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Wave Behaviour - Standing Waves and Resonance

Grade 11IBPhysics

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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Standing waves (stationary waves) are formed by the superposition of two waves of the same frequency, wavelength, and amplitude traveling in opposite directions.

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Unlike traveling waves, standing waves do not transfer energy; they store energy in the oscillating medium.

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Nodes are positions of constant zero displacement due to permanent destructive interference, while Antinodes are positions of maximum displacement due to constructive interference.

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The distance between two adjacent nodes or two adjacent antinodes is λ2\frac{\lambda}{2}.

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Resonance occurs when an external driving force is applied to a system at its natural frequency, leading to a maximum transfer of energy and a large increase in amplitude.

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For a string of length LL fixed at both ends or a pipe open at both ends, the wavelengths of the harmonics are given by λn=2Ln\lambda_n = \frac{2L}{n} where n=1,2,3,…n = 1, 2, 3, \dots.

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For a pipe closed at one end and open at the other, only odd harmonics exist (n=1,3,5,…n = 1, 3, 5, \dots) and the wavelengths are λn=4Ln\lambda_n = \frac{4L}{n}.

📐Formulae

v=fλv = f \lambda

fn=nf1=nv2Lf_n = n f_1 = \frac{nv}{2L}

λn=2Ln\lambda_n = \frac{2L}{n}

fn=nv4L(n=1,3,5,… )f_n = \frac{nv}{4L} \quad (n = 1, 3, 5, \dots)

v=Tμv = \sqrt{\frac{T}{\mu}}

💡Examples

Problem 1:

A guitar string of length L=0.64 mL = 0.64 \text{ m} is held under a tension of T=120 NT = 120 \text{ N}. If the mass per unit length is μ=2.0×10−3 kg m−1\mu = 2.0 \times 10^{-3} \text{ kg m}^{-1}, calculate the frequency of the first harmonic (f1f_1).

Solution:

  1. Calculate the wave speed: v=Tμ=1202.0×10−3=60000≈244.95 m s−1v = \sqrt{\frac{T}{\mu}} = \sqrt{\frac{120}{2.0 \times 10^{-3}}} = \sqrt{60000} \approx 244.95 \text{ m s}^{-1}.
  2. For the first harmonic (n=1n=1), λ1=2L=2×0.64=1.28 m\lambda_1 = 2L = 2 \times 0.64 = 1.28 \text{ m}.
  3. Calculate frequency: f1=vλ1=244.951.28≈191.4 Hzf_1 = \frac{v}{\lambda_1} = \frac{244.95}{1.28} \approx 191.4 \text{ Hz}.

Explanation:

The fundamental frequency is determined by the wave speed on the string and the boundary conditions (fixed at both ends), where the wavelength is twice the length of the string.

Problem 2:

An organ pipe is closed at one end and has a length of 0.85 m0.85 \text{ m}. Taking the speed of sound to be 340 m s−1340 \text{ m s}^{-1}, find the frequency of the third harmonic (n=3n=3).

Solution:

  1. For a closed pipe, the formula for harmonics is fn=nv4Lf_n = \frac{nv}{4L}.
  2. For the third harmonic, n=3n = 3.
  3. f3=3×3404×0.85=10203.4=300 Hzf_3 = \frac{3 \times 340}{4 \times 0.85} = \frac{1020}{3.4} = 300 \text{ Hz}.

Explanation:

In a pipe closed at one end, the first harmonic has a node at the closed end and an antinode at the open end, meaning L=14λL = \frac{1}{4}\lambda. The next available resonance (the 3rd harmonic) occurs at L=34λL = \frac{3}{4}\lambda.

Standing Waves and Resonance - Revision Notes & Key Formulas | IB Grade 11 Physics