Normal force as a function of time (oscillator)

[Luxucs]

Homework Statement

A spring with spring constant k is attached to a box of mass M in which is placed a small body of mass m. The system is displaced a distance A from equilibrium and released from rest. Find the normal force between the box and the small mass as a function of time. For what initial displacement A will the small mass just begin to lose contact with the box?

(Picture of the situation is attached if necessary.)

Homework Equations

[/B]
##\vec F = m\vec a##
##\vec F_{spring} = -k\vec y##
##ω_o = \sqrt{\frac k {m + M}}##
##y(t = 0) = A##
##v(t = 0) = 0##

Note that ##ω_o## is just the natural angular frequency here of the combined masses. We're also assuming there is no damping present here.

The Attempt at a Solution

I began by recognizing that while the blocks are in contact with one another, they will be moving together, and thus share a common velocity and acceleration. Thus, so I didn't have to initially worry about the normal forces, I wrote out Newton's 2nd Law for the combined masses as a whole (I'm taking the upwards direction to be positive here),

##-ky - (m + M)g = (m + M)\ddot y##

If we rearrange terms, we find that,

##\ddot y + {ω_o}^2y = -g## (1)

This differential equation has two solutions due to the presence of a constant on the RHS,

##y(t) = y_c + y_p## (2)

##y_p## is the particular solution, and ##y_c## is the complementary (homogeneous) solution, and it is rather well-known for the oscillator,

##y_c(t) = C\cos{ω_ot} + D\sin{w_ot}## (3)

Where C and D are constants to be determined by initial conditions. This now leaves us with ##y_p(t)## to find. Since the RHS of the differential equation is a constant, we guess that,

##y_p = B##

Where B is some constant. If we substitute this guess into (1), we observe,

##B = \frac {-g} {{ω_o}^2}## (4)

Substituting (3) and (4) into (2), we have the general solution for the motion of the system,

##y(t) = C\cos{ω_ot} + D\sin{w_ot} - \frac {g} {{ω_o}^2}## (5)

Taking the first time derivative of (5), we obtain,

##v(t) = -ω_oC\sin{ω_ot} + ω_oD\cos{ω_ot}## (6)

From initial conditions, we find from (5) and (6),

##C = A + \frac {g} {{ω_o}^2}##
##D = 0##

Thus, the complete solution to (1) is,

##y(t) = (A + \frac {g} {{ω_o}^2})\cos{ω_ot} - \frac {g} {{ω_o}^2}## (7)

If we then consider the forces acting exclusively on m, from Newton's 2nd Law we have, and knowing that the acceleration acting on m is the same as that on (m + M),

##n - mg = m\ddot y## (8)

From (7), taking the second time derivative,

##\ddot y = -{ω_o}^2(A + \frac {g} {{ω_o}^2})\cos{ω_ot}## (9)

Substituting (9) into (8), we obtain the normal force as a function of time,

##n(t) = m(-({ω_o}^2A + g)\cos{ω_ot} + g)## (10)

My questions mainly concern if I messed up anywhere on obtaining (10), whether it be the physics or the math. Everything seems right to me, but the negative sign in front of the cosine term has me a bit concerned. Furthermore, I absolutely have no idea on how to go about solving the second part of the question. If we take (10), equate it to 0, and argue that since we care about initial displacement, t = 0, we have,

##0 = m(-({ω_o}^2A + g) + g)##
##A = 0##

Which obviously isn't right. I have a feeling it has to do with ##{ω_o}##, as m is going to be in free fall when the normal force vanishes? Where am I messing up at?

 

[haruspex]

Equation 7 cannot be right. The displacement from equilibrium was A, so that must be the amplitude.
Looks like you defined y as spring extension. If so, what should be the value of y(0)?
It is generally simpler if you define it as displacement from equilibrium.

 

[Luxucs]

Equation 7 cannot be right. The displacement from equilibrium was A, so that must be the amplitude.
Looks like you defined y as spring extension. If so, what should be the value of y(0)?
It is generally simpler if you define it as displacement from equilibrium.

##y(0)## from (7) yields the expected displacement though, A. Are you saying in that the way that I have set up things, it should be instead yielding 2A? If so, would my new initial condition have the form ##y(t = 0) = 2A##? My understanding is that with Hooke's Law, y is the displacement from equilibrium, which is what was solved for above.

EDIT: I should clarify; I took y = 0 to be at equilibrium, where y is some distance from that equilibrium.

 

[BvU]

I took y = 0 to be at equilibrium

No: at ##y=0## you have ##\ddot y\ne 0##.

 

[haruspex]

No: at ##y=0## you have ##\ddot y\ne 0##.

Right, I was looking at that eqn (1) when I wrote

Looks like you defined y as spring extension

 

[Luxucs]

Alright, so I've stepped back and re-evaluated this problem from the beginning. I'm figuring that my issue mainly has to do with the way I've set up the coordinates here.

Let y = 0 be the relaxed position of the spring (i.e without any masses on it). Then, when the masses are added, the spring is displaced by -L. So, we have,

##\vec F_{spring} = -k \Delta \vec y = k \Delta \vec L##

So, our net force at equilibrium would then be,

## F_y = k \Delta L - (m + M)g = 0##
##\Delta L = \frac {(m+M)g} k##

Then, if we consider the problem when the masses are accelerating upward,

##-k\Delta y - (m + M)g = (m+M)\ddot y##
##\Delta y = y - \Delta L##

It reduces down to the following, eliminating the gravitational force,

##-ky = (m+M)\ddot y##

This is a homogeneous differential equation now, and the solution simply is,

##y(t) = A\cos{ω_ot}##

Where A is the initial amplitude given in the problem statement. From Newton's 2nd Law on m,

##n - mg = m\ddot y##
##n(t) = m(g - {ω_o}^2A\cos{ω_ot})##

For the initial displacement A such that n vanishes, we find,

##0 = m(g - {ω_o}^2A)##
##A = \frac g {{ω_o}^2}##

How's this look?

 

[haruspex]

How's this look?

Looks right.

 

[Luxucs]

Great, thanks for the help!

 

[Haroon Khan]

do you have solution manual for this book?

 

[haruspex]

do you have solution manual for this book?

The thread is six months old. Luxucs might not be around any more.

 

[Luxucs]

I'm still around. No, I don't have the solution manual for this book, sorry.