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Practice question

Question

A string of length 2.2 m fixed at both ends has a wave speed of 66 m/s. What is the frequency of its
fourth harmonic?

Options

Choose one · Correct answer highlighted

Explanation

**Relative motion** changes effective wavelength encountered. For moving source approaching stationary observer, wavelength ahead λ' = (v - v_s)/f, so f' = v/λ' = f·v/(v - v_s) > f, basis for calculating apparent pitch shift in sound. v_n = (n v/2L) . Fourth harmonic ( n = 4 ): v₄ = (4 × 66/2 × 2.2) = (264/4.4) = 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.

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