Max amplitude of a diving board that will not toss a pebble off?

[link223]

The FBD you drew for the plank is not nessecary, and it is wrong, because the board is attached to something at its left end.No, forget about the plank. You only need the forces acting on the pebble and what the acceleration of the pebble is, which is the same as the acceleration of the board when the pebble has contact with it. And since you know that the board is undergoing harmonic oscillation and you have its frequency, you are done.

right, so how do I relate the result for the acceleration to the hooke's law, if we have a normal force and the force of gravity (let me write it out real quick)

I sorry but I just don't get the translation from a mass-spring used in the textbook to the real world application where it's not attached.

 

[link223]

here it is (to add is it by knowing that we have harmonic osciallation we could simply apply this to situation of no mass-spring system acting in accordance to hooke's law or what?)

 

[malawi_glenn]

here it is

yeah, in magnitude that is.
This is a general property of acceleration due to harmonic motion, the magnitude of the acceleration is proportional to the magnitude displacement. The constant of proportionality is the angular velocity squared, and the direction of the acceleration is oppsoite to the displacement.
You can see it here as well, the position as a funtion of time is sinusodial $x(t) = A \sin (\omega t + \varphi)$ The acceleration is the second derivative w.r.t. time. By differenting twice, we get $a(t) = -\omega ^2 x(t)$ and the maximal acceleration is therefore (in magnitude) $\omega^2 A$

 

[link223]

yeah, in magnitude that is.
This is a general property of acceleration due to harmonic motion, the magnitude of the acceleration is proportional to the magnitude displacement. The constant of proportionality is the angular velocity squared, and the direction of the acceleration is oppsoite to the displacement.
You can see it here as well, the position as a funtion of time is sinusodial $x(t) = A \sin (\omega t + \varphi)$ The acceleration is the second derivative w.r.t. time. By differenting twice, we get $a(t) = -\omega ^2 x(t)$ and the maximal acceleration is therefore (in magnitude) $\omega^2 A$

oh I am so dumb... it is just $a_{max} = \omega^2 A_{max}$
The analysis of what I did will just show me what the ACCELERATION IS FROM THE KINETICS/DYNAMICS PART OF VIEW
BUT THEN GOING BACK TO THE KINEMATICS PART WE HAVE MOTION ACCORDING TO THE HARMONIC EQUATIONS, correct?

 

[link223]

Thanks for the help... I don't see where I was getting all of this from..
Because the acceleration essentially goes by $a = A\omega^2 cos(\omega t + \phi)$ and has nothing to do with the relation $F_s = ma (kx = mg)$

I guess I was trying too much to look at it as a mass-spring system and relate it to the board case which not the way to go (talking about incorrectly looking at the dynamics part of the mass spring system and relating that to the board-peddle system), but just a case of oscillation

 

[link223]

@malawi_glenn so from now on... looking at oscillations from other system, the thing only thing they have in common w/ the mass-spring system are the oscillation equations (that is the kinematics, nothing about their dynamics (besides the accelerations, right? )
just want to make sure. TIA

 

[Orodruin]

and the normal force on the pebble is 0 then because... (do not mention Newtons 2nd law)

If the board is acceleration down faster than the pebble then contact is lost and therefore no force.

 

[malawi_glenn]

Well, the fact that you get sinusodial motion does come from the equation of motion $\ddot x - \frac{k}{m}x = 0$... but you do not need to worry about those details here, you just need to know that the acceleration of the pebble when it has contact with the board is $\omega^2 x$ in magnitude.

 

[link223]

Well, the fact that you get sinusodial motion does come from the equation of motion $\ddot x - \frac{k}{m}x = 0$... but you do not need to worry about those details here, you just need to know that the acceleration of the pebble when it has contact with the board is $\omega^2 x$ in magnitude.

got it that's indeed just the derivation the 2nd order ODE, I was looking at it from the kinetics POV only.

 

[Orodruin]

The force of gravity on the board is not acting on the pebble, that is the point of FBD. You draw the forces acting on the object of interest.

To add to that, the board makes simple harmonic motion relative to its rest position, not relative to the position where it has no internal stresses. This will effectively cancel out the weight term from the equations of motion for the end if the board.

Since you have the result already, let me just mention how my argumentation would go:

As long as the board’s acceleration downwards is smaller than g, the pebble will stay (because it cannot fall theough the board). The limiting case is a = g. The amplitude of the board’s acceleration is $A\omega^2$ which means that the max downwards acceleration is precisely this. Thus, the limiting case becomes $g = A\omega^2$. Solve for $A$.