- #1

viciado123

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In the Beat not have friction force, correct ?

[tex]m \frac{d^2x}{dt^2} + kx = F_o cos(wt)[/tex]

We can write as

[tex]\frac{d^2x}{dt^2} + w_o^2 x = \frac{F_o}{m} cos(wt)[/tex]

If [tex] w \not= w_o[/tex]

Assuming (Particular solution)

[tex]x_p = acos(wt) + bsin(wt)[/tex] Why we have assuming this ?

How find [tex]a(w_o^2 - w^2)cos(wt) + b(w_o^2 - w^2)sin(wt) = \frac{F_o}{m} cos(wt)[/tex] ?

And why find that: [tex]x_p[/tex] is [tex]x_p = \frac{F_o}{m(w_o^2 -w^2)}cos(wt)[/tex] ?