E2.7 Analyse resonance conditions and applications

Ontario Grade 11 Physics

Waves and Sound

Ontario Grade 11 Physics — study topic E2.7

A swing can move farther when it is pushed at well-timed points in its motion. Push at other times, and the effect may be smaller. Resonance describes a frequency relationship that can make a system respond strongly to a repeated input. To analyse resonance, compare the input frequency with the system’s natural frequency. A frequency comparison can identify a possible resonance condition, but it does not by itself tell you the motion’s amplitude. This lesson defines the system and direction before using the model.

What you will learn

Describe the conditions for resonance.
Distinguish a system’s natural frequency from a driving frequency.
Use period and frequency to compare a system with a repeated input.
Explain useful and unwanted applications of resonance.

1. Prerequisite bridge: cycles, period, and frequency

A cycle is one complete repeat of a motion or wave pattern. For a swing, one cycle is a trip away from its resting position and back again. The period,

T

, is the time for one cycle. Period is a scalar, which means it has magnitude but no direction. Its SI unit is the second,

\mathrm{s}

.

Frequency,

f

, is the number of cycles completed each second. It is also a scalar. Its SI unit is the hertz,

\mathrm{Hz}

, where

1\ \mathrm{Hz}

means one cycle per second. Period and frequency are reciprocals. A short period corresponds to a high frequency.

An oscillating system repeats its motion around an equilibrium position. Equilibrium is the resting or balanced position. Displacement tells how far and in which direction the system is from equilibrium, so displacement is a vector. A vector has both magnitude and direction. Before describing displacement, choose a positive direction. For a swing, take forward from equilibrium as positive and backward as negative. The system is the object or group of objects whose motion is being considered.

f=\frac{1}{T}

Period is time per cycle; frequency is cycles per second.
Frequency and period are scalars. Displacement is a vector.
Choose a positive direction before describing displacement.

2. The resonance condition

A system’s natural frequency is the frequency at which it tends to oscillate after it is disturbed and then left without a repeated input. A driving force is a repeated push or pull from outside the system. The driving frequency is how often that input repeats.

Resonance occurs when the driving frequency equals or is close to the system’s natural frequency. In this condition, repeated input can add energy to the motion in a way that increases its amplitude. Amplitude is the greatest displacement from equilibrium. For a moving object, amplitude is measured in metres,

\mathrm{m}

.

There is no single frequency gap that counts as close for every system. Whether a particular difference is close enough depends on the system and its response. If no tolerance or system-specific information is given, compare the frequencies and report whether they are equal, which one is higher, and the size of the difference. Do not claim that a specific unequal pair is near resonance solely because the numbers seem close.

Real systems lose energy. This effect is called damping. Damping limits the amplitude, so resonance does not mean that motion grows without limit. The strength of the repeated input and the damping also affect the response. A frequency comparison identifies the resonance condition; it does not calculate the amplitude.

f_{\text{driving}}\approx f_{\text{natural}}

Natural frequency belongs to the system; driving frequency belongs to the repeated input.
Equal or sufficiently close frequencies are the condition for resonance.
How close is close enough depends on the system.
Damping limits the motion, and frequency alone does not determine amplitude.

3. Applications: useful and unwanted resonance

A person can build a swing’s motion by pushing at times that match the swing’s natural frequency. The swing’s displacement changes direction during each cycle, but the repeated pushes can be timed to increase the motion. Poorly timed pushes are less effective. The useful condition is the frequency match, not a particular direction for frequency; frequency has no direction.

Resonance is also useful in musical instruments. A vibrating part can cause nearby air to vibrate more strongly when the vibration matches a natural frequency of the instrument or air space. This can increase the sound produced. The example uses the same central idea: compare a repeated vibration with a natural frequency.

Resonance can also cause unwanted vibration. For example, a machine’s repeated vibration may drive a nearby part near one of that part’s natural frequencies. A designer may change the operating frequency, change the system, or add damping to reduce the response. These actions aim to avoid a frequency match or limit the resulting motion.

A comparison should be stated carefully. If two frequencies are equal, they match. If they are unequal, give their difference and say that deciding whether the difference is close enough requires information about the system. Do not infer a definite amplitude from the comparison alone.

Resonance can increase useful motion or sound.
The same frequency condition can cause unwanted vibration.
Changing the input or system, or adding damping, can reduce an unwanted response.

