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SauerKrauter
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I am greatly struggling with a homework assignment given out by my physics professor. It mostly differential equations but based on spring physics. I'll type out the first couple parts but will most likely need help with more as I get farther.
Y(t) : the y position of mass M on spring.
L0 : initial position of mass M.
K : the spring constant of spring.
B : the coefficient of friction.
show that Y(t) observes the differential equation Y''(t)+(b/m)y'(t)+(k/m)Y(t) = 0 and
find the general solution using Y(t) = const * eikt with k related to m, k, and b.
Fspring = mY''(t) = -kY(t)
Ffric = -BY'(t)
I'm thinking that the differential equation holds true if and only if the force of friction (and therefore Y'(t)) is equal to zero, because Y''(t) and (k/m)Y(t) are equal and opposite and therefore must add up to zero.
For solving the general solution of the differential equation I can get it the form Ceikt[i2a2+(b/m)ia+(k/m)]=0 but am not sure where to go from there. Clearly its this part of the question that's more difficult and I would love if someone could help me out with it, much more so than the above.
Homework Statement
Y(t) : the y position of mass M on spring.
L0 : initial position of mass M.
K : the spring constant of spring.
B : the coefficient of friction.
show that Y(t) observes the differential equation Y''(t)+(b/m)y'(t)+(k/m)Y(t) = 0 and
find the general solution using Y(t) = const * eikt with k related to m, k, and b.
Homework Equations
Fspring = mY''(t) = -kY(t)
Ffric = -BY'(t)
The Attempt at a Solution
I'm thinking that the differential equation holds true if and only if the force of friction (and therefore Y'(t)) is equal to zero, because Y''(t) and (k/m)Y(t) are equal and opposite and therefore must add up to zero.
For solving the general solution of the differential equation I can get it the form Ceikt[i2a2+(b/m)ia+(k/m)]=0 but am not sure where to go from there. Clearly its this part of the question that's more difficult and I would love if someone could help me out with it, much more so than the above.
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