Question

Frictionless plane M 1.) Consider the coupled system shown at the right. The mass M is free to slide on a frictionless surface and is connected to the wall with a spring of spring constant k. Mass M2 is 2000 attached to My with taut rope of length (it acts as a pendulum). The vertical line shows the equilibrium position when the spring is un- stretched (r = 0). The coordinates 21 and 12 denote the positions of the two masses. (a) Draw a free body diagram on each of the two masses below 2 (b) Assuming small oscillations so tan sin = (12 - 11/l, and equal masses M1 = M2 = M, show that applying Newton's 2nd Law to the the system leads to the equations of motion i = Au .6)= () -+679) (c) Suppose k = 15N/m, M = 1.0kg, and l = 0.98 metre. In this situation, one of the eigenvectors of A is i. Qualitatively describe (in words) the motion of the system in this case. ii. When the system is oscillating in the normal mode vi, what is the period of oscillation?

Answer 1

Consider the coupled system shown in the question, Here m, = m2 = m . & X, & Xz are their displacement * ho respectively le U₂ now drawing free body diagrams of rigo, we have frigo E → for m, i Tsino where I is tension in spring. kry m m. in Teoso infringine T Togo - my for Cose m₂ : : Tsino Tsino t a , resolving mg & T along the - coordinates. mgsino my singcoso Troso = mg & Tang, cost T for my , 1 (6) Now from the free body diagram, writing equations a equations of motion for m, & m₂. we have , Tsino - ku = m2, . for me, - Tsino = milz . ② . solving 0 * 4 = - + mg (22-11) , for small oscillation , Tang 2 sino =(22-x) RAM-NARINDORONTOTTOTONYA e ) : Solving © ¢ më, = -mg (2=X) + öz = tp x4 - 4X2 .--CO writing ② & 6 in matrix form , we have. (*) (*%) a 17 * lů = Au , fence proved. CS Scanned with CamScanner

(6) K= 15 N/m M= 1kg e=0.98 m - we have A=/- (6 + 9 10 g 0 g - 9.8 0.98 0.98 -25 LO A |-25 10 10 -40 To find eigen value, we will find (A-XI)=0 (bekominant) 1-257 10 l (2571) (1071) -100 =0 10 -10-11 12+357 +150 = 0 (quadratic equation) Solving a = -357/1225-600 = -35725 - 2 2 n=-5, -30 . for 15-5 7 , AX = XY et [A-AI]x = 0 -20 1015*, 12 + so ele eigenvector -20%, +10X2 = 0 + 12 = 2X, corresponding to get eigenvalue 1 =-57. . Similarily for 1=-30 [ 5 1 10 10 1 7 20 11 12 .. 54, = = 1022 & 2, = -2X₂ cs seemed t 1Prence proved. CamScanner ! WAT OTO KWANI