Standing Waves and Resonance

Nodes and antinodes; harmonics; strings and pipes; boundary conditions

# Standing Waves and Resonance — IB Physics

Standing waves form when two identical waves travel in opposite directions and superpose. They appear in musical instruments, microwave ovens, and many physical systems.


1. Formation

Two waves with the same frequency, amplitude, and wavelength travelling in opposite directions create a standing wave.

2. Nodes and Antinodes

Node: Point of zero displacement (destructive). Separation between nodes = λ/2\lambda/2. Antinode: Point of maximum displacement (constructive). Midway between nodes.

3. Standing Waves vs Progressive Waves

Progressive Standing
Energy transfer Yes No
Amplitude Same everywhere Varies (0 at node, max at antinode)
Phase Varies continuously Same between adjacent nodes

4. Strings (Fixed at Both Ends)

fn=nv2Lf_n = \frac{nv}{2L} (n=1,2,3,...n = 1, 2, 3, ...)

Fundamental (n=1n=1): L=λ/2L = \lambda/2, f1=v/(2L)f_1 = v/(2L)

v=T/μv = \sqrt{T/\mu} where TT = tension, μ\mu = mass per unit length.

5. Pipes

Open both ends: fn=nv/(2L)f_n = nv/(2L) — all harmonics. Antinodes at both ends.

Closed one end: fn=nv/(4L)f_n = nv/(4L) (n=1,3,5,...n = 1, 3, 5, ...) — odd harmonics only. Node at closed end, antinode at open end.

6. Resonance

Resonance occurs when the driving frequency matches a natural frequency of the system → maximum amplitude energy transfer.

Worked Example: Example 1

Problem

A string of length 60 cm vibrates at its fundamental frequency of 440 Hz. Find the wave speed. v=2Lf=2(0.6)(440)=528v = 2Lf = 2(0.6)(440) = 528 m/s

Solution

Worked Example: Closed Pipe

Problem

A closed pipe is 42.5 cm long. Speed of sound = 340 m/s. f1=v/(4L)=340/(4×0.425)=200f_1 = v/(4L) = 340/(4 \times 0.425) = 200 Hz Next harmonics: f3=600f_3 = 600 Hz, f5=1000f_5 = 1000 Hz.

Solution

8. Practice Questions

    1. Explain the difference between a node and an antinode. (2 marks)
    1. A string vibrates in its 4th harmonic. The string is 1.2 m long. Find the wavelength. (2 marks)
    1. A closed pipe produces its fundamental at 250 Hz. Find the next two resonant frequencies. (2 marks)

    Answers

    1. Node: point of zero displacement, formed by destructive interference. Antinode: point of maximum displacement, formed by constructive interference.
    1. 4th harmonic: L=4(λ/2)=2λL = 4(\lambda/2) = 2\lambda. λ=L/2=0.6\lambda = L/2 = 0.6 m.

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Summary

  • Standing wave: two opposite waves → nodes (zero) + antinodes (max)
  • String/open pipe: fn=nv/(2L)f_n = nv/(2L), all harmonics
  • Closed pipe: fn=nv/(4L)f_n = nv/(4L), odd harmonics only
  • Resonance: driving freq = natural freq → max amplitude

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