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A block of mass 2 kg hangs from a spring (spring constant 800 N/m) of negligible mass. The spring extends by 2.5 cm at equilibrium. The upper end of the spring is driven in SHM with amplitude 2 mm. The damping constant b (where the damping force = b * velocity) satisfies b/(2m) = 0.5 s⁻¹. Find the amplitude of forced oscillations (in cm) when the driving frequency equals the natural frequency.

1. (A) 1
2. (B) 2
3. (C) 3
4. (D) 4

Correct answer: (D) 4

Solution

With omega₀ = 20 rad/s, b = 2 kg/s, and F₀ = 800 * 0.002 = 1.6 N, the resonance amplitude is F₀/(b*omega₀) = 1.6/(2*20) = 0.04 m = 4 cm.

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