The block is slowly pulled from its equilibrium position to some position æinit > 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 æ (t) as a function of time in terms of w and æinit - It is known that a general solution for the displacement from equilibrium of a harmonic oscillator is æ(t) = C cos (wt) + S sin (wt), where C, S, and w are constants. (Figure 2) Your task, therefore, is to determine the values of C and S in terms of w and ¤init -
The block is slowly pulled from its equilibrium position to some position æinit > 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 æ (t) as a function of time in terms of w and æinit - It is known that a general solution for the displacement from equilibrium of a harmonic oscillator is æ(t) = C cos (wt) + S sin (wt), where C, S, and w are constants. (Figure 2) Your task, therefore, is to determine the values of C and S in terms of w and ¤init -
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Transcribed Image Text: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.
Constants I Periodic Table
Find the value of S using the given condition that the initial velocity of the block is zero: v(0) = 0.
The goal is to determine the position of the block x (t)
as a function of time in terms of w and xinit -
View Available Hint(s)
It is known that a general solution for the displacement
from equilibrium of a harmonic oscillator is
Hint 1. How to approach the problem
x(t) = C cos (wt) + S sin (wt),
Using the general equation x (t), obtain the expression for the block's velocity v (t) in terms of C, S, w,
and t. Then evaluate the general expression for v (t) when t = 0.
where C, S, and w are constants. (Figure 2)
Your task, therefore, is to determine the values of C and
S in terms of w and xinit.
Hint 2. Differentiating harmonic functions
Recall that
d cos wt
-w sin wt
dt
Figure
1 of 3
>
and
d sin wt
= w COS Wt.
dt
Note the negative sign in the first formula.
L
Xinit
Xinit tan(wt)
win-
Xinit w
Xinit
x = 0
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