The standing-wave pattern is the result of superposition, not a separate wave travelling down the string. At some positions the two opposite-travelling waves always cancel, forming nodes. At other positions they reinforce with maximum amplitude, forming antinodes. The allowed frequencies depend on the boundary conditions, which is why strings and pipes have harmonics.

A standing wave forms when two identical waves travel in opposite directions through the same medium and superpose.

This often happens when a progressive wave reflects at a boundary and overlaps with the incoming wave.

A stable pattern appears only at allowed frequencies where the reflected wave reinforces the incoming wave in a fixed pattern.

Nodes are points that remain at zero displacement; antinodes are points of maximum amplitude.

Unlike a progressive wave, a standing wave does not transfer energy along the medium as a travelling pattern.

Label the formation diagram for a standing wave on a reflected string.

Label

Labels

5

Describe how a standing wave can form on a string fixed at one end after a travelling wave reflects from the fixed end.

Stating only “waves reflect” without saying that identical opposite-travelling waves superpose to form fixed nodes and antinodes.

Describe how a standing wave can form on a string fixed at one end after a travelling wave reflects from the fixed end.

Choose

A standing-wave diagram is not just a shape to label. The envelope shows the maximum possible displacement at each point. Nodes stay fixed at zero displacement, while antinodes reach the largest amplitude. Between adjacent nodes, the particles rise and fall together. Across a node, the motion is reversed, so the two regions are 180 degrees out of phase.

A node is a point of zero displacement at all times because the two component waves always cancel there.

An antinode is a point of maximum amplitude because the component waves reinforce there.

Adjacent nodes are separated by λ/2, and adjacent antinodes are also separated by λ/2.

The distance from a node to the nearest antinode is λ/4.

All points between two adjacent nodes oscillate in phase, but points on opposite sides of a node are in antiphase.

Relative amplitude increases from zero at a node to maximum at an antinode.

Label the standing-wave diagram with node/antinode features, spacing, and phase relationships.

Label

Labels

6

A standing wave is shown on a string. State how to identify a node and an antinode, and describe the phase relationship of points on either side of a node.

Labelling the positions correctly but omitting the λ/2 spacing or phase relationship across a node.

A standing wave is shown on a string. State how to identify a node and an antinode, and describe the phase relationship of points on either side of a node.

Choose

Strings and pipes are boundary-condition problems. The ends decide where nodes and antinodes must be, and those positions decide which wavelengths fit. A fixed-fixed string or open-open pipe fits half-wavelength sections along its length. A closed-open pipe fits a node at one end and an antinode at the other, so the fundamental is a quarter wavelength and only odd harmonics fit.

For displacement waves, a fixed string end is a node and a closed pipe end is also a node.

An open pipe end is a displacement antinode.

For a string fixed at both ends, the allowed wavelengths satisfy L = nλ/2, so f_n = n v/2L.

A pipe open at both ends has antinodes at both ends and follows the same frequency pattern f_n = n v/2L.

A pipe closed at one end and open at the other fits only odd quarter wavelengths: L = mλ/4 and f = m v/4L for m = 1, 3, 5, ...

Harmonic problems are solved by choosing the pattern, linking L to λ, then using v = fλ.

Build the standing-wave frequency rules for strings and pipes from boundary conditions.

Formula

Target formula fixed-fixed or open-open: f_n = n v / (2L); closed-open: f_m = m v / (4L), m odd

f

allowed standing-wave frequency

Hz

n

harmonic number for fixed-fixed or open-open systems, n = 1, 2, 3, ...

m

odd harmonic index for closed-open pipes, m = 1, 3, 5, ...

v

wave speed in the string or air column

m s^-1

L

length of string or pipe

m

1Identify node/antinode conditions at the two ends.fixed or closed -> node; open -> antinode

2For fixed-fixed or open-open systems, fit half-wavelength sections.L = n lambda/2 -> f_n = n v/(2L)

3For closed-open pipes, fit odd quarter-wavelength sections.L = m lambda/4 -> f_m = m v/(4L), m = 1,3,5,...

4Use v = f lambda after choosing the allowed wavelength pattern.f = v/lambda

Build the formula

A pipe of length 0.85 m is closed at one end and open at the other. The speed of sound is 340 m s^-1. Calculate the fundamental frequency and the next allowed frequency.

Using f2 = 2f1 for a closed-open pipe, even though only odd harmonics are allowed.

A pipe of length 0.85 m is closed at one end and open at the other. The speed of sound is 340 m s^-1. Calculate the fundamental frequency and the next allowed frequency.

Choose

Resonance is a condition of forced oscillation. A driver can make an object oscillate at many frequencies, but the response is greatest when the driver frequency matches a natural frequency. Then the timing of the driving force adds energy in phase with the motion. In real systems, damping prevents the amplitude from growing without limit.

