Question

A 10 kg object is suspended from a spring causes the spring to lengthen by 5 cm down. The object is subjected to a force of 2cos (t / 4) N (newtons), and it moves in an environment where the viscosity force is 1N when the speed is 6 cm / s. If the object is set in motion from its position of equilibrium with an initial speed of 2 cm / s, formulate the initial value problem that describes the movement of the object.

1. Find the solution to this problem.

2. Determine the transient and permanent parts of the solution.

3. If the external force is replaced by a force of 4cosωt of frequency ω, find the value of ω for which the amplitude of the forced response is maximum.

Answer 1

forced damped vibrations mu" + yutku = f(+) General Case m= 10 kg k=? f (t) = 2 cos (4A) lets find spring constant 'k' ! 10 kg mass stretches it by 5cm KA2= mg K= mg 10x9.8 0.05 1960 N/m Δα Now damping coefficent ? fdamps Yx velocity 6cm/sec= 0.06 m/s r * 0.06 Y=%0.06 16.7 now w Equation to solve 10 u + 1960 u + 16.7 u = 2 cos (44) u(o)=0, u(O) = 0.02 m/s solution .( + Up (+) Wilay boshly + Asus lady where + Adidati + Bolsinjuts External Up (t) Ap cos(w t) + Bp Sin (wt) "force (kA - mw²A) casut + (kBe- mw²B) sin wt = 2 cos (k-mw?) Ą= 2 (1960 - 10x1) 4= 2 4= 1.02 x 10 WE frea of 14 2 cost/4) 3 Bao 0

Page : so Up (4) 1.02 cos(lt) 4 10-3 UdGADE for ų (A) we need to no External force is there solve the equation as if lou" of 16.7 u t 1960 u = 0 oots - 16.7 I v 16.72 - 4x1960 x 10 20 - 16.7 20 + 279.5 20 -0.835 + 14 i Geo -0.835t u(t) = Cos 14t -0.835€ + Ce sin 14t so solution U (A) = Up (H) + 4 (t) T Permanent "Transient UH) 1.02.*10-3 cos -0.885 as (Vyt) + Ge Cos 14t + Ge-o. 0-0.8356 sin cut u (0) 0 3 1.02810 + G O Ga 1.02 x 103

CIASSMAte Dato: Page : = 2 cm/s or 0.02 d ucts or ult) for to at 0.02 = 0.835G - 1462 G= - 1.02 X 10-3 C2 = 1.49810-3 Ult) Solutions: - 1.02x10-3 -0.835t e Cos 14t 1.49810 e 3-0.8856 sinlat Transient part 42 1.02x10-3 + cas (%t) Permanent part at w which "due to Resonance Amplitude is maximum tohen w is equal natural frequency) wo Wo= W-14 1196 K m 14 rad/sec