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#harmonics

13 public questions tagged with this topic.

Sound Waves, Reflection and Characteristics

A pipe closed at one end has a length of 0.25 m. What is the frequency of its third harmonic if the speed of sound is 34

**Reflection at boundaries** follows phase change rules: rigid boundary (fixed end) introduces π phase shift, inverting displacement y → -y, while free boundary reflects without phase change. Reflected wave derived by reversing propagation direction kx → -kx and applying phase shift, preserving k = 2π/λ and ω = 2πf. For closed pipe: v_n = (n + (1/2)) (v/2L) , n = 0, 1, 2, ldots . Third harmonic: n = 2 . v₂ = (2 + (1/2)) (340/2 × 0.25) = 2.5 × (340/0.5) = 1700 Hz . Using v = fλ and standing-wave condition fₙ = n v/(2L) or v/(4L) as applicable, calculation yields

Ref: NCERT > Physics Book > Waves > Sound Waves, Reflection and Characteristics

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Doppler Effect

A pipe 0.5 m long, open at both ends, resonates with a 340 Hz source. What is the harmonic number if the speed of sound

**Frequency shift** proportional to source speed relative to wave speed v. Understanding sign convention for approaching versus receding is key, with approaching increasing frequency and receding decreasing, central to Doppler applications. For open pipe: v_n = (n v/2L) . Given: v = 340 m/s , L = 0.5 m , v_n = 340 Hz . 340 = (n × 340/2 × 0.5) ⇒ 340 = 340n ⇒ n = 1 . First harmonic (fundamental mode). Using v = fλ and standing-wave condition fₙ = n v/(2L) or v/(4L) as applicable, calculation yields 1, illustrating frequency-length-speed interdependence and quantization by boundaries.

Ref: NCERT > Physics Book > Waves > Doppler Effect

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Beats Phenomenon

A string of length 2.4 m fixed at both ends has a wave speed of 72 m/s. What is the frequency of its fifth harmonic?

**Superposition of nearly equal frequencies** creates resultant y = 2A cos(Δω·t/2) sin(ω_avg·t), envelope frequency Δf/2, beat frequency Δf. Total beats heard in Δt is f_beat·Δt, explaining counting over seconds. v_n = (n v/2L) . Fifth harmonic ( n = 5 ): v₅ = (5 × 72/2 × 2.4) = (360/4.8) = 75 Hz . Using v = fλ and standing-wave condition fₙ = n v/(2L) or v/(4L) as applicable, calculation yields 75 Hz, illustrating frequency-length-speed interdependence and quantization by boundaries.

Ref: NCERT > Physics Book > Waves > Beats Phenomenon

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Superposition and Interference of Waves

A string of length 1.2 m fixed at both ends has a wave speed of 48 m/s. What is the frequency difference between its fou

**Interference of waves** produces enhancement or cancellation based on phase. Two equal amplitude waves out of phase by π cancel completely, A = 0, while in-phase superposition doubles amplitude to 2a, demonstrating energy redistribution without violation of conservation. v_n = (n v/2L) . Second harmonic ( n = 2 ): v₂ = (2 × 48/2 × 1.2) = (96/2.4) = 40 Hz . Fourth harmonic ( n = 4 ): v₄ = (4 × 48/2 × 1.2) = (192/2.4) = 80 Hz . Difference: v₄ - v₂ = 80 - 40 = 40 Hz . Using v = fλ and standing-wave condition fₙ = n

Ref: NCERT > Physics Book > Waves > Superposition and Interference of Waves

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Superposition and Interference of Waves

A string of length 0.9 m fixed at both ends has a wave speed of 45 m/s. What is the frequency difference between its sec

**Wave addition** governed by phase difference determines resultant intensity ∝ A². Phase arises from path difference Δ = (2π/λ)·Δx, and resultant formula captures interference condition quantitatively for NCERT problems. v_n = (n v/2L) . First harmonic ( n = 1 ): v₁ = (45/2 × 0.9) = 25 Hz . Second harmonic ( n = 2 ): v₂ = (2 × 45/2 × 0.9) = 50 Hz . Difference: v₂ - v₁ = 50 - 25 = 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 > Superposition and Interference of Waves

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Vibrations of Air Columns - Open and Closed Pipes

A pipe open at both ends has a length of 0.51 m and a speed of sound of 340 m/s. What is the frequency of its third harm

