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#string tension

26 public questions tagged with this topic.

Doppler Effect

A string of length 2 m and mass 0.01 kg is under a tension of 100 N. What is the time taken by a transverse pulse to tra

**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. Linear mass density: μ = (mass/length) = (0.01/2) = 0.005 kg/m . Speed: v = √((T/μ)) = √((100/0.005)) = √(20000) ≈ 141.4 m/s . Time: t = (length/v) = (2/141.4) ≈ 0.014 s . Using v = fλ and standing-wave condition fₙ = n v/(2L) or v/(4L) as applicable, calculation yields 0.014 s, illustrating frequency-length-speed interdependence and quantization by boundaries.

Ref: NCERT > Physics Book > Waves > Doppler Effect

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Doppler Effect

Two strings produce beats of 4 Hz. One has a frequency of 320 Hz. When the tension in the second string is increased, th

**Doppler effect** describes apparent frequency shift due to relative motion between source and observer, f' = f·v/(v ∓ v_s) for source motion, f' = f·(v ± v_o)/v for observer motion, upper signs for approach increasing observed frequency. Motion towards observer compresses wavelength raising f'. Let v₂ be the original frequency. |320 - v₂| = 4 ⇒ v₂ = 316 Hz or 324 Hz . Increasing tension increases frequency. If v₂ = 316 , new v₂’ > 316 , beat = 320 - v₂’ < 4 , becomes 2 Hz ( v₂’ = 318 ), consistent. If v₂ = 324 , beat increases, contradicts. So, v₂

Ref: NCERT > Physics Book > Waves > Doppler Effect

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Doppler Effect

A string of length 3 m and mass 0.06 kg is under a tension of 150 N. What is the speed of a transverse wave on the strin

**Relative motion** changes effective wavelength encountered. For moving source approaching stationary observer, wavelength ahead λ' = (v - v_s)/f, so f' = v/λ' = f·v/(v - v_s) > f, basis for calculating apparent pitch shift in sound. Linear mass density: μ = (0.06/3) = 0.02 kg/m . Speed: v = √((T/μ)) = √((150/0.02)) = √(7500) ≈ 86.6 m/s . Using v = fλ and standing-wave condition fₙ = n v/(2L) or v/(4L) as applicable, calculation yields 86.6 m/s, illustrating frequency-length-speed interdependence and quantization by boundaries.

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

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 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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Doppler Effect

Which factor explains why a transverse wave’s speed is independent of its frequency in a uniform string?

**Doppler effect** describes apparent frequency shift due to relative motion between source and observer, f' = f·v/(v ∓ v_s) for source motion, f' = f·(v ± v_o)/v for observer motion, upper signs for approach increasing observed frequency. Motion towards observer compresses wavelength raising f'. The speed v = √((T/μ)) depends on tension and mass density, properties of the medium, not frequency, which is source-dependent. Using v = fλ and standing-wave condition fₙ = n v/(2L) or v/(4L) as applicable, calculation yields Medium properties, illustrating frequency-length-speed interdependence and quantization by boundaries.

Ref: NCERT > Physics Book > Waves > Doppler Effect

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Superposition and Interference of Waves

A string of length 2 m and mass 0.04 kg is under a tension of 100 N. What is the speed of a transverse wave on the strin

**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. Linear mass density: μ = (0.04/2) = 0.02 kg/m . Speed: v = √((T/μ)) = √((100/0.02)) = √(5000) ≈ 70.71 m/s . Using v = fλ and standing-wave condition fₙ = n v/(2L) or v/(4L) as applicable, calculation yields 70.7 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 > 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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Wave Speed, Energy and Power

A string of length 1.8 m and mass 0.045 kg has a fundamental frequency of 40 Hz. What is the tension in the string?

**Transverse wave velocity** depends on medium not frequency alone. For string under tension, v ∝ √(T/μ), calculation requires μ from mass and length, then square root evaluation, giving v in m/s, then f = v/λ for given wavelength. μ = (0.045/1.8) = 0.025 kg/m . v₁ = (v/2L) ⇒ 40 = (v/2 × 1.8) ⇒ v = 40 × 3.6 = 144 m/s . v = √((T/μ)) ⇒ 144 = √((T/0.025)) ⇒ 144² = (T/0.025) . T = 20736 × 0.025 = 518.4 N . Using v = fλ and standing-wave condition fₙ = n v/(2L) or v/(4L) as applicable, calculation yields 518 N, illustrating

Ref: NCERT > Physics Book > Waves > Wave Speed, Energy and Power

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Wave Speed, Energy and Power

A string of length 4 m and mass 0.016 kg is under a tension of 64 N. What is the speed of a transverse wave on the strin

**Energy transport** in waves scales with amplitude squared A² and frequency squared ω². Wave speed determines propagation rate, and understanding T and μ allows quantitative prediction of v and associated frequencies. Linear mass density: μ = (0.016/4) = 0.004 kg/m . Speed: v = √((T/μ)) = √((64/0.004)) = √(16000) ≈ 126.5 m/s . Using v = fλ and standing-wave condition fₙ = n v/(2L) or v/(4L) as applicable, calculation yields 126 m/s, illustrating frequency-length-speed interdependence and quantization by boundaries.

Ref: NCERT > Physics Book > Waves > Wave Speed, Energy and Power

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Wave Speed, Energy and Power

A string of length 5 m and mass 0.05 kg is under a tension of 80 N. How long does a transverse pulse take to travel its

**Wave speed on string** is v = √(T/μ), T tension (N), μ = m/L linear mass density (kg/m). Higher T increases restoring force raising speed, heavier μ lowers speed. Frequency follows f = v/λ, linking mechanical properties to wave dynamics and harmonic series. Linear mass density: μ = (0.05/5) = 0.01 kg/m . Speed: v = √((T/μ)) = √((80/0.01)) = √(8000) ≈ 89.44 m/s . Time: t = (length/v) = (5/89.44) ≈ 0.056 s . Using v = fλ and standing-wave condition fₙ = n v/(2L) or v/(4L) as applicable, calculation yields 0.056 s, illustrating frequency-length-speed interdependence and quantization by boundaries.

Ref: NCERT > Physics Book > Waves > Wave Speed, Energy and Power

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