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Waves - Reflection of Waves and Standing Waves

Grade 11CBSEPhysics

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

🔑Concepts

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Reflection at a Fixed Boundary: When a wave is reflected from a rigid boundary (denser medium), it undergoes a phase change of π\pi radians (180∘180^\circ). The reflected wave is inverted.

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Reflection at a Free Boundary: When a wave is reflected from a non-rigid boundary (rarer medium), there is no phase change (phase change is 00 radians). The reflected wave is not inverted.

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Standing Waves (Stationary Waves): Formed by the superposition of two identical waves traveling in opposite directions with the same frequency and amplitude. Unlike progressive waves, they do not transport energy.

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Nodes and Antinodes: Nodes are positions of zero displacement (amplitude is zero), while Antinodes are positions of maximum displacement (amplitude is 2A2A).

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Standing Waves in Strings: For a string of length LL fixed at both ends, the allowed frequencies (harmonics) are integral multiples of the fundamental frequency: fn=nf1f_n = n f_1, where n=1,2,3...n = 1, 2, 3....

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Standing Waves in Air Columns: In an open pipe, both even and odd harmonics are present. In a pipe closed at one end, only odd harmonics are present (1st,3rd,5th...1^{st}, 3^{rd}, 5^{th}...).

📐Formulae

yr(x,t)=−Asin⁡(kx+ωt) (Reflection at fixed boundary)y_r(x, t) = -A \sin(kx + \omega t) \text{ (Reflection at fixed boundary)}

yr(x,t)=Asin⁡(kx+ωt) (Reflection at free boundary)y_r(x, t) = A \sin(kx + \omega t) \text{ (Reflection at free boundary)}

y(x,t)=[2Asin⁡(kx)]cos⁡(ωt) (Standing wave equation)y(x, t) = [2A \sin(kx)] \cos(\omega t) \text{ (Standing wave equation)}

fn=n2LTμ (Frequencies in a stretched string)f_n = \frac{n}{2L} \sqrt{\frac{T}{\mu}} \text{ (Frequencies in a stretched string)}

fn=nv2L (Frequencies in an open pipe, n=1,2,3...)f_n = \frac{nv}{2L} \text{ (Frequencies in an open pipe, } n=1, 2, 3...)

fn=(2n−1)v4L (Frequencies in a closed pipe, n=1,2,3...)f_n = \frac{(2n-1)v}{4L} \text{ (Frequencies in a closed pipe, } n=1, 2, 3...)

💡Examples

Problem 1:

A steel wire 0.72 m0.72 \text{ m} long has a mass of 5.0×10−3 kg5.0 \times 10^{-3} \text{ kg}. If the wire is under a tension of 60 N60 \text{ N}, what is the speed of transverse waves on the wire and its fundamental frequency?

Solution:

Mass per unit length μ=ML=5.0×10−30.72≈6.94×10−3 kg/m\mu = \frac{M}{L} = \frac{5.0 \times 10^{-3}}{0.72} \approx 6.94 \times 10^{-3} \text{ kg/m}. Speed v=Tμ=606.94×10−3≈93 m/sv = \sqrt{\frac{T}{\mu}} = \sqrt{\frac{60}{6.94 \times 10^{-3}}} \approx 93 \text{ m/s}. Fundamental frequency f1=v2L=932×0.72≈64.6 Hzf_1 = \frac{v}{2L} = \frac{93}{2 \times 0.72} \approx 64.6 \text{ Hz}.

Explanation:

The speed of the wave depends on the tension TT and linear mass density μ\mu. The fundamental frequency for a string fixed at both ends occurs when the length LL equals λ2\frac{\lambda}{2}.

Problem 2:

A pipe 20 cm20 \text{ cm} long is closed at one end. Which harmonic mode of the pipe is resonantly excited by a 430 Hz430 \text{ Hz} source? (Speed of sound v=340 m/sv = 340 \text{ m/s})

Solution:

Length L=0.2 mL = 0.2 \text{ m}. Fundamental frequency for closed pipe f1=v4L=3404×0.2=425 Hzf_1 = \frac{v}{4L} = \frac{340}{4 \times 0.2} = 425 \text{ Hz}. Since 430 Hz430 \text{ Hz} is approximately equal to 425 Hz425 \text{ Hz}, it is the first harmonic (fundamental mode).

Explanation:

In a closed pipe, resonance occurs at frequencies fn=(2n−1)f1f_n = (2n-1)f_1. Here, the source frequency matches the calculated fundamental frequency (n=1n=1).

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