A string fixed at both ends has length 0.75 m and wave speed 128 m s⁻¹. It vibrates in harmonic n = 3. Determine the wavelength and frequency.

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0.5 m; 256 Hz Both fixed ends are displacement nodes, so harmonic n = 3 fits 3 half-wavelengths into the string: L = nλ/2. Rearranging and substituting gives λ = 2L/n = 2(0.75)/3 = 0.5 m. Wave speed, frequency and wavelength obey v = fλ, so f = v/λ = 128/0.5 = 256 Hz. The equivalent resonance relation fₙ = nv/(2L) produces the same frequency, checking that the stated harmonic has been counted correctly.

String resonance 1

A string fixed at both ends has length 0.75 m and wave speed 128 m s⁻¹. It vibrates in harmonic n = 3. Determine the wavelength and frequency.

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