4. A method for analysing resonance

First name the system and the repeated input. If motion direction matters, define a positive direction. Next list the known information and the unknown. A period may be given for the system, while the input may be described by a driving frequency.

If a period is given, convert it to frequency using the reciprocal relationship. Keep the period in seconds; the result is in hertz. Then compare the natural and driving frequencies in the same units. If they are equal, state that they match. If they differ, state which is higher and calculate the size of the gap when useful.

Use system-specific information if the task asks whether unequal frequencies are close enough for resonance. Without such information, do not apply an invented cutoff. Report that the frequencies differ and that closeness cannot be decided from the values alone. Finish by checking units and whether the conclusion follows from the comparison. Do not claim a numerical amplitude without information about the input and damping.

f=\frac{1}{T}

Convert period to frequency before comparing, if needed.
Compare frequencies in hertz and report any difference clearly.
For unequal frequencies, the system determines whether the gap is close enough.
Check units and avoid claiming an amplitude from frequency data alone.

Worked example

Finding a swing’s natural frequency

A swing completes one full cycle in

2.0\ \mathrm{s}

. A person pushes it once every

2.0\ \mathrm{s}

. Do the frequencies match?

Define the system and direction
The system is the swing and rider. Take forward displacement from the resting position as positive and backward displacement as negative. The unknown is whether the driving frequency matches the natural frequency.
Calculate the natural frequency
The swing’s period is $2.0\ \mathrm{s}$ . Frequency is the reciprocal of period, because it counts cycles per second.
$f_{\text{natural}}=\frac{1}{2.0\ \mathrm{s}}=0.50\ \mathrm{Hz}$
Find and compare the driving frequency
One push every $2.0\ \mathrm{s}$ corresponds to a driving frequency of $0.50\ \mathrm{Hz}$ . The frequencies are equal, so the input matches the swing’s natural frequency.
$f_{\text{driving}}=\frac{1}{2.0\ \mathrm{s}}=0.50\ \mathrm{Hz}$

Answer: The frequencies match at

0.50\ \mathrm{Hz}

. This is the resonance condition; well-timed pushes can increase the swing’s amplitude.

Check: The reciprocal of seconds has units of hertz. A period of two seconds means half a cycle per second, which is reasonable. Frequency has no direction; the swing’s displacement direction was defined separately.

Worked example

Comparing unequal frequencies

A model oscillator has a natural frequency of

1.5\ \mathrm{Hz}

and is driven at

1.4\ \mathrm{Hz}

. What can you conclude about resonance from these values?

Define the system and direction
The system is the model oscillator. If describing its displacement, choose one direction as positive and the opposite direction as negative. The known quantities are two frequencies in hertz.
Compare the frequencies
Subtract the driving frequency from the natural frequency to find the size of their difference. The values are unequal. Without information about this oscillator’s resonance range, the difference alone does not establish whether they are close enough for resonance.
$|1.5\ \mathrm{Hz}-1.4\ \mathrm{Hz}|=0.1\ \mathrm{Hz}$

Answer: The natural frequency is higher by

0.1\ \mathrm{Hz}

. The values are not equal. Whether they are close enough for resonance depends on the system, so these values alone do not settle that question.

Check: Both inputs to the subtraction are in hertz, so the difference is also in hertz. The positive result is the size of the gap, not a direction. No unsupported cutoff is used to classify the unequal frequencies.

Worked example

Converting a period before comparing

A machine part has a natural period of

0.40\ \mathrm{s}

. The machine produces a repeated vibration at

2.0\ \mathrm{Hz}

. Compare the frequencies and state what the values establish.

Define the system and direction
The system is the machine part. Choose one direction of its motion as positive and the opposite direction as negative. The natural period is given; the driving frequency is already in hertz.
Calculate the natural frequency
Use the reciprocal relationship to convert the period to natural frequency. Since the period is in seconds, the result is in hertz.
$f_{\text{natural}}=\frac{1}{0.40\ \mathrm{s}}=2.5\ \mathrm{Hz}$
Compare and qualify the result
The natural frequency is $2.5\ \mathrm{Hz}$ , while the driving frequency is $2.0\ \mathrm{Hz}$ . The natural frequency is higher by $0.5\ \mathrm{Hz}$ . The frequencies are unequal; deciding whether the gap is close enough for resonance requires information about the part’s response.
$|2.5\ \mathrm{Hz}-2.0\ \mathrm{Hz}|=0.5\ \mathrm{Hz}$

Answer: The natural frequency is

2.5\ \mathrm{Hz}

and the driving frequency is

2.0\ \mathrm{Hz}

. They are not equal. The values alone do not determine whether the part is near resonance.

