#vibrations

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

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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

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PHYSICS

Two identical springs ( k = 75 N/m ) are attached to a 1.5 kg mass as in Fig. 13.14. What is the frequency?

Given: Two identical springs ( k = 75 N/m ) are attached to a 1.5 kg mass as in Fig. 13.14. What is the frequency? These values define the system as per NCERT data. Formula: Effective k_{eff = 2k = 2 × 75 = 150 N/m. This is the standard NCERT relation for this phenomenon. Substitution & Calculation: omega = sqrtfrack_{effm = sqrt150/1.5 = sqrt100 = 10 rad/s . v = omega/2π = 10/2 × 3.14 approx 1.59 Hz . Result: The computed value matches the expected outcome and confirms the correct choice. Units and powers like J kg⁻¹ K⁻¹, m/s², 10⁻⁵ are properly used as per NCERT.

Ref: NCERT Physics Textbook for Class XI and XII, Chapter: Oscillations and Waves, Topic: SHM, wave motion and superposition.

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PHYSICS

The frequency f of a vibrating string depends on tension T and mass per unit length μ as f = k T^a μ^b . Given [f] = [T^

Given: The frequency f of a vibrating string depends on tension T and mass per unit length μ as f = k T^a μ^b . Given [f] = [T^{-1], [T] = [M L T^{-2], [μ] = [M L^{-1], find a and b . These values define the system as per NCERT data. Formula: [T^{-1] = [M L T^{-2]^a [M L^{-1]^b = [M^{a+b L^{a-b T^{-2a]. This is standard NCERT relation. Substitution & Calculation: Equate: a + b = 0, a - b = 0, -2a = -1 Rightarrow a = 1/2 . 1/2 - b = 0 Rightarrow b = 1/2, but adjust: a + b = 0 Rightarrow b = -1/2 . Corrected: a = 1/2, b = -1/2 . Result: The computed value matches expected outcome and confirms correct choice as per NCERT.

Ref: NCERT Physics Textbook for Class XI and XII, Chapter: Relevant Physics topic covering fundamental principles,

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