#fixed string

Doppler Effect

A string fixed at both ends has a length of 1.5 m and a fundamental frequency of 50 Hz. What is the speed of the wave on

**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. Fundamental frequency: v₁ = (v/2L) . 50 = (v/2 × 1.5) ⇒ 50 = (v/3) ⇒ v = 150 m/s . Using v = fλ and standing-wave condition fₙ = n v/(2L) or v/(4L) as applicable, calculation yields 150 m/s, illustrating frequency-length-speed interdependence and quantization by boundaries.

Ref: NCERT > Physics Book > Waves > Doppler Effect

View Details

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

View Details

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

View Details

Stationary Waves and Standing Waves in Strings

A string fixed at both ends has a length of 2 m and a wave speed of 90 m/s. What is the frequency of its third harmonic?

**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) . Third harmonic ( n = 3 ): v₃ = (3 × 90/2 × 2) = (270/4) = 67.5 Hz . Using v = fλ and standing-wave condition fₙ = n v/(2L) or v/(4L) as applicable, calculation yields 67.5 Hz, illustrating frequency-length-speed interdependence and quantization by boundaries.

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

View Details

Stationary Waves and Standing Waves in Strings

In a standing wave formed on a string fixed at both ends, what is the condition for the position of nodes?

**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. In a standing wave, nodes occur where the displacement is zero, which happens when sin(kx) = 0 . This implies kx = nπ , where k = (2π/λ) , so x = (nλ/2) (n = 0, 1, 2, ..). Using v = fλ and standing-wave condition fₙ = n v/(2L) or v/(4L) as applicable, calculation yields Displacement is zero, illustrating frequency-length-speed interdependence and quantization by boundaries.

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

View Details