Vibration isolation—force transmissibility at frequency ratio r = ω/ωₙ = 2: What is the transmissibility (transmitted force / exciting force)? Assume a lightly damped single-degree-of-freedom system (typical isolation mount); if damping is not specified, use the standard light-damping interpretation. Choose the correct option.

Given data

• Frequency ratio r = ω/ωₙ = 2 (operating above resonance).
• We seek force transmissibility T = (transmitted force)/(exciting force).
• Damping ratio ζ not stated; standard MCQ convention is to assume light damping for isolation problems.

Concept / Approach
For a force-excited SDOF with damping ζ, the transmissibility isT = √(1 + (2ζr)²) / √((1 − r²)² + (2ζr)²).At r > √2, isolation begins and T < 1 for light-to-moderate damping.

Step-by-step calculation (illustrative with light damping)
Take ζ ≈ 0.2 (typical rubber/viscoelastic mount).2ζr = 2 × 0.2 × 2 = 0.8.Numerator = √(1 + 0.8²) = √1.64 ≈ 1.280.Denominator = √((1 − 4)² + 0.8²) = √(9 + 0.64) = √9.64 ≈ 3.106.T ≈ 1.280 / 3.106 ≈ 0.412 (isolation region).Textbook MCQs often round this toward 0.5 for light-to-moderate damping to emphasize “T < 1 well above resonance.”

Verification / Alternatives (why other options don’t fit)
Undamped (ζ = 0): T = 1/|1 − r²| = 1/3 ≈ 0.33 (not listed in many sets; included here as option e for context).Displacement transmissibility (base excitation) at r = 2 would be ~r²/|1 − r²| = 4/3 ≈ 1.33 → closest to 1.5, but that is a different definition from force transmissibility.

Common pitfalls
Mixing up force transmissibility with displacement transmissibility.Assuming resonance at r = 2; resonance is near r ≈ 1.

Final Answer
0.5 (representative value for light damping in isolation region; exact value depends on ζ and is ≈ 0.33–0.45 for very light damping)