Resonance Examples in Physics

Start with the recap, study the fully worked examples, then use the practice problems to check your understanding of Resonance.

This page combines explanation, solved examples, and follow-up practice so you can move from recognition to confident problem-solving in Physics.

Concept Recap

Resonance occurs when a system is driven at or near one of its natural frequencies, causing the amplitude of its oscillation to grow strongly.

Push at just the right rhythm and the vibration builds up much more than usual.

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How to Use These Examples

  • Read the first worked example with the solution open so the structure is clear.
  • Try the practice problems before revealing each solution.
  • Use the related concepts and background knowledge badges if you feel stuck.

What to Focus On

Core idea: Resonance asks what oscillates, what travels, and which wave quantity is being measured.

Common stuck point: Students often know a formula related to resonance but skip the recognition step: Am I describing a repeating disturbance using wavelength, frequency, amplitude, speed, medium, or superposition? That leads to a correct-looking substitution attached to the wrong physical model.

Sense of Study hint: Ask: Am I describing a repeating disturbance using wavelength, frequency, amplitude, speed, medium, or superposition?

Worked Examples

Example 1

medium
A simple pendulum has length L=1.0 mL = 1.0\text{ m}. Take g=9.8 m/s2g = 9.8\text{ m/s}^2. Find its natural frequency and the corresponding resonant driving frequency.

Answer

f0≈0.498 Hzf_0 \approx 0.498 \text{ Hz}

First step

1
T=2πL/g=2π1.0/9.8≈2π(0.3194)≈2.007T = 2\pi\sqrt{L/g} = 2\pi\sqrt{1.0/9.8} \approx 2\pi(0.3194) \approx 2.007 s.

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Example 2

hard
A closed resonance tube shows successive resonances at L1=0.15 mL_1 = 0.15\text{ m} and L2=0.55 mL_2 = 0.55\text{ m} with a 425 Hz425\text{ Hz} fork. Find the wavelength and speed of sound.

Example 3

hard
Two driven oscillators have the same natural frequency f0=10 Hzf_0 = 10\text{ Hz}, one lightly damped and one heavily damped. Compare their peak amplitudes and bandwidth.

Example 4

challenge
A closed tube of length L=0.34 mL = 0.34\text{ m} resonates at fundamental. With v=340 m/sv = 340\text{ m/s}, find the next two resonant frequencies above the fundamental.

Practice Problems

Try these problems on your own first, then open the solution to compare your method.

Example 1

easy
Resonance occurs when a system is driven at which frequency?

Example 2

easy
At resonance, what happens to the amplitude of the oscillation?

Example 3

easy
A pendulum's natural frequency is 0.5 Hz. To resonate it, at what frequency should you push?

Example 4

easy
What limits how large the amplitude grows at resonance?

Example 5

easy
Does pushing a swing at a frequency far from its natural frequency cause large oscillations?

Example 6

easy
A tuning fork at 440 Hz is held near a guitar string also tuned to 440 Hz. What happens to the string?

Example 7

easy
Is resonance more about matching frequency or matching amplitude of the driving force?

Example 8

easy
A wine glass shatters when a singer holds a note. What note frequency causes this?

Example 9

medium
A closed air column resonates with sound when its length matches λ4\frac{\lambda}{4}. If the sound is 340 Hz (v = 340 m/s), what is the shortest resonant length?

Example 10

medium
A mass on a spring has natural frequency f0=12πkmf_0 = \frac{1}{2\pi}\sqrt{\frac{k}{m}}. With k=200k = 200 N/m and m=0.5m = 0.5 kg, find the driving frequency for resonance.

Example 11

medium
A resonance tube (closed at the water surface) first resonates at length 0.20 m for a 425 Hz fork. Estimate the speed of sound. (L=λ4L = \frac{\lambda}{4}.)

Example 12

medium
A child on a swing of natural period 2 s is pushed every 2 s and the amplitude grows. If the pushing period changed to 1 s, would resonance still occur?

Example 13

medium
An open tube of length 0.5 m (v = 340 m/s) resonates at its fundamental. What driving sound frequency causes this resonance? (f=v2Lf = \frac{v}{2L}.)

Example 14

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A bridge has a natural frequency of 2 Hz. Soldiers march across at 2 steps per second. Why is this dangerous?

Example 15

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A closed resonance tube shows successive resonances at lengths 0.20 m and 0.60 m for the same fork. Find the wavelength. (Successive closed-tube resonances differ by λ2\frac{\lambda}{2}.)

Example 16

medium
A driven oscillator with little damping has a sharp resonance peak; a heavily damped one has a broad, low peak. Which responds with greater amplitude exactly at resonance?

