Practice question
Question
A string of length 2.8 m fixed at both ends has a wave speed of 84 m/s. What is the frequency of its
fourth harmonic?
Explanation
**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. v_n = (n v/2L) . Fourth harmonic ( n = 4 ): v₄ = (4 × 84/2 × 2.8) = (336/5.6) = 60 Hz . Using v = fλ and standing-wave condition fₙ = n v/(2L) or v/(4L) as applicable, calculation yields 60 Hz, illustrating frequency-length-speed interdependence and quantization by boundaries.
Discussion
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