8791 9561 Ch 14 HW 1 Harmonic Oscillator Kinematics ?) 20112 (> ConstantsPart A

Question

8791 9561 Ch 14 HW 1 Harmonic Oscillator Kinematics ?) 20112 (> ConstantsPart A Learning Goal: To understand the application of the general harmonic equation to the kinematics of a spring oscillator Using the general equation for ar(t) given in the problem introduction, express the initial position of the block t in terms of C, S and u (Greek letler omega) One end of a spring with spring constant k is atached to the wall. The other end is attached to a block of mass m The block rests on a frictionless horizontal surface. The equilibrium o be z 0 The length of the relaxed spring is L (Figure 1) View Available Hints) of the left side of the block is defined The block is slowly pulled from its equalbrium position to some position T0 along the x axis. At timet0 the block is released with zero intial velocity The goal is to determine the position of the block z(t) as a function of time in terms of w and Submit It is known that a general solution for the displacement from e uitnum of a harm ??? oscilator X Incorrect: Try Again;5 attempts remaining where C. S, and ? are constants. (Figure 2) Part B This question wil be shown after you complete previous question(s) Your task, therefore, is to determine the values of C and Snterms of w and z Part C This question will be shown ater you complete previcus question(s) Now, imagine that we have exactly the same physical situation but that the x axis 3) The iniial position of the block is the same as before, but in the new coordinate system the block's starting posiion is given by t 0)-L Figure 1 of 3 s translated, so that the postion of the wall is now defined to be z 0 (Figure Part D (t) in the new coordinate Find the equation for the block's position system Express your answer in terms of L(Greek lecter omega), and t View Available Hinns)

Explanation / Answer

Part A

We know that,

at t=0, x=xinitial

Just, by putting t=0 in the equation,

xinital=C

Part D,

When the eequation is,

x=L+xinital

So,

xinital=L+C

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