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Question

A string of length 0.8 m is fixed at both ends. If the speed of the wave is 40 m/s, what is the
frequency of the second harmonic?

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Explanation

**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) . Second harmonic: n = 2 . v₂ = (2 × 40/2 × 0.8) = (80/1.6) = 50 Hz . Using v = fλ and standing-wave condition fₙ = n v/(2L) or v/(4L) as applicable, calculation yields 50 Hz, illustrating frequency-length-speed interdependence and quantization by boundaries.

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