Question

L Xinit x = 0

Learning Goal:

To understand the application of the general harmonic equation to the kinematics of a spring oscillator.

One end of a spring with spring constant k is attached to the wall. The other end is attached to a block of mass m. The block rests on a frictionless horizontal surface. The equilibrium position of the left side of the block is defined to be x=0. The length of the relaxed spring is L. (Figure 1)

The block is slowly pulled from its equilibrium position to some position xinit>0 along the x axis. At time t=0 , the block is released with zero initial velocity.

The goal is to determine the position of the block x(t) as a function of time in terms of ω and xinit.

It is known that a general solution for the displacement from equilibrium of a harmonic oscillator is

x(t)=Ccos(ωt)+Ssin(ωt),

where C, S, and ω are constants. (Figure 2)

1. Using the general equation for x(t) given in the problem introduction, express the initial position of the block xinit in terms of C, S, and ω (Greek letter omega).

xinit= ___________________

4. Find the equation for the block's position xnew(t) in the new coordinate system. Express your answer in terms of L, xinit, ω (Greek letter omega), and t.

xnew(t)= _______________

Answer 1

2 A Colt Here force exerted by spring F=-kr and according to Newtow's second law Fama or marka or m dx =-kis lor dx I on ta x=0 at How multiply false) in the equation we will gett 2 de dx + 2 x t =0 Now by integrating this equation, (de) + x² = c. here to an awe and c=oximit For (de) + x or dx two x = (int) at at at t=0 for sint x) = wtto x = Kinnt kinit or xft) = Gut Sin (wt+d hence &= 2 / 2 or rett) - tinit Costwt) or alt = C Cos (wit) at tro xf = 6