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Question

A string fixed at both ends has a length of 2 m and a wave speed of 60 m/s. What is the frequency of
its first harmonic?

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Choose one · Correct answer highlighted

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. For fixed ends: v_n = (n v/2L) . First harmonic ( n = 1 ): v₁ = (60/2 × 2) = (60/4) = 15 Hz . Using v = fλ and standing-wave condition fₙ = n v/(2L) or v/(4L) as applicable, calculation yields 15 Hz, illustrating frequency-length-speed interdependence and quantization by boundaries.

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