E3.2 Explain the components and conditions of resonance

Ontario Grade 11 Physics

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Ontario Grade 11 Physics — E3.2: Explain the components and conditions of resonance

A small push can make a swing move. Repeated pushes can make it move much farther, but only when the pushes arrive at the right times. This is resonance. It occurs in many vibrating systems, from a playground swing to a musical instrument. To explain resonance, we need to identify what is moving, what is driving it, and how the timing of the driving compares with the system’s own preferred timing.

First, recall two ideas. A vibration is repeated motion about a resting position. Frequency tells how many complete cycles occur in one second. Its SI unit is the hertz, written as

\mathrm{Hz}

, where one hertz means one cycle per second. Period is the time for one complete cycle, measured in seconds. Frequency and period are related by

f=1/T

.

What you will learn

Identify the parts of a resonance situation: an oscillator, a periodic driving force, and energy transfer.
Explain how resonance depends on the driving frequency matching the oscillator’s natural frequency.
Describe how damping affects the size of the response without changing the basic condition for resonance.
Use frequency and period to decide whether a system is being driven at resonance.

1. The parts of a resonance situation

Define the physical system as the object or set of objects whose motion we are studying. For a swing, the system is the swing and rider. For a ruler clamped to a desk, the system is the ruler’s vibrating end. A reference frame is the viewpoint used to describe motion. Here, use the room as the reference frame, and call the system’s resting position zero. Choose one direction from that position as positive; the opposite direction is negative. The motion is back and forth, so its direction changes. Frequency and period, by contrast, are scalar quantities: they have size but no direction.

An oscillator is a system that can vibrate. If an oscillator is disturbed and then allowed to move without continued repeated pushes, it vibrates at its natural frequency. The natural frequency is the frequency at which that system tends to vibrate. It depends on features of the system, such as its design and physical properties. It is not the same as the frequency of a force that someone applies.

A driving force is an external force that repeatedly acts on an oscillator. If the force repeats at regular time intervals, it is a periodic driving force. The frequency of these repeated pushes is the driving frequency. The driver supplies energy to the oscillator. Some energy is also transferred away, for example through friction or air resistance. The loss of energy that reduces the motion over time is called damping.

Resonance is a greatly increased response when a periodic driving force acts at or near the oscillator’s natural frequency. The driver and oscillator are then timed so that energy is transferred effectively over repeated cycles. In a swing, for example, well-timed pushes add motion rather than cancelling it. The key components are the oscillator, a repeated driver, and energy transfer. The key condition is a close match between driving frequency and natural frequency.

f_d\approx f_n

The oscillator has a natural frequency, $f_n$ , measured in hertz.
The repeated driver has a driving frequency, $f_d$ , measured in hertz.
Resonance occurs when $f_d$ is equal or close to $f_n$ .
Damping removes energy and limits the response.

2. What happens at resonance

Think of one complete cycle as the motion from a position, through the other side, and back to the starting position moving in the same direction. For resonance, the driver must repeat in step with this cycle. If pushes are much too frequent or too widely spaced, they do not add motion effectively each time. At or near the natural frequency, the repeated input can build a larger vibration.

The size of the vibration is called its amplitude. Amplitude is the greatest displacement from the resting position. It is a distance, so its SI unit is the metre. At resonance, amplitude can become much larger than it is when the driving frequency is far from the natural frequency. Resonance does not mean the system moves in only one direction; it continues to vibrate back and forth.

The response is not unlimited. Damping transfers energy away from the oscillator, so it prevents the amplitude from growing without bound. A system with stronger damping generally has a smaller resonant response than the same system with weaker damping. Damping does not change the basic idea that the driving frequency must match, or nearly match, the natural frequency for resonance.

A useful comparison is a swing. The swing is the oscillator. A person pushing at regular intervals is the driver. The swing’s repeated cycle has a natural frequency. If the pushes arrive at the right point in each cycle, they transfer energy to the motion. If they arrive at poorly timed points, they may add little motion or oppose the existing motion. This explains why the presence of a repeated force alone is not enough: the timing matters.

T=\frac{1}{f}

Amplitude describes the size of the oscillation, not its frequency.
At resonance, repeated energy input is timed to reinforce the oscillation.
Stronger damping usually means a smaller resonant amplitude.
A repeated force can act without producing resonance if its frequency is not close to the natural frequency.

