# Calculating Oscillation Amplitude & Period of Deli Plate

• WY
In summary: PE = zero at the stopping point. What about the spring?In summary, the conversation is about a physics problem involving a plate, a spring, and slices of ham. The goal is to find the amplitude and period of the oscillation of the scale after the ham lands on the plate. The correct approach involves using conservation of energy and taking into account the potential energy of the ham and the plate. The final step is to solve a quadratic equation to find the amplitude.
WY
Hi
I'm attempting this question and I'm wondering if this is the correct way of going about it or if I'm completely off track:
For lunch you and your friends decide to stop at the nearest deli and have a sandwich made fresh for you with 0.300 kg of Italian ham. The slices of ham are weighed on a plate of mass 0.400 kg placed atop a vertical spring of negligible mass and force constant of 200 N/m. The slices of ham are dropped on the plate all at the same time from a height of 0.250 m. They make a totally inelastic collision with the plate and set the scale into vertical simple harmonic motion (SHM). You may assume that the collision time is extremely small.

What is the amplitude of oscillation A of the scale after the slices of ham land on the plate?
What is the period of oscillation T of the scale?

I started using conservation of GPE and equating it to the energy of the spring
mgh = 1/2*k*A^2
and made A the subject to find the amplitude - would this give me the right answer?

And then for the second part I used (2*pi)/sqrt(k/m)=T and m = the mass of ham - should i include the mass of the plate?

No, the 'X' in hooke's law represents the total displacement, which is actually a multiple of the amplitude.

At a glance the second part looks fine. You need to include the mass of the scale since it is also part of the SHM.

a multiple of the amplitude - so would i have to divide it by 2 to find the amplitude?

WY said:
I started using conservation of GPE and equating it to the energy of the spring
mgh = 1/2*k*A^2
and made A the subject to find the amplitude - would this give me the right answer?
No. Use conservation of energy to find out how far the scale depresses from its initial position after the downward motion stops. Don't forget to include how far the scale depresses in calculating the GPE. Once you've found the lowest point, compare that to the equilibrium position of the system to figure out the amplitude. (You'll have to find the new equilibrium position.)

I still don't understand how to get it - can you explain it in detail?

I thought I just did?

(1) Call the initial vertical position of the plate 0 and its lowest point (after the ham hits it) x. Now write an expression for conservation of energy to find x.

(2) Before the ham hits the plate, the equilibrium of the system is at position 0. How far below zero is the new equilibrium position of the "plate + ham" system?

(3) Combine the answers from 1 and 2 to figure out the amplitude of the motion about the new equilibrium position.

Sorry can you show how to find x - the lowest position (after the ham hits it)?

I describe how to do that as step 1 above. Try it! It's just conservation of energy. The ham starts with gravitational PE which gets transformed into spring PE. Note that the ham doesn't just lower a distance h = 0.250 m. (Measure the change in height of the ham from its initial position above the plate to the lowest point of the motion.)

But then if mgh = 1/2kA^2
that means for the ham:
0.3*9.8*0.25 = 1/2*200*A^2?
I don't understand? I've done that before but it doesn't work?
But what you are saying is to work out how far the ham goes past the 0.25m? I still don't understand how to get that?

ussrasu said:
But then if mgh = 1/2kA^2
that means for the ham:
0.3*9.8*0.25 = 1/2*200*A^2?
That is incorrect, as I stated in my first post in this thread. It is incorrect for two reasons:

(1) The potential energy of the ham is not mgh, where h is the height above the plate. Measure the height of the ham from the lowest position. (The lowest position is what I call "x" below the original position. So... if the ham starts at a height "h" above the plate, and ends up "x" below the plate... what's the total change in height?)

(2) The displacement from the original position is not the amplitude! The spring potential energy is given by 1/2 k x^2, where x = how far the plate lowers from its original position.

Given these tips, see if you can write the correct statement of conservation of energy.

So then U = mg(h+x)
therefore: 0.3*9.8*(0.25+x) = 1/2*200*x^2
So then do you solve for x, using the quadratic formula?
I get x = 0.455 or x = -0.16
Which one would then be the displacement?
I got these answers from: 100x^2 - 29.4x -7.35
Am i doing the right thing now?

Im not sure if I am doing the right thing here? Can anyone show me how this question is solved?

Have you used the hints? Please don't tell me you're one of those people who refuse to use the hints =P

ussrasu said:
So then U = mg(h+x)
therefore: 0.3*9.8*(0.25+x) = 1/2*200*x^2
So then do you solve for x, using the quadratic formula?
I get x = 0.455 or x = -0.16
Which one would then be the displacement?
I got these answers from: 100x^2 - 29.4x -7.35
Am i doing the right thing now?

