Beats Phenomenon

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.

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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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Beats Phenomenon

A string of length 2.5 m and mass 0.025 kg has a fundamental frequency of 40 Hz. What is the tension in the string?

**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. μ = (0.025/2.5) = 0.01 kg/m . v₁ = (v/2L) ⇒ 40 = (v/2 × 2.5) ⇒ v = 40 × 5 = 200 m/s . v = √((T/μ)) ⇒ 200 = √((T/0.01)) ⇒ 200² = (T/0.01) . T = 40000 × 0.01 = 400 N . Using v = fλ and standing-wave condition fₙ = n v/(2L) or v/(4L) as applicable, calculation yields 400 N, illustrating frequency-length-speed interdependence and quantization by boundaries.

Ref: NCERT > Physics Book > Waves > Beats Phenomenon

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Beats Phenomenon

A stationary wave on a string fixed at both ends has a wavelength of 0.6 m and a frequency of 100 Hz. What is the wave s

**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. Speed: v = v λ = 100 × 0.6 = 60 m/s . Using v = fλ and standing-wave condition fₙ = n v/(2L) or v/(4L) as applicable, calculation yields 60 m/s, 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

A stationary wave on a string is given by \( y = 0.08 \sin (2\pi x) \cos (100\pi t) \), where \( x \) and \( y \) are in

**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. Form: y = A sin (kx) cos (ω t) , k = 2π rad/m . Wavelength: λ = (2π/k) = (2π/2π) = 1 m . Distance between nodes: (λ/2) = (1/2) = 0.5 m . Using v = fλ and standing-wave condition fₙ = n v/(2L) or v/(4L) as applicable, calculation yields 0.5 m, illustrating frequency-length-speed interdependence and quantization by boundaries.

Ref: NCERT > Physics Book > Waves > Beats Phenomenon

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Beats Phenomenon

A transverse wave on a string has a tension of 225 N and a linear mass density of 0.025 kg/m. What is the wavelength if

**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. Speed: v = √((T/μ)) = √((225/0.025)) = √(9000) ≈ 94.87 m/s . Wavelength: λ = (v/v) = (94.87/30) ≈ 3.16 m . Using v = fλ and standing-wave condition fₙ = n v/(2L) or v/(4L) as applicable, calculation yields 3.16 m, illustrating frequency-length-speed interdependence and quantization by boundaries.

Ref: NCERT > Physics Book > Waves > Beats Phenomenon

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Beats Phenomenon

When two waves interfere constructively, what happens to the resultant intensity if their amplitudes are equal?

**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. For constructive interference with equal amplitudes a , the resultant amplitude is 2a . Intensity is proportional to the square of amplitude, so I ∝ (2a)² = 4a² , quadrupling the individual intensity. Using v = fλ and standing-wave condition fₙ = n v/(2L) or v/(4L) as applicable, calculation yields It quadruples, 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 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

A transverse wave on a string has a speed of 18 m/s and a frequency of 6 Hz. What is its wavelength?

**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. Speed: v = v λ . Wavelength: λ = (v/v) = (18/6) = 3 m . Using v = fλ and standing-wave condition fₙ = n v/(2L) or v/(4L) as applicable, calculation yields 3 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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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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Beats Phenomenon

A stationary wave is given by \( y = 0.05 \sin (\frac{\pi x}{2}) \cos (100\pi t) \), where \( x \) and \( y \) are in me

**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. Form: y = A sin (kx) cos (ω t) , k = (π/2) rad/m . Wavelength: λ = (2π/k) = (2π/(π/2)) = 4 m . Distance between node and antinode: (λ/4) = (4/4) = 1 m . Using v = fλ and standing-wave condition fₙ = n v/(2L) or v/(4L) as applicable, calculation yields 1 m, 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 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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