3. Comparing frequencies in practice

To check for resonance, identify or determine the system’s natural frequency and compare it with the driving frequency. If a period is given instead of a frequency, use the relationship between period and frequency. Keep units consistent: seconds for period and hertz for frequency. The frequency tells the number of cycles per second; it does not specify a direction.

A frequency match supports the condition for resonance, but it does not by itself tell the exact amplitude. The amount of damping and how energy is supplied also matter. Therefore, do not claim that matching frequencies always produce the same large motion in every system. The reliable conclusion is that matching or nearly matching frequencies allow a strong resonant response, while damping limits that response.

The examples below use stated values to practise the frequency comparison. They are calculations, not reports of experiments. For each one, the system is the named oscillator, the room is the reference frame, and the positive direction is defined from the resting position toward the named positive side. Since the calculations compare scalar frequencies, that direction does not change the numerical result.

f=\frac{1}{T}

Use $f=1/T$ when a period is provided.
Compare driving frequency with natural frequency in the same unit.
Frequency matching indicates resonance conditions; damping affects response size.

Worked example

1. A swing pushed at its natural rate

A swing has a natural period of

2.0\ \mathrm{s}

. A person pushes it once every

2.0\ \mathrm{s}

. Determine whether the driving condition is resonance. The system is the swing and rider, viewed from the room. Let positive displacement point forward from the resting position.

Find the natural frequency
The known natural period is $T_n=2.0\ \mathrm{s}$ . The unknown is the natural frequency. Use the period-frequency relationship because the period gives the time for one cycle.
$f_n=\frac{1}{T_n}=\frac{1}{2.0\ \mathrm{s}}=0.50\ \mathrm{Hz}$
Find the driving frequency and compare
One push every $2.0\ \mathrm{s}$ means a driving period of $2.0\ \mathrm{s}$ . Convert it to frequency, then compare the two frequencies. Both are scalar quantities, so neither has a direction.
$f_d=\frac{1}{2.0\ \mathrm{s}}=0.50\ \mathrm{Hz}$

Answer: The driving frequency equals the natural frequency:

f_d=f_n=0.50\ \mathrm{Hz}

. The frequency condition for resonance is met.

Check: The reciprocal of seconds has units of per second, or hertz. The two frequencies match, so the conclusion is reasonable. The actual amplitude cannot be calculated from these values because damping and energy input are not specified.

Worked example

2. A ruler driven too slowly

A ruler’s natural frequency is

4.0\ \mathrm{Hz}

. A repeated push drives it at

2.0\ \mathrm{Hz}

. Is it being driven at resonance? The system is the ruler’s vibrating end, viewed from the room. Let positive displacement point upward from its resting position.

Identify the known and unknown values
The natural frequency is $f_n=4.0\ \mathrm{Hz}$ , and the driving frequency is $f_d=2.0\ \mathrm{Hz}$ . The unknown is whether these frequencies meet the resonance condition.
$f_n=4.0\ \mathrm{Hz}, f_d=2.0\ \mathrm{Hz}$
Compare the frequencies
Resonance requires the driving frequency to equal or be close to the natural frequency. Here, the driver repeats at half the natural frequency, so the frequencies are not close.
$\frac{f_d}{f_n}=\frac{2.0\ \mathrm{Hz}}{4.0\ \mathrm{Hz}}=0.50$

Answer: The ruler is not being driven at resonance. The driving frequency is

2.0\ \mathrm{Hz}

, well below the natural frequency of

4.0\ \mathrm{Hz}

.

Check: The ratio is dimensionless because hertz cancels. A ratio of

0.50

means the driving frequency is half the natural frequency, which is not a close match. The comparison does not predict an exact amplitude.

Worked example

3. Finding a resonant push interval

A simple oscillator has a natural frequency of

3.0\ \mathrm{Hz}

. At what interval should a repeated driver act to meet the resonance condition? The system is the oscillator, viewed from the room. Let positive displacement point right from its resting position.