**Resonance in pipes** occurs when length accommodates standing wave pattern. Closed pipe L = (2n-1)λ/4, so f₁ = v/(4L). Given f₁ and v, length follows L = v/(4f₁), enabling length calculation from measured resonance frequency and sound speed 330-340 m/s. For open pipe: v_n = (n v/2L) . Third harmonic ( n = 3 ): v₃ = (3 × 340/2 × 0.51) = (1020/1.02) = 1000 Hz . Using v = fλ and standing-wave condition fₙ = n v/(2L) or v/(4L) as applicable, calculation yields 1000 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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Vibrations of Air Columns - Open and Closed Pipes

In a pipe closed at one end, which harmonics are absent compared to a pipe open at both ends?

**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. A pipe closed at one end has frequencies v_n = (2n - 1) (v/4L) , producing only odd harmonics (1, 3, 5, ..). A pipe open at both ends has all harmonics (1, 2, 3, ..), so even harmonics are absent in the closed pipe. Using v = fλ and standing-wave condition fₙ = n v/(2L) or v/(4L) as

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

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Vibrations of Air Columns - Open and Closed Pipes

A pipe closed at one end has a length of 0.8 m and resonates at its third harmonic with a speed of sound of 360 m/s. Wha

**Organ pipe modes** illustrate boundary influence. Closed pipe odd harmonic series contrasts with open pipe full series, affecting timbre. Frequency scales inversely with length, explaining pitch variation with pipe length. For closed pipe: v_n = (n + (1/2)) (v/2L) , n = 2 for third harmonic. v₂ = (2 + (1/2)) (360/2 × 0.8) = 2.5 × (360/1.6) = 2.5 × 225 = 562.5 Hz . Using v = fλ and standing-wave condition fₙ = n v/(2L) or v/(4L) as applicable, calculation yields 562.5 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 pipe open at both ends has a length of 0.6 m and a wave speed of 300 m/s. What is the difference in frequency between

**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. v_n = (n v/2L) . First harmonic ( n = 1 ): v₁ = (300/2 × 0.6) = 250 Hz . Third harmonic ( n = 3 ): v₃ = (3 × 300/2 × 0.6) = 750 Hz . Difference: v₃ - v₁ = 750 - 250 = 500 Hz . Using v = fλ and standing-wave condition fₙ = n v/(2L) or

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 2.5 m and a wave speed of 100 m/s. What is the frequency of its fourth harmo

**Standing wave in fixed string** has nodes at ends, quantizing modes. Fundamental n=1 has λ₁ = 2L, higher harmonics multiples of fundamental f₁. Third harmonic n=3 has three half-wavelengths in length L, f₃ = 3v/(2L), illustrating standing wave condition and boundary enforcement. For fixed ends: v_n = (n v/2L) . Fourth harmonic ( n = 4 ): v₄ = (4 × 100/2 × 2.5) = (400/5) = 80 Hz . Using v = fλ and standing-wave condition fₙ = n v/(2L) or v/(4L) as applicable, calculation yields 80 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

What is the relationship between the wavelengths of the fundamental mode and the second harmonic in a string fixed at bo

**Standing wave in fixed string** has nodes at ends, quantizing modes. Fundamental n=1 has λ₁ = 2L, higher harmonics multiples of fundamental f₁. Third harmonic n=3 has three half-wavelengths in length L, f₃ = 3v/(2L), illustrating standing wave condition and boundary enforcement. For a string fixed at both ends, fundamental wavelength is λ₁ = 2L , and second harmonic is λ₂ = L . Thus, λ₁ = 2 λ₂ , or λ₂ = λ₁ / 2 . Using v = fλ and standing-wave condition fₙ = n v/(2L) or v/(4L) as applicable, calculation yields Second harmonic is half the fundamental, 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 of length 2.4 m fixed at both ends has a wave speed of 96 m/s. What is the frequency difference between its thi

**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. v_n = (n v/2L) . First harmonic ( n = 1 ): v₁ = (96/2 × 2.4) = (96/4.8) = 20 Hz . Third harmonic ( n = 3 ): v₃ = (3 × 96/2 × 2.4) = (288/4.8) = 60 Hz . Difference: v₃ - v₁ = 60 - 20 = 40 Hz . Using v = fλ and standing-wave condition fₙ = n v/(2L) or v/(4L) as applicable, calculation yields 40

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

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