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#frequency calculation

17 public questions tagged with this topic.

Equations of SHM, Phase and Angular Frequency

A simple pendulum has a frequency of \( 0.25 \, \text{Hz} \) on Earth (\( g = 9.8 \, \text{m/s}^2 \)). What is its lengt

**Angular frequency** ω = 2πf = 2π/T characterizes rapidity, independent of amplitude for SHM. Phase constant φ shifts sine/cosine, allowing any initial condition, e.g., x(0)=0 requires φ=0 for sine form. Displacement, velocity, acceleration share ω but differ in phase by 90° and 180°. Period: T = (1/v) = (1/0.25) = 4 s . T = 2π √((L/g)) ⇒ 4 = 2π √((L/9.8)) . √((L/9.8)) = (4/2π) ≈ 0.637 ⇒ (L/9.8) = (0.637)² ⇒ L ≈ 3.98 m . Applying x = A cos(ωt + φ), v = -ωA sin(ωt + φ), a = -ω²x and E = ½kA² = ½mω²A², result 3.98 m follows, reflecting

Ref: NCERT > Physics Book > Oscillations > Equations of SHM, Phase and Angular Frequency

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Beats Phenomenon

Two strings produce beats of 8 Hz. One has a frequency of 440 Hz. When the tension in the second string is increased, th

**Beats arise** when two waves of slightly different frequencies f₁ and f₂ superpose, producing amplitude modulation at beat frequency f_beat = |f₁ - f₂| (Hz). Intensity waxes and wanes periodically, number of beats in interval Δt equals f_beat·Δt, loudness variation audible when f_beat < 10 Hz. Let v₂ be the original frequency. |440 - v₂| = 8 ⇒ v₂ = 432 Hz or 448 Hz . Increasing tension increases frequency. If v₂ = 432 , new v₂’ > 432 , beat = 440 - v₂’ < 8 , becomes 6 Hz ( v₂’ = 434 ), consistent. If v₂ = 448 , beat increases, contradicts.

Ref: NCERT > Physics Book > Waves > Beats Phenomenon

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Sound Waves, Reflection and Characteristics

A pipe closed at one end has a length of 0.6 m and resonates at its second harmonic with a speed of sound of 360 m/s. Wh

**Sound wave reflection** at rigid wall behaves like string fixed end, displacement inverted. Equation of reflected wave includes sign change and direction reversal, amplitude unchanged but sign may flip, explaining standing wave formation with incident wave. For closed pipe: v_n = (n + (1/2)) (v/2L) , n = 1 for second harmonic. v₁ = (1 + (1/2)) (360/2 × 0.6) = 1.5 × (360/1.2) = 1.5 × 300 = 450 Hz . Using v = fλ and standing-wave condition fₙ = n v/(2L) or v/(4L) as applicable, calculation yields 450 Hz, illustrating frequency-length-speed interdependence and quantization by boundaries.

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

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Beats Phenomenon

A stationary wave on a string fixed at both ends has a frequency of 90 Hz and a wave speed of 36 m/s. What is the length

**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. Fundamental: v₁ = (v/2L) . 90 = (36/2L) ⇒ 2L = (36/90) ⇒ 2L = 0.4 ⇒ L = 0.2 m . Using v = fλ and standing-wave condition fₙ = n v/(2L) or v/(4L) as applicable, calculation yields 0.2 m, illustrating frequency-length-speed interdependence and quantization by boundaries. This aligns with NCERT Class 11 treatment, emphasizing conservation, symmetry and dimensional consistency useful for CBSE, NEET and CUET.

Ref: NCERT > Physics Book > Waves > Beats Phenomenon

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Sound Waves, Reflection and Characteristics

A pipe open at both ends has a length of 0.25 m and a speed of sound of 340 m/s. What is the frequency of its second har

**Sound wave reflection** at rigid wall behaves like string fixed end, displacement inverted. Equation of reflected wave includes sign change and direction reversal, amplitude unchanged but sign may flip, explaining standing wave formation with incident wave. For open pipe: v_n = (n v/2L) . Second harmonic ( n = 2 ): v₂ = (2 × 340/2 × 0.25) = (680/0.5) = 1360 Hz . Using v = fλ and standing-wave condition fₙ = n v/(2L) or v/(4L) as applicable, calculation yields 1360 Hz, illustrating frequency-length-speed interdependence and quantization by boundaries.

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

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Superposition and Interference of Waves

A string of length 1.5 m fixed at both ends has a wave speed of 45 m/s. What is the frequency of its fifth harmonic?

**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) . Fifth harmonic ( n = 5 ): v₅ = (5 × 45/2 × 1.5) = (225/3) = 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 > Superposition and Interference of Waves

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Beats Phenomenon

Two strings produce beats of 7 Hz. One has a frequency of 392 Hz. When the tension in the second string is increased, th

**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. Let v₂ be the original frequency. |392 - v₂| = 7 ⇒ v₂ = 385 Hz or 399 Hz . Increasing tension increases frequency. If v₂ = 385 , new v₂’ > 385 , beat = 392 - v₂’ < 7 , becomes 5 Hz ( v₂’ = 387 ), consistent. If v₂ = 399 , beat increases, contradicts. So, v₂ = 385 Hz . Using v = fλ and standing-wave condition fₙ = n v/(2L)

Ref: NCERT > Physics Book > Waves > Beats Phenomenon

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Vibrations of Air Columns - Open and Closed Pipes

A pipe open at both ends has a length of 0.3 m and a speed of sound of 330 m/s. What is the frequency of its second 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) . Second harmonic ( n = 2 ): v₂ = (2 × 330/2 × 0.3) = (660/0.6) = 1100 Hz . Using v = fλ and standing-wave condition fₙ = n v/(2L) or v/(4L) as applicable, calculation yields 1100 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

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

A pipe closed at one end has a length of 0.28 m and a speed of sound of 336 m/s. What is the frequency of its fourth har

**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 closed pipe: v_n = (n + (1/2)) (v/2L) , n = 3 for fourth harmonic. v₃ = (3 + (1/2)) (336/2 × 0.28) = 3.5 × (336/0.56) = 3.5 × 600 = 2100 Hz . Using v = fλ and standing-wave condition fₙ = n v/(2L) or v/(4L) as applicable, calculation yields 2100 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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Beats Phenomenon

Two strings produce beats of 3 Hz. One has a frequency of 400 Hz. When the tension in the second string is slightly incr

**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. Let v₂ be the original frequency. |400 - v₂| = 3 ⇒ v₂ = 397 Hz or 403 Hz . Increasing tension increases frequency. If v₂ = 397 , new v₂’ > 397 , beat = 400 - v₂’ < 3 , becomes 2 Hz, consistent ( v₂’ = 398 ). If v₂ = 403 , beat increases, contradicts. So, v₂ = 397 Hz . Using v = fλ and standing-wave condition fₙ = n v/(2L)

Ref: NCERT > Physics Book > Waves > Beats Phenomenon

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Stationary Waves and Standing Waves in Strings

A string of length 0.8 m is fixed at both ends. If the speed of the wave is 40 m/s, what is the frequency of the second

**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) . Second harmonic: n = 2 . v₂ = (2 × 40/2 × 0.8) = (80/1.6) = 50 Hz . Using v = fλ and standing-wave condition fₙ = n v/(2L) or v/(4L) as applicable, calculation yields 50 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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