State the governing relationship
The driver should have the same frequency as the oscillator’s natural frequency. The unknown is the time between successive pushes, which is the driving period.
$f_d\approx f_n$
Convert the matched frequency to a period
For the resonance condition, use $f_d=3.0\ \mathrm{Hz}$ . The period is the reciprocal of frequency. Preserve the hertz unit in the substitution.
$T_d=\frac{1}{f_d}=\frac{1}{3.0\ \mathrm{s^{-1}}}=0.33\ \mathrm{s}$

Answer: The driver should act about once every

0.33\ \mathrm{s}

to match the natural frequency and meet the condition for resonance.

Check: The reciprocal of

\mathrm{s^{-1}}

is seconds, as required for a period. The two-significant-figure result is reasonable: about three cycles occur each second, so one cycle takes about one third of a second. This gives the matching timing, not a prediction of amplitude.

Common mistakes and how to avoid them

Treating natural frequency and driving frequency as the same thing by definition.

Correction: Natural frequency belongs to the oscillator. Driving frequency belongs to the repeated external force. Compare them to check the resonance condition.

Saying that any repeated push produces resonance.

Correction: The driving frequency must be equal or close to the natural frequency for resonance.

Assuming resonance means motion grows without limit.

Correction: Damping transfers energy away and limits the amplitude.

Confusing amplitude with frequency.

Correction: Amplitude is the greatest displacement from rest. Frequency is the number of cycles per second.

Lesson summary

Resonance involves an oscillator and a periodic driving force.
The oscillator has a natural frequency; the driver has a driving frequency.
A strong resonant response occurs when the driving frequency is equal or close to the natural frequency.
Energy supplied by the driver can build the motion when the timing matches.
Damping limits the amplitude by transferring energy away.

Check your understanding

Question 1

A system has a natural frequency of

5.0\ \mathrm{Hz}

. Which driving frequency best meets the resonance condition?

$1.0\ \mathrm{Hz}$
$5.0\ \mathrm{Hz}$
$10\ \mathrm{Hz}$
$0.20\ \mathrm{Hz}$

Show answer and explanation

5.0\ \mathrm{Hz}

The driving frequency should equal or be close to the natural frequency. The exact match is

5.0\ \mathrm{Hz}

.

Question 2

What does damping do to a resonant vibration?

It removes energy and limits the amplitude.
It makes the driving frequency equal to the natural frequency.
It changes amplitude into frequency.
It guarantees that the system stops immediately.

Show answer and explanation

It removes energy and limits the amplitude.

Damping transfers energy away from the oscillator and usually reduces the size of its response. It does not itself set the driving frequency.

Question 3

A driver acts once every

0.50\ \mathrm{s}

. What is its frequency?

$0.50\ \mathrm{Hz}$
$1.0\ \mathrm{Hz}$
$2.0\ \mathrm{Hz}$
$5.0\ \mathrm{Hz}$

Show answer and explanation

2.0\ \mathrm{Hz}

Use

f=1/T

. The reciprocal of

0.50\ \mathrm{s}

is

2.0\ \mathrm{Hz}

.

Key terms

Amplitude: The greatest displacement of a vibrating system from its resting position.
Damping: The transfer of energy away from an oscillator, which reduces or limits its motion.
Driving frequency: The number of cycles per second of a periodic external force.
Natural frequency: The frequency at which a system tends to vibrate when it is disturbed and then allowed to move.
Oscillator: A system that can vibrate.
Period: The time for one complete cycle, measured in seconds.
Resonance: A greatly increased response when a periodic driving force acts at or near an oscillator’s natural frequency.

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About this lesson and its review

Published by DoAssignment. This AI-assisted lesson follows Ontario Grade 11 Physics (SPH3U), expectation E3.2. It is a study resource, not an official curriculum publication.

Before publication, the draft is checked for structure, mathematical or chemical notation, calculations, course boundaries, and readability, and then requires administrator approval. Errors can still occur, so corrections are welcomed.

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E3.2 · Explain the components and conditions of resonance

Ontario Grade 11 Physics

Ontario Grade 11 Physics — E3.2: Explain the components and conditions of resonance

What you will learn

1. The parts of a resonance situation

2. What happens at resonance

3. Comparing frequencies in practice

1. A swing pushed at its natural rate

2. A ruler driven too slowly

3. Finding a resonant push interval

Common mistakes and how to avoid them

Lesson summary

Check your understanding

Question 1

Question 2

Question 3

Key terms

Continue through SPH3U

About this lesson and its review