Question

could you provide a detailed solution for this question. Like and comment are rewarded, thanks

2. Consider the system shown in the figure below. y m1 k,1 Mass mi moves horizontally along the x axis and its position is given by coordinate x1. It is attached to mass m2 by a light spring of spring constant k and natural length 1. The spring is constrained to oscillate in the r-y plane. Let the angle between the spring and the negative y-direction be 0 and the distance between the masses be r. The gravitational field strength is g in the negative vertical direction. You can ignore air resistance and friction. Show that the Lagrangian of the system is тi + m2 (** + r*") + mgin (i sin(@) + rô cos(0)] т2 2 + m2gr cos(0) – (r – 1)?.

Answer 1

. Т), $ 0 0 0 0 0 0 0 0 0 0 krsino ля

- along only x axis so so As my moves along only x axis Pic' į miez M2 moves along x-axis has and pendulum motion Kinetic energy due to motion ad for my a spring vibration along x -axis

Position of ma is x, trsine ^2 - xy + y sing in = x + (ésine tre cose) äŽ = 22 + • 2 sino +182 2 cas29+2 jersine cose + Sż ci sine tre cose) Sla yz = r cose ý, = i cose - rý sing y2 2 = 8² case + 8 282 sinze - arré sine cose Toe 284 m [*+ j}] = {ma [ 832 + 32(šim?© +6030) +72 62 (sin?8+60320) + DXi (isine +re ceses] = į mok? + y mira + 0252) + me xi (isine +rèroses Total kinetic energy of system T= Titta = + m + m2) s t + 1 m2 (8² +8262) + m2 %, (i sinet recase) - RA Potential energy v=-ma greuset K(8-1)2 So Lagrangian L = T-V = į mi + m2) *? + 4 m2 (32+02 62) + m2, (o sinetrecase) + m go cose MALU -IK (-8) 2 59