A weight with a total mass, m, lies on a frictionless surface. The weight is als

Question

A weight with a total mass, m, lies on a frictionless surface. The weight is also connected to a spring with spring constant k and then the weight is driven on by an outer force F(t) = F0*sin(omega*t).
In the spring we also have a friction -bv(t).

With the help of the 2. law of newton the movement of the weight can be described with the equation

Now we have a theory that

So the real question is here

Put equation (2) into equation (1) and find the coefficients for A and B as function of m, b k and omega.

please explain with step by step for quick lifesaver response

Explanation / Answer

you are given x(t).
Thus, dx/dt = -Asin(t) + Bcos(t).

And d^2x/dt^2 = -A^2cos(t) - B^2sin(t).

Substitute these into the differential equation and you get:

m [-A^2cos(t) - B^2sin(t)] + b[-Asin(t) + Bcos(t)] + k[Acos(t) + Bsin(t)] = Fsin(t)

Expand these out and you are left with:

-mA^2cos(t) - Bm^2sin(t) - bAsin(t) + bBcos(t) + kAcos(t) + kBsin(t) = Fsin(t)

group the sin and cos together to get:

sin(t) [-Bm^2 - bA + kB] + cos(t) [ -mA^2 + bB + kA] = F sin(t)

The coefficient of sin should be equal to F and the coefficient of cos should be 0 because there is no cos term on the right hand side.

Thus, F = [-Bm^2 - bA + kB] ....Eq1

0 = [ -mA^2 + bB + kA].......Eq2

Thus, you have 2 simultaneous equations.

From 2nd equation, A (k - m^2) + Bb = 0.

Rearranging, A(m^2 - k) = Bb.

Thus, B = A(m^2 - k)/b........Eq 3

Substitute this into equation 1 to get A in terms of other constants. Then substitute that A value into the above equation 3 to get B.

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