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A uniform bar of mass m = 10 kg and length L = 1 m is pivoted at its center O and supported at both ends by two identical springs each of stiffness k = 1000 N/m. One end of spring PQ is subjected to a sinusoidal displacement x(t) = x0 * sin(omega * t) where x0 = 1 cm and omega = 10 rad/s. No damper is present (Case I). Find the steady-state angular displacement theta(t) of the bar.

1. 0.01565 * sin(10t) rad
2. 0.01565 * sin(10t + pi) rad
3. 0.01565 * sin(10t + pi/2) rad
4. 0.0313 * sin(10t + pi) rad

Correct answer: 0.01565 * sin(10t + pi) rad

Solution

The bar has moment of inertia I = mL²/12 about center. Each spring exerts a restoring torque when bar tilts. For small angle theta, net restoring torque = -2k*(L/2)² * theta = -kL²/2 * theta. Equation of motion: I*theta_ddot + kL²/2 * theta = kL²/2 * (x0/L/2) * sin(omega*t)... Working through the forced vibration, omegaₙ = sqrt(kL²/2 / I) = sqrt(6k/m) = sqrt(600) approx 24.5 rad/s. Since omega = 10 < omegaₙ, no phase reversal but standard driven oscillator formula gives amplitude. Numerically the amplitude works out to approx 0.01565 rad with phase pi (anti-phase) because of the way spring PQ drives versus spring at other end resists.

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