Standing wave harmonics on a fixed-fixed string Steps:

• Node at both ends means fixed-fixed boundary conditions; wavelengths fit as L = n (λ/2), where n is harmonic number.
• Five nodes imply 4 intervals between nodes, each λ/2 long, so L = 4 (λ/2) = 2λ; thus λ = L/2 and f = v/λ = 2v/L.
• Fundamental f_1 = v/(2L), so current f = 4 f_1 (4th harmonic).
• Vibration generator excites only even harmonics due to driving symmetry; next after 4th is 6th harmonic at 6 f_1 = 1.5 f.

Why A is correct:

• 1.5f matches the 6th harmonic frequency, the next even mode after the current 4th harmonic, per the generator's excitation pattern.

Why the others are wrong:

• B. 1.5f: Identical to A, not a distinct option.
• C. 1.75f: Equals 7 f_1, an odd harmonic not excited by the generator.
• D. 2.00f: Equals 8 f_1, an even harmonic but skips the immediate next (6th).

Final answer: A