude at complete overlap.

At this instant, the two triangular profiles are perfectly aligned.

Apply superposition:

\[ y_R=y_A+y_B \]

\[ y_R=3.0+(-2.0) \]

\[ \boxed{y_R=+1.0\text{ cm}} \]

The resultant retains the triangular shape but has a smaller upward amplitude.

Step 5: Consider what happens afterwards.

The pulses continue through one another.

In the ideal linear model, pulse A continues travelling right with its original shape and amplitude. Pulse B continues travelling left with its original shape and amplitude.

The smaller resultant during overlap does not represent a permanent loss of either pulse.

Question 3

Two coherent sources produce continuous sinusoidal waves that begin in phase.

At observation point P:

• The first wave travels 1.25 m.
• The second wave travels 1.65 m.
• Both waves have a wavelength of 0.80 m.
• Both waves have an amplitude of 3.0 mm.

(a) Calculate the path difference.

(b) Calculate the phase difference in degrees.

(c) Determine whether constructive or destructive interference occurs, and calculate the resultant amplitude.

(d) Would a path difference of half a wavelength automatically explain cancellation between two isolated triangular pulses? Explain.

Solution 3

The continuous waves arrive 180° out of phase, producing complete destructive interference at P. Their resultant amplitude is zero.

However, the same phase-difference argument cannot automatically be applied to two isolated triangular pulses.

Step 1: Calculate the path difference.

Path difference is the absolute difference between the distances travelled:

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

Substitute:

\[ \Delta r=|1.25-1.65| \]

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

Step 2: Calculate the phase difference.

For sources that begin in phase:

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

Substitute:

\[ \Delta\phi= \frac{360^\circ\times0.40}{0.80} \]

\[ \boxed{\Delta\phi=180^\circ} \]

The waves arrive half a cycle apart.

Step 3: Determine the interference.

A phase difference of 180° means a crest from one wave meets a trough from the other.

Because the amplitudes are equal:

\[ A_R=|A_1-A_2| \]

\[ A_R=|3.0-3.0| \]

\[ \boxed{A_R=0\text{ mm}} \]

Complete destructive interference occurs at P.

Step 4: Consider isolated pulses.

The half-wavelength rule describes the relative phases of periodic waves.

An isolated triangular pulse does not have a repeating cycle with a single wavelength in the same sense.

For two isolated pulses, you must compare their actual positions and displacement profiles. You cannot conclude that they cancel simply because someone has assigned a path difference of half a wavelength.

The distinction is between predicting the relative phase of continuous waves and calculating the actual resultant displacement of two pulses.

Question 4

An upward pulse with an amplitude of +2.0 cm approaches the end of a stretched string.

Consider two separate situations:

• Situation A: The string is attached to a rigid fixed end.
• Situation B: The string has an ideal free end that can move vertically.

Assume ideal reflection without energy loss.

(a) Describe the orientation of the reflected pulse in each situation.

(b) Calculate the resultant displacement at the endpoint when the incident pulse reaches it, using the idealised incident and reflected pulse contributions.

(c) Explain why the results differ.

Solution 4

At the fixed end, the reflected pulse is inverted and the endpoint displacement is zero. At the free end, the reflected pulse retains its upward orientation, giving an ideal resultant displacement of +4.0 cm at the endpoint.

Step 1: Determine the reflected pulse at the fixed end.

A fixed endpoint cannot move vertically.

Therefore, the reflected pulse must have the opposite displacement sign to the incident pulse.

\[ y_{\text{reflected}}=-2.0\text{ cm} \]

Step 2: Calculate the resultant at the fixed endpoint.

Apply superposition:

\[ y_R=y_{\text{incident}}+y_{\text{reflected}} \]

\[ y_R=2.0+(-2.0) \]

\[ \boxed{y_R=0\text{ cm}} \]

This satisfies the fixed-end condition.

Step 3: Determine the reflected pulse at the free end.

A free endpoint can move vertically.

The reflected pulse is not inverted, so:

\[ y_{\text{reflected}}=+2.0\text{ cm} \]

Step 4: Calculate the resultant at the free endpoint.

\[ y_R=2.0+2.0 \]

\[ \boxed{y_R=+4.0\text{ cm}} \]

The important point is that the boundary determines the reflected pulse’s orientation.

The direction of travel alone does not determine whether a pulse is upward or downward.

Question 5

A string is clamped at \(x=8.0\) m.

At a particular instant, two triangular pulses completely overlap between \(x=3.0\) m and \(x=5.0\) m.

Their properties are:

Pulse B is a reflected pulse from the fixed end. It has already reflected and is travelling away from the clamp.

Assume an ideal uniform string, negligible damping, and linear superposition.

A student makes three claims:

1. The string is stationary because the resultant displacement is zero everywhere during complete overlap.
2. The pulses are opposite because they travel in opposite directions.
3. If the fixed end were replaced by a free end, the pulses would still cancel because their directions of travel would remain unchanged.

Evaluate all three claims.

Then calculate the centres of the two pulses 0.50 s later. Determine the resultant displacements at \(x=3.7\) m, \(x=4.0\) m, and \(x=4.3\) m at that time.

Solution 5

All three claims are incorrect.

