Wave Mechanics Examples: Resonance

This companion article provides exercises for WM11, wave mechanics: Resonance. The exercises are stated first so they can be attempted without seeing the answers. Complete worked solutions follow in Part II.

WM11 distinguished three related ideas:

(1)

For an ideal string of length L fixed at both ends,

(2)

A periodic driver produces a strong response when its frequency lies near one of the natural frequencies,

(3)

For one driven damped mode, WM11 also introduced the standard response-amplitude model

(4)

The exercises below use these results without deriving the driven-oscillator differential equation. Standard treatments of resonance and normal modes can be found in French, Crawford, OpenStax, Feynman, and MIT 8.03 [1, 2, 3, 4, 5, 6, 7].

How to use this problem set

Attempt every exercise in Part I before consulting Part II. In each problem, first identify whether the question concerns an allowed free mode, a natural frequency, a driving frequency, detuning, damping, or a forced response amplitude. A normal mode and a resonance are closely connected, but they are not the same thing.

Part I: Exercises

Exercise 1: Rebuild the natural-frequency sequence

A uniform string of length L is fixed at both ends and supports waves with speed v.

1. Starting from the fixed-end condition, write the allowed wavelengths λ_n.
2. Use f = v∕λ to obtain f_n.
3. Show that f_n = nf₁.
4. State the physical meaning of f₁.

Exercise 2: Calculate the first four resonant frequencies

A fixed-fixed string has

(5)

and wave speed

(6)

Find

1. f₁,
2. f₂,
3. f₃,
4. f₄.

If the string is driven at 241 Hz, which mode is nearest resonance?

Exercise 3: Identify a mode from its shape

The figure below shows three fixed-end standing-wave patterns. Their labels A, B, and C are intentionally not mode numbers.

1. Identify the mode number n for patterns A, B, and C.
2. Rank their natural frequencies from lowest to highest.
3. If pattern A has frequency 45 Hz, find the frequencies of B and C.
4. Which pattern has the shortest wavelength?

Exercise 4: Driving frequency and mode selection

An ideal fixed-fixed string has fundamental frequency

(7)

A periodic driver is operated successively at

(8)

For each driving frequency:

1. identify the nearest ideal-string natural frequency,
2. identify the corresponding mode number,
3. calculate the absolute detuning |f_d − f_n|.

Which drive is closest to an exact natural frequency?

Exercise 5: Harmonics and overtones

A system has the ideal-string natural frequencies

(9)

For each frequency, state both its harmonic number and, where applicable, its overtone number.

Then explain why the phrase “third overtone” does not mean the same thing as “third harmonic.”

Exercise 6: Read a resonance curve

The following schematic shows two resonance curves for systems with the same natural frequency but different damping.

Use the figure to answer:

1. Which curve represents weaker damping?
2. Which curve has the taller resonance peak?
3. Which curve has the broader resonance response?
4. Which marked driving frequency, f_d1 or f_d2, is expected to produce the larger response?
5. Why does WM11 describe the resonance as occurring near the natural frequency rather than making an absolute statement about the exact peak location for every measured response quantity?

Exercise 7: Compare detuning

A mode has natural frequency

(10)

Three drives are applied:

(11)

1. Find the absolute detuning of each drive.
2. Rank the drives from nearest to farthest from resonance.
3. Assuming comparable coupling and the same damping, which drive should produce the largest steady response?
4. Can the exact response amplitudes be found from detuning alone? Explain.

Exercise 8: Use the driven-oscillator response formula

A single damped mode is modeled by

(12)

Use

(13)

1. Find the undamped natural angular frequency ω₀.
2. Calculate Q at ω_d = 10 rad/s.
3. Calculate Q at ω_d = 20 rad/s.
4. Calculate Q at ω_d = 30 rad/s.
5. Which of the three driving angular frequencies gives the largest response?

Exercise 9: Effect of damping

Two otherwise identical driven oscillators have the same m, κ, and F₀, but damping coefficients

(14)

Without calculating a full response curve, answer:

1. Which system is expected to have the taller resonance peak?
2. Which system is expected to have the broader resonance peak?
3. Which system loses energy more rapidly through damping?
4. Why does damping keep a real resonance peak finite?

Exercise 10: Read a multi-mode frequency sweep

A frequency sweep of a bounded wave system gives the response shown below.

1. Estimate the four resonance frequencies from the graph.
2. Are the peaks consistent with an ideal fixed-fixed string whose harmonics are integer multiples of a fundamental?
3. Estimate the fundamental frequency.
4. Which peak corresponds to the third harmonic?
5. If the drive is set to 130 Hz, which resonance is it closest to?

Exercise 11: A system whose modes are not harmonic

A different physical system has measured natural frequencies

(15)

1. Are these frequencies exact integer multiples of the lowest frequency?
2. Should they automatically be called the first, second, and third harmonics of an ideal string? Explain.
3. Which one is the fundamental natural frequency of this measured set?
4. If the system is driven at 180 Hz, which measured mode is nearest resonance?