Check: A period of

0.40\ \mathrm{s}

gives

2.5

cycles per second, which is reasonable. The difference has units of hertz. The direction of the comparison is clear: the natural frequency is higher. No amplitude or resonance range can be calculated from the information given.

Common mistakes and how to avoid them

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

Correction: Natural frequency describes the system. Driving frequency describes the repeated input. Compare them to analyse resonance.

Calling any unequal frequencies near resonance because their numbers look close.

Correction: There is no universal cutoff for closeness. Use system-specific information, or state the difference and say that the values alone do not decide.

Saying resonance makes amplitude grow without limit.

Correction: Damping limits motion in real systems. Frequency matching can increase amplitude, but does not specify its size.

Using period as though it were frequency, or giving frequency a direction.

Correction: Period is measured in seconds and frequency in hertz; convert using the reciprocal relationship. Frequency is a scalar, while displacement has direction.

Lesson summary

A cycle is one complete repeat. Period is time per cycle, and frequency is cycles per second.
Resonance occurs when the driving frequency equals or is sufficiently close to the system’s natural frequency.
There is no universal closeness threshold; the system’s response matters.
Resonance can be useful, such as for a swing or musical instrument, or unwanted, such as machine vibration.
Damping limits amplitude. A frequency comparison alone does not determine amplitude.

Check your understanding

Question 1

A system has a natural frequency of

3.0\ \mathrm{Hz}

. Which listed driving frequency is numerically closest to it?

$0.30\ \mathrm{Hz}$
$2.9\ \mathrm{Hz}$
$6.0\ \mathrm{Hz}$
$30\ \mathrm{Hz}$

Show answer and explanation

2.9\ \mathrm{Hz}

The difference between

3.0\ \mathrm{Hz}

and

2.9\ \mathrm{Hz}

is

0.1\ \mathrm{Hz}

, smaller than the differences for the other options. This identifies the closest listed value, but closeness alone does not establish resonance; that depends on the system.

Question 2

A system completes one cycle in

0.25\ \mathrm{s}

. What is its frequency?

$0.25\ \mathrm{Hz}$
$2.5\ \mathrm{Hz}$
$4.0\ \mathrm{Hz}$
$25\ \mathrm{Hz}$

Show answer and explanation

4.0\ \mathrm{Hz}

Frequency is the reciprocal of period:

1/(0.25\ \mathrm{s})=4.0\ \mathrm{Hz}

. The reciprocal of seconds is cycles per second, or hertz.

Question 3

What does damping do in an oscillating system?

It makes the driving frequency equal to the natural frequency.
It limits the amplitude of the system’s motion.
It changes frequency into displacement.
It guarantees that the system stops immediately.

Show answer and explanation

It limits the amplitude of the system’s motion.

Damping is energy loss that reduces or limits oscillation. It does not set the driving frequency and does not imply that the system stops immediately.

Key terms

Amplitude: The greatest displacement from equilibrium during an oscillation.
Damping: Energy loss that reduces or limits the motion of an oscillating system.
Driving frequency: The frequency of a repeated outside input, such as repeated pushes or vibrations.
Equilibrium: The resting or balanced position around which a system oscillates.
Frequency: The number of complete cycles per second, measured in hertz.
Natural frequency: The frequency at which a system tends to oscillate after it is disturbed and left without a repeated input.
Period: The time required for one complete cycle, measured in seconds.
Resonance: The condition in which a repeated input matches or is sufficiently close to a system’s natural frequency.

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Published by DoAssignment. This AI-assisted lesson follows Ontario Grade 11 Physics (SPH3U), expectation E2.7. 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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E2.7 · Analyse resonance conditions and applications

Ontario Grade 11 Physics

Ontario Grade 11 Physics — study topic E2.7

What you will learn

1. Prerequisite bridge: cycles, period, and frequency

2. The resonance condition

3. Applications: useful and unwanted resonance

4. A method for analysing resonance

Finding a swing’s natural frequency

Comparing unequal frequencies

Converting a period before comparing

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