Question

(help pls)A wooden rod is placed on top of the piston, performs a vertical simple harmonic motion. find 1) and 2)?

1) Let the period of the simple harmonic motion be 1.0s. What amplitude of the motion can the wooden rod and piston be separated?
2) When the motion's amplitude is 5.0cm, what is the maximum frequency needed for the block and the piston to keep the motion while still being attached to each other?

Answer 1

1) For amplitudes of `approx 25 " cm"` or greater, the rod will separate from the piston.

2) The rod and piston will remain in contact for angular frequencies up to `approx 14 \ rad s^(-1)`.

Explanation:

1)

For SHM in the vertical direction, with co-ordinate `y` being positive in upward direction, governing DE, and an idealised solution, are:

• `ddot y = - omega^2 y`

• `y = Y sin omega t`

The tendency of the rod, absent the piston, would be to freefall with acceleration `- g`. Rod and piston will, therefore, come apart if the piston out-accelerates the rod in the downward direction.

Max downward acceleration of piston occurs at `y = Y` such that:

• `ddot y = - omega^2 Y`

The piston and rod will separate if:

• `abs( - omega^2 Y) gt abs(- g) qquad implies Y gt g/omega^2 qquad square`

Here: `T = 1 = (2 pi)/omega`, so:

• `Y gt g/( 4 pi^2) [approx bb( 0.25" m")]`, with `g approx 9.81 \ m s^(-2)`

So for amplitudes of `approx 25 " cm"` or greater, the rod will separate from the piston.

2)

Re-writing `square` for angular frequency, piston and rod will separate if:

• `omega gt sqrt(g/Y)`

With same approximate value for `g`:

• `sqrt(9.81/0.05) approx 14 \ rad \ s^(-1)`

Bearing in mind the rounding and approximate values, the rod and piston will remain in contact for angular frequencies up to `approx 14 \ rad \ s^(-1)`.

[To convert to oscillation frequency `f`, use `omega = 2 pi f`]