Example 17

medium
A closed tube resonates at its fundamental for a 170 Hz fork (v = 340 m/s). Find the tube length. (L=λ4L = \frac{\lambda}{4}.)

Example 18

challenge
A closed resonance tube first resonates at 0.18 m and next at 0.55 m for a fork. Find the speed of sound if the fork is 460 Hz. (Use the half-wavelength difference.)

Example 19

challenge
A spring-mass system has f0=4f_0 = 4 Hz. If the mass is quadrupled, find the new resonant driving frequency. (f0∝1mf_0 \propto \frac{1}{\sqrt{m}}.)

Example 20

challenge
An open pipe (v = 340 m/s) must resonate at exactly 850 Hz in its fundamental. What length should it be? (f1=v2Lf_1 = \frac{v}{2L}.)

Example 21

easy
A pendulum has natural period 1.0 s1.0\text{ s}. At what driving frequency should you push to cause resonance?

Example 22

easy
A closed (one-end-closed) tube of length 0.5 m0.5\text{ m} resonates fundamentally when L=λ/4L = \lambda/4. Find the wavelength.

Example 23

easy
An open tube of length 1.0 m1.0\text{ m} resonates fundamentally when L=λ/2L = \lambda/2. Find the fundamental wavelength.

Example 24

easy
A child on a swing of period 3.0 s3.0\text{ s} is pushed every 3.0 s3.0\text{ s}. Does the amplitude grow?

Example 25

easy
A tuning fork at 256 Hz256\text{ Hz} resonates with an air column. The column resonates because its natural frequency matches what?

Example 26

medium
A closed tube fundamental resonates at f=500 Hzf = 500\text{ Hz} with v=340 m/sv = 340\text{ m/s}. Find the tube length.

Example 27

medium
A mass on a spring has k=400 N/mk = 400\text{ N/m} and m=1.0 kgm = 1.0\text{ kg}. Find the resonant driving frequency.

Example 28

medium
A closed resonance tube first resonates at L1=0.10 mL_1 = 0.10\text{ m} and next at L2=0.30 mL_2 = 0.30\text{ m} for the same fork. Find the wavelength.

Example 29

medium
A bridge has a natural frequency of 1.5 Hz1.5\text{ Hz}. Soldiers march at 9090 steps per minute. Will resonance occur?

Example 30

medium
An open tube of length 0.4 m0.4\text{ m} (v = 320 m/s320\text{ m/s}) resonates at fundamental. Find the frequency.

Example 31

medium
A closed tube has L=0.25 mL = 0.25\text{ m}. With v=340 m/sv = 340\text{ m/s}, find the fundamental resonant frequency.

Example 32

medium
A guitar string is tuned to 440 Hz440\text{ Hz}. A nearby tuning fork at 440 Hz440\text{ Hz} is struck. What happens to the string?

Example 33

medium
An open organ pipe is 0.85 m0.85\text{ m} long; sound speed is 340 m/s340\text{ m/s}. Find its fundamental resonance frequency.

Example 34

medium
A driven oscillator's amplitude is highest at f=50 Hzf = 50\text{ Hz}. What is the system's natural frequency?

Example 35

medium
A closed tube resonates fundamentally at 850 Hz850\text{ Hz} with v=340 m/sv = 340\text{ m/s}. Find the tube length.

Example 36

medium
A wine glass has a natural frequency of 660 Hz660\text{ Hz}. A singer holds a 660 Hz660\text{ Hz} note loudly. What can happen?

Example 37

hard
A spring-mass system has f0=5 Hzf_0 = 5\text{ Hz}. If the mass is doubled, find the new resonant frequency.

Example 38

hard
A spring-mass system has f0=4 Hzf_0 = 4\text{ Hz}. The spring constant is quadrupled. Find the new resonant frequency.

Example 39

hard
A closed tube has length L=0.5 mL = 0.5\text{ m}; v=340 m/sv = 340\text{ m/s}. Find the third harmonic resonance frequency.

Example 40

hard
An open tube of length 0.6 m0.6\text{ m} resonates with v=360 m/sv = 360\text{ m/s}. Find the second harmonic frequency.

Example 41

hard
A pendulum has L=0.25 mL = 0.25\text{ m} and g=9.8 m/s2g = 9.8\text{ m/s}^2. Find the resonant driving frequency.

Example 42

challenge
A circuit has natural (resonant) frequency f0=1/(2πLC)f_0 = 1/(2\pi\sqrt{LC}) with L=1.0 mHL = 1.0\text{ mH} and C=10 nFC = 10\text{ nF}. Find f0f_0.
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Related Concepts

Standing WavesHarmonics

Background Knowledge

These ideas may be useful before you work through the harder examples.

standing wavesharmonics