Exercise 12: Normal mode versus resonance

For each statement, decide whether it describes a normal mode, a resonance, both, or neither.

1. “An allowed free-oscillation pattern satisfying the system boundaries.”
2. “A large forced response produced when a periodic drive lies near a natural frequency.”
3. “A property determined by the system even when no external driver is applied.”
4. “Its amplitude depends strongly on damping and on how the system is driven.”
5. “For a fixed-fixed ideal string, its associated frequency can be nv∕(2L).”

Exercise 13: Resonance and energy

A student says:

Explain why this statement is incorrect. Your answer should identify the energy source and explain the role of damping in a steady resonant response.

Exercise 14: Synthesis – design a string resonance experiment

A string has length

(16)

and wave speed

(17)

A frequency sweep will be performed from 50 Hz to 450 Hz.

1. Calculate the natural frequencies that lie within the sweep range.
2. For each resonance in the range, identify its mode number.
3. What standing-wave pattern should be associated with the 300 Hz resonance?
4. Suppose the measured resonance peaks are broader than expected. Which physical effect discussed in WM11 is a likely explanation?
5. If the driver is operated at 295 Hz, calculate its detuning from the nearest natural frequency.

Part II: Complete Worked Solutions

Solution 1: Rebuild the natural-frequency sequence

For a string fixed at both ends, an integer number of half wavelengths must fit into the length:

(18)

Therefore

(19)

Using f = v∕λ,

fn	=	(20)
	=	(21)
	= .	(22)

For n = 1,

(23)

Hence

(24)

The frequency f₁ is the lowest natural frequency, called the fundamental frequency.

Solution 2: Calculate the first four resonant frequencies

The fundamental is

f1	=	(25)
	=	(26)
	= 80 Hz .	(27)

Therefore

f2	= 160 Hz ,	(28)
f3	= 240 Hz ,	(29)
f4	= 320 Hz .	(30)

The drive 241 Hz is only 1 Hz from f₃ = 240 Hz, so it is nearest the

(31)

Solution 3: Identify a mode from its shape

Count the number of half-wave lobes between the fixed ends.

1. Pattern A has one lobe, so

   n = 1 . Pattern B has two lobes, so

   n = 2 . Pattern C has three lobes, so

   n = 3 .

2. Since f_n = nf₁ for the ideal string,

   (32)

3. If f_A = 45 Hz, then

   fB	= 2fA = 90 Hz ,	(33)
   fC	= 3fA = 135 Hz .	(34)

4. Because λ_n = 2L∕n, the largest n has the shortest wavelength. Therefore pattern

   C has the shortest wavelength.

Solution 4: Driving frequency and mode selection

The ideal natural frequencies are

(35)

Thus:

• 54 Hz is nearest 55 Hz: mode n = 1, detuning

  1 Hz .

• 111 Hz is nearest 110 Hz: mode n = 2, detuning

  1 Hz .

• 168 Hz is nearest 165 Hz: mode n = 3, detuning

  3 Hz .

• 205 Hz is nearest 220 Hz: mode n = 4, detuning

  15 Hz .

The first two drives are tied for closest to an exact natural frequency, each with 1 Hz detuning.

Solution 5: Harmonics and overtones

For an ideal harmonic sequence:

• 80 Hz is the

  firstharmonic and the fundamental. It is not called an overtone.

• 160 Hz is the

  secondharmonic and the

  firstovertone .

• 240 Hz is the

  thirdharmonic and the

  secondovertone .

• 320 Hz is the

  fourthharmonic and the

  thirdovertone .

The terminology differs because harmonic counting includes the fundamental as harmonic number 1, whereas overtone counting starts with the first frequency above the fundamental. Therefore the third overtone is the fourth harmonic in this ideal sequence.

Solution 6: Read a resonance curve

1. The taller and narrower curve represents

   weakerdamping .

2. The weak-damping curve has the taller resonance peak.
3. The strong-damping curve has the broader response.
4. The marked frequency f_d1 lies much closer to the resonance peak than f_d2, so

   f_d1 is expected to produce the larger response.

5. Damping can shift the exact location of the maximum slightly, and the precise peak can depend on which response quantity is measured. WM11 therefore emphasizes that the strongest response occurs near the natural frequency rather than claiming one universal exact equality for every damped measurement.

Solution 7: Compare detuning

The absolute detunings are

|124 − 125|	= 1 Hz ,	(36)
|118 − 125|	= 7 Hz ,	(37)
|150 − 125|	= 25 Hz .	(38)

Therefore the ranking from nearest to farthest is

(39)

Assuming comparable coupling and unchanged damping, f_d1 = 124 Hz should produce the largest response because it is closest to the natural frequency.

The exact response amplitudes cannot be found from detuning alone. One also needs information such as damping and driving strength, as shown by the response-amplitude formula in WM11.

