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**

## We’ll help your grades soar

Start your 48-hour free trial and unlock all the summaries, Q&A, and analyses you need to get better grades now.

- 30,000+ book summaries
- 20% study tools discount
- Ad-free content
- PDF downloads
- 300,000+ answers
- 5-star customer support

Already a member? Log in here.

Are you a teacher? Sign up now