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#first harmonic

3 public questions tagged with this topic.

Vibrations of Air Columns - Open and Closed Pipes

A pipe open at both ends has a length of 0.85 m and a speed of sound of 340 m/s. What is the frequency of its first harm

**Air column vibrations** depend on end conditions. Pipe closed at one end has displacement node at closed end and antinode at open, allowing only odd harmonics, fundamental f₁ = v/(4L). Open pipe has antinodes at both ends, fₙ = n·v/(2L), all harmonics present, v sound speed. For open pipe: v_n = (n v/2L) . First harmonic ( n = 1 ): v₁ = (340/2 × 0.85) = (340/1.7) = 200 Hz . Using v = fλ and standing-wave condition fₙ = n v/(2L) or v/(4L) as applicable, calculation yields 200 Hz, illustrating frequency-length-speed interdependence and quantization by boundaries.

Ref: NCERT > Physics Book > Waves > Vibrations of Air Columns - Open and Closed Pipes

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Stationary Waves and Standing Waves in Strings

A string fixed at both ends has a length of 1.6 m and a wave speed of 80 m/s. What is the frequency of its first harmoni

**Stationary waves** form when identical progressive waves traveling opposite directions interfere, y = 2A sin(kx) cos(ωt), nodes where sin(kx)=0, antinodes where |sin(kx)|=1. For string fixed at both ends, allowed wavelengths λₙ = 2L/n, frequencies fₙ = n·v/(2L), n = 1,2,3… harmonic number. For fixed ends: v_n = (n v/2L) . First harmonic ( n = 1 ): v₁ = (80/2 × 1.6) = (80/3.2) = 25 Hz . Using v = fλ and standing-wave condition fₙ = n v/(2L) or v/(4L) as applicable, calculation yields 25 Hz, illustrating frequency-length-speed interdependence and quantization by boundaries.

Ref: NCERT > Physics Book > Waves > Stationary Waves and Standing Waves in Strings

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Stationary Waves and Standing Waves in Strings

A string fixed at both ends has a length of 2 m and a wave speed of 60 m/s. What is the frequency of its first harmonic?

**Quantization due to boundaries** leads to discrete harmonic series. Frequency difference between harmonics is f₁, so f₃ - f₁ = 2f₁ = v/L. Understanding node-antinode pattern explains resonance and overtones in strings. For fixed ends: v_n = (n v/2L) . First harmonic ( n = 1 ): v₁ = (60/2 × 2) = (60/4) = 15 Hz . Using v = fλ and standing-wave condition fₙ = n v/(2L) or v/(4L) as applicable, calculation yields 15 Hz, illustrating frequency-length-speed interdependence and quantization by boundaries.

Ref: NCERT > Physics Book > Waves > Stationary Waves and Standing Waves in Strings

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