The resultant displacement is zero at the instant of complete overlap, but the string is not stationary. Opposite displacement does not follow from opposite directions of travel, and changing the boundary from fixed to free would reverse the reflected pulse’s orientation, changing cancellation into reinforcement.

Step 1: Determine the resultant at complete overlap.

The pulses have identical profiles but opposite displacements.

At every horizontal position:

\[ y_R=y_A+y_B \]

\[ y_R=2.0+(-2.0) \]

The profiles cancel completely:

\[ \boxed{y_R=0\text{ cm}} \]

The string lies along equilibrium at that instant.

However, this describes the string’s position, not whether its particles are moving.

Step 2: Explain why the string is not stationary.

A snapshot tells you where the string is, but not how quickly its points are moving.

Immediately after complete overlap, the pulses continue travelling in their original directions. The resultant profile therefore begins changing again.

The string can be momentarily straight while its particles still have transverse velocity.

In the ideal string model, the disturbance’s energy has not disappeared. At this instant, the energy is associated with transverse motion rather than displacement of the string away from equilibrium.

This is a subtle but important distinction: zero displacement does not necessarily mean zero velocity.

Step 3: Explain why opposite directions do not guarantee cancellation.

The direction of travel and the direction of displacement are separate properties.

For example, two upward pulses could travel towards each other. They would reinforce each other when their profiles overlap, despite travelling in opposite directions.

In this question, the pulses have opposite displacements because one is upward and the other is downward.

The downward orientation of Pulse B comes from its reflection at the fixed boundary.

Step 4: Calculate the positions of the pulse centres after 0.50 s.

Use the relationship:

\[ \Delta x=vt \]

Substitute:

\[ \Delta x=0.60\times0.50 \]

\[ \boxed{\Delta x=0.30\text{ m}} \]

Pulse A travels right, so its centre moves to:

\[ x_A=4.0+0.30 \]

\[ \boxed{x_A=4.30\text{ m}} \]

Pulse B travels left, so its centre moves to:

\[ x_B=4.0-0.30 \]

\[ \boxed{x_B=3.70\text{ m}} \]

The centres are now separated by:

\[ 4.30-3.70=0.60\text{ m} \]

The pulses have started passing through one another.

Step 5: Calculate the resultant displacement at \(x=3.7\) m.

Pulse B is now at its maximum downward displacement:

\[ y_B=-2.0\text{ cm} \]

Pulse A has its centre at 4.30 m, so the distance from its centre to this point is:

\[ 4.30-3.70=0.60\text{ m} \]

Each triangular pulse extends 1.0 m from its centre to its base edge. Because the displacement changes linearly:

\[ y_A=2.0\left(1-\frac{0.60}{1.0}\right) \]

\[ y_A=+0.80\text{ cm} \]

Apply superposition:

\[ y_R=0.80+(-2.0) \]

\[ \boxed{y_R=-1.20\text{ cm}} \]

Step 6: Calculate the resultant displacement at \(x=4.0\) m.

This position is 0.30 m from each pulse centre.

For Pulse A:

\[ y_A=2.0\left(1-\frac{0.30}{1.0}\right) \]

\[ y_A=+1.40\text{ cm} \]

For Pulse B:

\[ y_B=-2.0\left(1-\frac{0.30}{1.0}\right) \]

\[ y_B=-1.40\text{ cm} \]

Therefore:

\[ y_R=1.40+(-1.40) \]

\[ \boxed{y_R=0\text{ cm}} \]

The pulses still cancel at this particular position, even though their profiles no longer completely overlap.

Step 7: Calculate the resultant displacement at \(x=4.3\) m.

Pulse A is now at its maximum upward displacement:

\[ y_A=+2.0\text{ cm} \]

Pulse B is 0.60 m from its centre, so:

\[ y_B=-2.0\left(1-\frac{0.60}{1.0}\right) \]

\[ y_B=-0.80\text{ cm} \]

Therefore:

\[ y_R=2.0+(-0.80) \]

\[ \boxed{y_R=+1.20\text{ cm}} \]

The resultant profile now has a negative region on the left and a positive region on the right.

Step 8: Reconsider the free-end claim.

If the original pulse had reflected from an ideal free end instead of a fixed end, the reflected pulse would not be inverted.

Pulse B would therefore be upward rather than downward.

At complete overlap:

\[ y_R=2.0+2.0 \]

\[ \boxed{y_R=+4.0\text{ cm}} \]

The pulses would reinforce rather than cancel.

The student’s third claim is therefore incorrect. The boundary condition changes the reflected pulse’s displacement sign, even though its direction of travel is still away from the boundary.

This question brings together three separate ideas:

• Superposition determines the resultant displacement.
• Pulse direction describes the direction of propagation, not the sign of displacement.
• Boundary conditions determine whether a reflected pulse is inverted.

08What to remember when sketching opposite pulses

When you encounter an HSC question involving two opposite pulses, use this decision process.

The most common mistake is to treat opposite pulses as if they annihilate. They don’t. Superposition describes their combined displacement at a particular instant.

Once you can sketch individual pulse interactions, the next useful step is understanding how continuous waves combine to produce stationary nodes and antinodes. The relationship between phase difference, superposition, and node spacing is covered in How to Find Node Spacing from Wavelength in HSC Physics.