Practice question
Question
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 second and first harmonics?
Explanation
**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.
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