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Oscillations and Waves - Reflection of Waves and Standing Waves

Grade 11ICSEPhysics

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

🔑Concepts

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Reflection of waves occurs when a wave hits a boundary. At a rigid/fixed boundary, the wave undergoes a phase change of π\pi radians (180∘180^\circ). At a free/open boundary, the wave reflects without any phase change.

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The Principle of Superposition states that when two or more waves overlap in a medium, the resultant displacement at any point is the vector sum of the individual displacements: y=y1+y2y = y_1 + y_2.

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Standing waves (stationary waves) are formed by the superposition of two identical waves traveling in opposite directions with the same frequency and amplitude. Unlike progressive waves, standing waves do not transport energy.

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Nodes are points in a standing wave where the amplitude is always zero (Anet=0A_{net} = 0). Antinodes are points where the amplitude is maximum (Anet=2AA_{net} = 2A).

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The distance between two consecutive nodes or two consecutive antinodes is λ2\frac{\lambda}{2}. The distance between a node and its adjacent antinode is λ4\frac{\lambda}{4}.

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In a stretched string of length LL fixed at both ends, the possible wavelengths are given by λ=2Ln\lambda = \frac{2L}{n}, where nn is the harmonic number (n=1,2,3,…n = 1, 2, 3, \dots).

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Organ pipes: An open pipe (open at both ends) produces all harmonics (ratios 1:2:3…1:2:3\dots), while a closed pipe (closed at one end) produces only odd harmonics (ratios 1:3:5…1:3:5\dots).

📐Formulae

y=(2Asin⁡kx)cos⁡ωty = (2A \sin kx) \cos \omega t

v=Tμv = \sqrt{\frac{T}{\mu}}

fn=n2LTμf_n = \frac{n}{2L}\sqrt{\frac{T}{\mu}}

fopen=nv2Lf_{open} = \frac{nv}{2L}

fclosed=(2n−1)v4Lf_{closed} = \frac{(2n-1)v}{4L}

k=2πλk = \frac{2\pi}{\lambda}

ω=2πf\omega = 2\pi f

💡Examples

Problem 1:

A string of length 0.5 m0.5 \text{ m} and mass 0.01 kg0.01 \text{ kg} is stretched with a tension of 400 N400 \text{ N}. Calculate the fundamental frequency of the transverse vibration.

Solution:

  1. Calculate mass per unit length: μ=mL=0.01 kg0.5 m=0.02 kg/m\mu = \frac{m}{L} = \frac{0.01 \text{ kg}}{0.5 \text{ m}} = 0.02 \text{ kg/m}.
  2. Calculate wave velocity: v=Tμ=4000.02=20000=1002≈141.42 m/sv = \sqrt{\frac{T}{\mu}} = \sqrt{\frac{400}{0.02}} = \sqrt{20000} = 100\sqrt{2} \approx 141.42 \text{ m/s}.
  3. Calculate fundamental frequency (n=1n=1): f1=v2L=141.422×0.5=141.42 Hzf_1 = \frac{v}{2L} = \frac{141.42}{2 \times 0.5} = 141.42 \text{ Hz}.

Explanation:

The fundamental frequency for a string fixed at both ends corresponds to the first harmonic, where the length L=λ2L = \frac{\lambda}{2}.

Problem 2:

Find the ratio of the frequency of the third harmonic of a closed organ pipe to the frequency of the second harmonic of an open organ pipe of the same length.

Solution:

  1. For a closed pipe, the harmonics are f1,3f1,5f1...f_1, 3f_1, 5f_1.... The 'third harmonic' is f3=3v4Lf_3 = \frac{3v}{4L}.
  2. For an open pipe, the harmonics are f1,2f1,3f1...f_1, 2f_1, 3f_1.... The second harmonic is f2′=2v2L=vLf'_2 = \frac{2v}{2L} = \frac{v}{L}.
  3. Ratio: f3f2′=3v/4Lv/L=34=0.75\frac{f_3}{f'_2} = \frac{3v/4L}{v/L} = \frac{3}{4} = 0.75.

Explanation:

A closed pipe only supports odd harmonics. The third harmonic is the second resonant mode. An open pipe supports all harmonics.

Reflection of Waves and Standing Waves Revision - Class 11 Physics ICSE