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 a medium 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 (10 points).

2. Determine the transient and permanent parts of the solution (5 points).

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. (5
points).

Answer 1

M = 10 kg ka mg 444 = 5.cm Дх ECH) 2 cos (1) N Fv at IN = vo 6 cm/sec Vlt=0) 2 cm/s ka tox9.8 oros = 1960 W/m By Newton's I law : Fext 2 cos + € -1960n da 0.06 dt d² dt² b + m da dt tka 2 cos 2. cos (9) 10 st ma b = 16.67 į k = 1960 2 da +1-667 dx dt + 1960 x = 0.2cos th otz alt=0) = 0 wit=0 = 0.02

Bunk Pages O first let us find complementary solution axe at² F 5 da 이야 F1g6t 3. mt mt 196 = 0 m = Ii N703) 6 0 60 X,6) = A sin 17031 t + B cos 703) O + 3 at Now let us find particular solution : dup 5 dep +796-Xpe şco (4) dt² Let Up (t) a sin ( + C2 cos ( 1 ) substitute 24p(t) in in ☺ & com efficients of sin (tn) & cos(ty) comparing Xp (t) 90288 88454 425 88454 425 1921 unin sin (#) + coset 2:171 xcoʻsimo) + 10021 x16 Smy

Bunk Pages -5446 А е o alt) A sin W7031 t +B cos +B conto37 + 2.1718106 sin -3 +10021 x 10 cos -I initial Using conditions alto] =0 Vits) 0.02 → B ť 1:0 21 X 16 -3 7 B 3 -1.02 Xlo v (t =o = 0.02 SB 03) + V7031 o ܝ -5 +5.128 x 10 A 6 -3 A = 1.37 x 10 -548 3 x (t) 1:37x10sin (13.975+ + 1.021816-3 cos 13.975 t) 45 t) ] 7.2.171. x16 sint + 1.021, X103 cos е и

④ transient part -578- -3 1:37x10 sin (T3. 9251 e + • 02/XD COM3,975t Permanent part 5_2-12] K10 'S ( + 1.0 21X153 se Resonance frequency .. given by वह वेज nिakar) + * m से 1960 %C ala - 196 - 14 0 50 , - 10 13.95