A natural frequency is a frequency at which a system oscillates freely when disturbed.

A periodic driving force transfers energy to the oscillator.

Resonance occurs when the driving frequency equals or is very close to a natural frequency.

At resonance the amplitude becomes large because energy is transferred efficiently each cycle.

Damping removes energy and limits the maximum amplitude, so real resonance peaks are finite.

Choose which driving condition produces the largest resonance response.

Decision

The driver frequency is much lower than the natural frequency.

The driver frequency matches the natural frequency.

The driver frequency is much higher than the natural frequency.

A mass-spring system is driven by a periodic force. Explain why the amplitude is largest when the driving frequency is close to the natural frequency.

Writing “the force is bigger” instead of explaining the frequency match and efficient energy transfer.

A mass-spring system is driven by a periodic force. Explain why the amplitude is largest when the driving frequency is close to the natural frequency.

Choose

A resonance curve shows how strongly a system responds to different driving frequencies. With little damping, the system stores energy efficiently near its natural frequency, so the peak is high and sharp. With more damping, energy is removed faster, so the maximum amplitude is smaller and the response is spread over a wider frequency range.

Damping removes energy from an oscillating system, reducing amplitude unless energy is supplied by a driver.

On an amplitude-frequency graph, resonance appears as a peak near the natural frequency.

Light damping produces a tall, narrow peak because little energy is lost per cycle.

Heavier damping produces a lower, broader peak because more energy is removed from the system.

Damping limits the amplitude at resonance and can be useful when large oscillations are dangerous.

Interpret how damping changes the amplitude-frequency resonance curve.

Graph

The graph plots oscillation amplitude against driving frequency for three systems with different damping strengths. One curve has a tall narrow peak, one is moderate, and one is low and broad.

1Identify the curve with the lowest and broadest peak.

2Compare its maximum amplitude with the lightly damped curve.

3Explain the change using energy removal by damping.

Which curve corresponds to the greatest damping, and what does damping do to the resonance peak?

Sketch or describe how the amplitude-frequency graph of a driven oscillator changes as damping is increased.

Saying only “amplitude decreases” without describing the broader resonance peak or the role of energy loss.

Sketch or describe how the amplitude-frequency graph of a driven oscillator changes as damping is increased.

Choose

The damping type is best read from a displacement-time graph. If the system crosses equilibrium repeatedly while the envelope decays, it is lightly damped. If it reaches equilibrium fastest without crossing, it is critically damped. If it does not oscillate but takes longer to settle, it is heavily damped. Applications choose the damping type depending on whether oscillation or slow return is more harmful.

Damping removes energy from an oscillating system, reducing amplitude over time.

Light damping allows oscillations to continue, but their amplitude decreases over many cycles.

Critical damping returns the system to equilibrium as quickly as possible without oscillation or overshoot.

Heavy damping, or overdamping, also avoids oscillation but returns to equilibrium more slowly than critical damping.

In resonance contexts, more damping reduces the maximum amplitude and broadens the resonance peak.

Sort each statement into the correct damping type.

Sort

Unsorted

6

light damping

0

critical damping

0

heavy damping

0

A damped oscillator returns to equilibrium without crossing it, but takes longer than another system that also does not overshoot. Identify the damping type and justify your answer.

Saying “critical damping” for any non-oscillatory return without checking whether it is the fastest return.

A damped oscillator returns to equilibrium without crossing it, but takes longer than another system that also does not overshoot. Identify the damping type and justify your answer.

Choose

C.4 is secure when the student chooses the correct representation first. Standing-wave formation starts with superposition of opposite-travelling waves. Node and antinode diagrams give spacing and phase. Strings and pipes then add boundary conditions and harmonic formulas. Resonance and damping shift the focus from spatial patterns to amplitude response over frequency or time.

Standing waves form from identical waves travelling in opposite directions through the same medium and superposing.

Nodes have zero displacement at all times; antinodes have maximum amplitude; adjacent nodes are λ/2 apart.

Points between adjacent nodes oscillate in phase, while points on opposite sides of a node are in antiphase.

Boundary conditions in strings and pipes decide whether the ends are nodes or antinodes and therefore which harmonics fit.

Resonance occurs when driving frequency matches or is close to a natural frequency, giving a large amplitude limited by damping.

More damping lowers and broadens the resonance peak; light, critical, and heavy damping have different time responses.

Match each C.4 retrieval cue to the response it should trigger.

Match

Reasons

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IB Physics HL C.4 Standing waves and resonance Notes

Model Standing wave formation

Nodes and antinodes

Strings and pipes

Model Resonance

Model Damping and resonance

Types of damping

Retrieve the C.4 Standing waves and resonance Model