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#beats

19 public questions tagged with this topic.

Doppler Effect

Two waves of frequencies 510 Hz and 514 Hz interfere. How many beats are heard in 15 seconds?

**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. Beat frequency: vbₑₐt = 514 - 510 = 4 Hz . Beats in 15 s: 4 × 15 = 60 . Using v = fλ and standing-wave condition fₙ = n v/(2L) or v/(4L) as applicable, calculation yields 60, 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 > Doppler Effect

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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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Beats Phenomenon

Two waves of frequencies 480 Hz and 486 Hz interfere. How many beats are heard in 12 seconds?

**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. Beat frequency: vbₑₐt = 486 - 480 = 6 Hz . Beats in 12 s: 6 × 12 = 72 . Using v = fλ and standing-wave condition fₙ = n v/(2L) or v/(4L) as applicable, calculation yields 72, 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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Beats Phenomenon

Two waves of frequencies 475 Hz and 480 Hz interfere. How many beats are heard in 20 seconds?

**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. Beat frequency: vbₑₐt = 480 - 475 = 5 Hz . Beats in 20 s: 5 × 20 = 100 . Using v = fλ and standing-wave condition fₙ = n v/(2L) or v/(4L) as applicable, calculation yields 100, illustrating frequency-length-speed interdependence and quantization by boundaries.

Ref: NCERT > Physics Book > Waves > Beats Phenomenon

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Beats Phenomenon

Two waves of frequencies 500 Hz and 504 Hz interfere to produce beats. How many beats are heard in 10 seconds?

**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. Beat frequency: vbₑₐt = v₁ - v₂ = 504 - 500 = 4 Hz . Beats in 10 s: 4 × 10 = 40 . Using v = fλ and standing-wave condition fₙ = n v/(2L) or v/(4L) as applicable, calculation yields 40, illustrating frequency-length-speed interdependence and quantization by boundaries.

Ref: NCERT > Physics Book > Waves > Beats Phenomenon

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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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Beats Phenomenon

Two waves of frequencies 465 Hz and 470 Hz interfere. How many beats are heard in 12 seconds?

**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. Beat frequency: vbₑₐt = 470 - 465 = 5 Hz . Beats in 12 s: 5 × 12 = 60 . Using v = fλ and standing-wave condition fₙ = n v/(2L) or v/(4L) as applicable, calculation yields 60, 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

Which condition is necessary for the formation of beats between two waves?

**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. Beats require two waves with slightly different frequencies to produce periodic interference, resulting in amplitude modulation. Equal amplitudes or specific wavelengths are not necessary. Using v = fλ and standing-wave condition fₙ = n v/(2L) or v/(4L) as applicable, calculation yields Slightly different frequencies, illustrating frequency-length-speed interdependence and quantization by boundaries.

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

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Superposition and Interference of Waves

What is the physical basis for the formation of beats in sound waves?

**Superposition principle** states resultant displacement equals algebraic sum of individual waves, y = y₁ + y₂. For coherent waves with phase difference φ, resultant amplitude A = √(a₁² + a₂² + 2a₁a₂ cosφ), equal amplitudes give A = 2a cos(φ/2), constructive when φ = 2nπ, destructive when φ = (2n+1)π. Beats arise from the superposition of two waves with slightly different frequencies, causing periodic constructive and destructive interference, perceived as amplitude variation. Using v = fλ and standing-wave condition fₙ = n v/(2L) or v/(4L) as applicable, calculation yields Superposition, illustrating frequency-length-speed interdependence and quantization by boundaries.

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

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Superposition and Interference of Waves

Which phenomenon explains the periodic waxing and waning of sound intensity when two sound waves of slightly different f

**Superposition principle** states resultant displacement equals algebraic sum of individual waves, y = y₁ + y₂. For coherent waves with phase difference φ, resultant amplitude A = √(a₁² + a₂² + 2a₁a₂ cosφ), equal amplitudes give A = 2a cos(φ/2), constructive when φ = 2nπ, destructive when φ = (2n+1)π. Beats occur when two waves of slightly different frequencies superpose, causing constructive and destructive interference periodically, resulting in alternating loudness (waxing and waning). Using v = fλ and standing-wave condition fₙ = n v/(2L) or v/(4L) as applicable, calculation yields Beats, 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 4 Hz. If one has a frequency of 256 Hz and the tension in the other is increased, the beat

**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 original frequency of second string = v₂ . |256 - v₂| = 4 ⇒ v₂ = 252 Hz or 260 Hz . Tension increase raises frequency. If v₂ = 252 , new v₂ > 252 , beat = 256 - v₂’ < 4 , contradicts. If v₂ = 260 , new v₂ > 260 , beat = v₂’ - 256 = 6 ⇒ v₂’ = 262 Hz , consistent. So, v₂ = 260 Hz .

Ref: NCERT > Physics Book > Waves > Beats Phenomenon

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Beats Phenomenon

Two sitar strings produce beats of 5 Hz. If one has a frequency of 320 Hz and tension in the other is reduced, the beat

**Beat formation** is interference in time with time-varying amplitude. Frequencies close together generate slow modulation, count in given duration obtained by multiplying beat frequency by duration, e.g., 6 Hz × 5 s = 30 beats. Let v₂ be the original frequency. |320 - v₂| = 5 ⇒ v₂ = 315 Hz or 325 Hz . Reducing tension decreases frequency. If v₂ = 325 , new v₂’ < 325 , beat = 325 - 320 = 5 or increases, so v₂’ = 313 , beat = 320 - 313 = 7 , consistent. Thus, v₂ = 325 Hz . Using v = fλ and standing-wave condition

Ref: NCERT > Physics Book > Waves > Beats Phenomenon

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