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Wave Behaviour - Standing Waves and Resonance

Grade 12IBPhysics

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

🔑Concepts

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A standing (stationary) wave is formed by the superposition of two progressive waves of the same frequency and amplitude traveling in opposite directions.

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Nodes are positions along a standing wave where the displacement is always zero due to destructive interference (A=0A = 0).

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Antinodes are positions where the amplitude of the oscillation is at its maximum due to constructive interference (A=AmaxA = A_{max}).

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In a standing wave, energy is stored and localized within the system; there is no net transfer of energy through the medium.

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Boundary Conditions: For a string of length LL fixed at both ends, or a pipe open at both ends, the allowed wavelengths are λn=2Ln\lambda_n = \frac{2L}{n} where n=1,2,3...n = 1, 2, 3....

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Boundary Conditions: For a pipe closed at one end and open at the other, only odd harmonics exist, and λn=4L2n−1\lambda_n = \frac{4L}{2n-1} where n=1,2,3...n = 1, 2, 3....

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Resonance occurs when the frequency of an external driving force matches the natural frequency of a system, resulting in oscillations with maximum amplitude.

📐Formulae

v=fλv = f \lambda

fn=nv2L(Fixed strings or Open pipes)f_n = n \frac{v}{2L} \quad (\text{Fixed strings or Open pipes})

fn=(2n−1)v4L(Closed pipes)f_n = (2n-1) \frac{v}{4L} \quad (\text{Closed pipes})

v=Tμ(Wave speed on a string)v = \sqrt{\frac{T}{\mu}} \quad (\text{Wave speed on a string})

💡Examples

Problem 1:

A guitar string of length L=0.65 mL = 0.65\text{ m} is tuned to a fundamental frequency of 440 Hz440\text{ Hz}. Calculate the speed vv of the wave on the string and the frequency of the third harmonic f3f_3.

Solution:

  1. For the fundamental frequency (n=1n=1), f1=v2Lf_1 = \frac{v}{2L}. Rearranging for vv: v=f1⋅2L=440 Hz⋅2⋅0.65 m=572 m s−1v = f_1 \cdot 2L = 440\text{ Hz} \cdot 2 \cdot 0.65\text{ m} = 572\text{ m s}^{-1}.
  2. For the third harmonic: f3=3⋅f1=3⋅440 Hz=1320 Hzf_3 = 3 \cdot f_1 = 3 \cdot 440\text{ Hz} = 1320\text{ Hz}.

Explanation:

In a string fixed at both ends, the nn-th harmonic is an integer multiple of the fundamental frequency (fn=nf1f_n = nf_1). The fundamental occurs when the string length equals half a wavelength (L=λ2L = \frac{\lambda}{2}).

Problem 2:

An organ pipe is closed at one end and has a length of 0.85 m0.85\text{ m}. If the speed of sound is v=340 m s−1v = 340\text{ m s}^{-1}, find the frequency of the first overtone.

Solution:

  1. For a closed pipe, the first overtone corresponds to the next available harmonic, which is the 3rd harmonic (n=2n=2 in the sequence 2n−12n-1).
  2. Formula: f=3v4Lf = \frac{3v}{4L}.
  3. Calculation: f=3⋅3404⋅0.85=10203.4=300 Hzf = \frac{3 \cdot 340}{4 \cdot 0.85} = \frac{1020}{3.4} = 300\text{ Hz}.

Explanation:

In a pipe closed at one end, only odd harmonics are possible (1st,3rd,5th...1^{st}, 3^{rd}, 5^{th}...). The first overtone is the 3rd3^{rd} harmonic. The boundary conditions require a node at the closed end and an antinode at the open end.

Standing Waves and Resonance - Revision Notes & Key Formulas | IB Grade 12 Physics