Solution 8: Use the driven-oscillator response formula

The undamped natural angular frequency is

ω0	=	(40)
	=	(41)
	=	(42)
	= 20 rad/s .	(43)

At ω_d = 10 rad/s,

Q	=	(44)
	=	(45)
	≈ 6.61 × 10−3 m .	(46)

At ω_d = 20 rad/s,

Q	=	(47)
	=	(48)
	= 2.50 × 10−2 m .	(49)

At ω_d = 30 rad/s,

Q	=	(50)
	=	(51)
	≈ 3.89 × 10−3 m .	(52)

Among these three values, the largest response occurs at

(53)

which equals the undamped natural angular frequency for this example.

Solution 9: Effect of damping

1. The system with b₁ = 1.0 N s/m has the

   taller resonance peak because it is less strongly damped.

2. The system with b₂ = 5.0 N s/m has the

   broader resonance response.

3. The larger damping coefficient b₂ corresponds to more rapid energy loss through damping.
4. Damping continually removes energy from the oscillation. In steady state, the driver must replenish those losses, which prevents the idealized unbounded growth associated with an undamped system driven exactly at resonance.

Solution 10: Read a multi-mode frequency sweep

From the plotted peaks, the resonances occur at approximately

(54)

These are integer multiples of 50 Hz, so they are consistent with the ideal fixed-fixed string relation f_n = nf₁.

Thus the fundamental is

(55)

The third harmonic is the peak at

(56)

A drive at 130 Hz is 20 Hz from 150 Hz and 30 Hz from 100 Hz, so it is closest to the third-harmonic resonance near 150 Hz.

Solution 11: A system whose modes are not harmonic

Divide the measured frequencies by the lowest one:

(57)

These are not exact integers, so the frequencies are not an ideal harmonic series based on 70 Hz.

Therefore they should not automatically be labeled the first, second, and third harmonics of an ideal string. They are simply measured natural frequencies of this system unless additional physics justifies harmonic terminology.

The lowest measured natural frequency is

(58)

so it is the fundamental natural frequency of the measured set.

A drive at 180 Hz is only 1 Hz from the 181 Hz mode, so that mode is nearest resonance.

Solution 12: Normal mode versus resonance

1. An allowed free-oscillation pattern satisfying the boundaries describes a

   normalmode .

2. A large forced response near a natural frequency describes

   resonance .

3. A property determined by the system even without external driving describes a

   normalmode and its natural frequency.

4. Strong dependence on damping and the manner of driving describes the

   resonantresponse .

5. The relation f_n = nv∕(2L) gives natural frequencies of the fixed-fixed string, so it is associated with

   normalmodes ; those same frequencies are where resonances can occur when the system is driven.

Solution 13: Resonance and energy

Resonance does not create energy. The energy comes from the external periodic driver, which performs work on the system repeatedly.

Near resonance, the timing of the driving force allows energy to be transferred efficiently into the oscillation. The amplitude can therefore become large even when the applied force is modest.

In a real damped system, energy is also continually removed by friction, material loss, radiation, or other mechanisms. In steady state, the average energy supplied by the driver balances the average energy lost through damping. The large resonant amplitude reflects efficient energy transfer, not energy creation.

Solution 14: Synthesis – design a string resonance experiment

The fundamental is

f1	=	(59)
	=	(60)
	= 100 Hz .	(61)

Therefore the natural frequencies are

(62)

Within the sweep range from 50 to 450 Hz, the observable ideal resonances are

(63)

They correspond respectively to

(64)

The 300 Hz resonance is the third mode, so its standing-wave pattern has three half-wave lobes between the fixed ends and two interior nodes.

Broader-than-expected resonance peaks are consistent with

strongerdamping or other loss mechanisms discussed in WM11.

For a drive at 295 Hz, the nearest natural frequency is 300 Hz, giving detuning

(65)

Common mistakes

• Mistake: treating a normal mode and a resonance as synonyms. A normal mode is an allowed free pattern; resonance is a forced response near its natural frequency.
• Mistake: assuming the driver creates the natural frequencies. The natural frequencies belong to the system; the driver probes or excites them.
• Mistake: assuming every large response occurs at exactly f_d = f_n under all definitions. Damping can shift the exact peak slightly, depending on the measured response quantity.
• Mistake: confusing harmonic and overtone numbering. The second harmonic is the first overtone.
• Mistake: assuming all bounded systems have f_n = nf₁. That integer relation is specific to systems such as the ideal fixed-fixed string.
• Mistake: claiming resonance creates energy. The external driver supplies the energy; damping removes it.

What WM11E1 reinforces

These exercises reinforce the sequence

(66)

For the ideal fixed-fixed string,

(67)

A drive near one of those frequencies can produce a large response, while damping controls the height and breadth of the resonance. The exercises also reinforce the distinction between harmonic mode structure, detuning, and the forced response of an individual damped mode.

References

References