Survey
* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
Solution to Problem 11, p. 202, Boyce-DpPrima, 8th Ed. 11. A spring is stretched 10 cm by a force of 3 newtons. A mass of 2 kg is hung from the spring and is also attached to a viscous damper that exerts a froce of 3 newtons when the velocity of mass is 5 m/sec. If the mass is pulled down 4 cm below its equilibrium position and given an initial downward velocity of 10 cm/sec, determine its position u at any time t. Find the quasi-frequency µ and the ratio of µ to the natural frequency of the corresponding undamped motion. Solution: The equation of motion is mü + γ u̇ + ku = 0. (1) The characteristic equation is mr2 + γr + k = 0 with roots √ −γ ± γ 2 − 4km r= . 2m It will turn out that the quantity γ 2 − 4km is negative, so the roots are complex of the form √ γ 4km − γ 2 r=− ±i . 2m 2m The quasi-frequency is √ µ= 4km − γ 2 . 2m The natural frequency of the undamped motion is s ω0 = k . m The general solution is then, γ u(t) = e− 2m t (c1 cos(µt) + c2 sin(µt)) . (2) We only need to find m, γ, k, and determine c1 , c2 from the initial conditions. Also, we express all terms in units of meters and kilograms. Determinations of k, m, γ: The spring is stretched 10 cm = 0.1 m by a force of 3 newtons. Hooke’s law gives k × (distance stretched) = f orce. So, we get k ×0.1 = 3. This gives k = 30. The mass is 2 kg, so m = 2. The magnitude of the force due to damping is γ| u̇ | = 3, so γ × 5 = 3 or γ = 53 . This gives √ √ 4km − γ 2 5991 3 µ= = = 3.87008, γ= = 0.15 2m 20 20 The initial conditions give: u(0) = 5cm = 1 m, 20 u̇(0) = 10cm/sec = 0.1m/sec. Using (2) we get 1 0.05 = u(0) = c1 , 0.1 = u̇(0) = −0.15c1 + c2 µ. So, c2 = 0.1 + (0.15)(0.05) = 0.0277772. 3.87008 We can also write this as u(t) = e−0.15t Rcos(µt − δ) q where R = c21 + c22 ∼ 0.0571977, and γ = arctan(c2 /c1 ) ∼ 0.50701. Quasi-frequency: µ ∼ 3.87008, q √ k Undamped frequency: ω 0 = m = 15. √ ∼ 0.99925. Ratio: √µ15 ∼ 3.87008 15 2