The minimum mass of A (m) for B (M) just to lift off is
m = M
The figure with the forces acting on mass A is below.
The equation of motion for mass A is
`F =M*a`
`M*g -k*x =M*a`
`M*g -k*x =M*(d^2x)/(dt^2)`
that can be rearranged in the form
`(d^2x)/dt^2 + (k/M)*x = g`
and by defining `omega = sqrt(k/M)` we have
`(d^2x)/(dt^2) +omega^2*x =g`
with the general solution
`x(t) = X0*cos(omega*t) + C`
by inserting the general solution back into the equation we have
`-omega^2*X0*cos(omega*t) + omega^2*X0*cos(omega*t) +omega^2*C =g`
Hence `C = g/omega^2`
At `t =0` we have `x(0) =0`
`0 = X0 + g/omega^2`
Hence `X0 =-g/omega^2`
Therefore
`x(t) =-(g/omega^2)*cos(omega*t) +g/omega^2`
The speed of mass A is
`v(t) = dx/dt = ((g*omega)/omega^2)*sin(omega*t)`
and its maximum value is
`v_max = g/omega =g*sqrt(M/k)`
Further Reading
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