A linear oscillator moves with the following acceleration: a = 2-sin(omega t) -2

Question

A linear oscillator moves with the following acceleration: a = 2-sin(omega t) -22x-1.2v. Modify the Euler-Cromer program in each of the following ways: change the argument list of the second-derivative function to include the first derivative, and swap the update positions of the velocity and position. [See the note below.]Then, use the program to analyze the motion of this oscillator given an initial position of '1' and an initial velocity of '0' for the following angular frequencies of the driving function. 2, 3.5, 4.2, 4, 4.7, 5, and 6. In each case plot position vs. time for 1, 700 steps and identify the steady -state oscillations. What observations can you make about what you see here? In particular, what do you see happening to the amplitude of the steady-state oscillations? sol_ec(x_0, y_0, dy_0, f, Delta x, N): = ||[x_0 y_0 w_0] rightarrow [x_0 y_0 dy_0] for i elementof 1..N || x_i leftarrow x_i -1 + Delta x y_i leftarrow y_i - 1 + w_i - 1middot Delta x w_i leftarrow w_i - 1 + f(x_i, y_i) middot Delta x augment (x, y, w)

Explanation / Answer

Solution:-

The relation here is :-

ec(xo, yo, dyo, f , delta x, N) = (xo, yo, wo) <-- (xo, yo, dyo) for i belong 1.... N

argument (x, y, w)

So in this relation:-

xi <--x(i-1) + delta x

yi <-- y (i-1) + w(i-1) .delta x

w <--- w (i-1) + f (xi,yi). delta x

Hence argument (x, y, w)

And also a = 2 sinwt - 22x -1.2v

So da/dt = 2 Coswt .w - 22 dx/dt- 1.2 dv/dt

Hence the second derivative d2a/dt2 = - 2 Sin wt. w^2 - 22 d^2x/dt^2 - 1.2 d^2v/dt^2

When v = 0

Then the second derivative d2a/dt2 = - 2 Sin wt. w^2 - 22 d^2x/dt^2

Hence the observation from the above results are shown below:-

The motion of this osciallator is dependent upon the x an w instead of v.

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