Looks like you now have the potential energy of the ham relative to PE = zero at the stopping point. What about the plate?

ussrasu said:
So then U = mg(h+x)
therefore: 0.3*9.8*(0.25+x) = 1/2*200*x^2
So then do you solve for x, using the quadratic formula?
Exactly!
I get x = 0.455 or x = -0.16
Which one would then be the displacement?
I got these answers from: 100x^2 - 29.4x -7.35
Am i doing the right thing now?
Almost. You made an error when you plugged in the numbers. I think the quadratic that you need to solve should be: 100x^2 - 2.94x - 0.735 = 0

You'll get two answers, but only one is relevant to this problem (the positive one).

OlderDan said:
Looks like you now have the potential energy of the ham relative to PE = zero at the stopping point. What about the plate?
Representing the energy of the "plate + spring" system as $1/2 k x^2$ where x is measured from the equilibrium position (the original position before the ham drops) automatically includes the change in PE of the plate.

Doc Al said:
-Representing the energy of the "plate + spring" system as $1/2 k x^2$ where x is measured from the equilibrium position (the original position before the ham drops) automatically includes the change in PE of the plate.

Sorry. I lost track of what you meant by x, but even if it was the amplitude I would have been off by a term for failing to include the change in equilibrium position as part of the height of the ham relative to final equilibrium.

I've been having some problem resolving your statement, which I know to be true for a mass hanging on a spring. I keep getting inconsistent results depending on how I write the displacements. At first I thought it had to be an algebra mistake, but now I'm thinking there is a fundamental problem with the approach.

The problem clearly states that the collision is inelastic and that the collision time is very small (can be neglected). To me that suggests that the collision needs to be treated as an instantaneous conservation of momentum problem to find an initial velocity of the combined mass/spring system at a position above final equilibrium. Exteral forces can be neglected during the collison. After the collision, energy will be conserved, but not for the whole process. Does that make sense??

OlderDan said:
The problem clearly states that the collision is inelastic and that the collision time is very small (can be neglected).
D'oh! I misread the problem as saying "totally elastic collision".

To me that suggests that the collision needs to be treated as an instantaneous conservation of momentum problem to find an initial velocity of the combined mass/spring system at a position above final equilibrium. Exteral forces can be neglected during the collison. After the collision, energy will be conserved, but not for the whole process. Does that make sense??
Yep.

Good catch, OlderDan. Thanks!

Allow me to correct the misinformation from my earlier post (post #6 in this thread).
Doc Al said:
(1) Call the initial vertical position of the plate 0 and its lowest point (after the ham hits it) x. Now write an expression for conservation of energy to find x.
As OlderDan points out, first calculate the speed of the "ham + plate" after the collision using conservation of momentum. Then you can apply conservation of energy. There are several ways to track the energy. The easiest may be to treat the "plate + spring" as a system initially at its equilibrium point.

Thus, the initial energy of the system (after the collision) is the sum of: (a) KE of the ham, (b) gravitational PE of the ham, (c) KE of the "plate + spring" system (just the KE of the plate), and (d) spring PE of the "plate + spring" system (which equals zero, since it's at equilibrium).

The final energy of the system (as it reaches the lowest point) is the sum of those same terms. Of course, the KE terms are zero, since the system is momentarily at rest at the lowest point.

Note that by measuring the spring PE of the "plate + spring" system from its equilibrium point (which is its initial position) as $1/2 k x^2$, the gravitational PE of the plate is already accounted for.

(2) Before the ham hits the plate, the equilibrium of the system is at position 0. How far below zero is the new equilibrium position of the "plate + ham" system?
Still valid.

(3) Combine the answers from 1 and 2 to figure out the amplitude of the motion about the new equilibrium position.
Still valid.

So is this positive x value now the displacement after the ham hits the plate? Is it also the amplitude or is this just the new equilibrium position from where the plate and ham are at rest?

Using the answer from the quadratic is obviously not the amplitude - it was incorrect. So where do i go from here? What does the speed of the plate have to do with the amplitude?
The total mechanical energy of the scale at the point of maximum displacement is given by 1/2*k*A^2 , where k is the force constant and A is the amplitude of oscillation. Since the total energy in SHM is constant, then the energy of the scale immediately after the collision must also equal 1/2*k*A^2 . What is E_tot, the total energy of the system (spring, plate, and ham) immediately after the collision?

I've opened the first hint and it asks what the displacement on the plate from it's equilibrium position is before the energy conservation stuff. How would you do this?

Thanks.

I'm even stuck on the 1st hint as well! "How far from the new equilibrium position is the plate when the slices of ham land on it?"

The string's compression force opposes the force of gravity (weight of the system on it). Less force is required to keep only the plate level than to keep both the plate and ham level. The spring is compressed more with both on top, the new equilibrium position is lower than the original.

So at the new point of equilibrium will the force of the spring equal the force of gravity? Would that mean that kx = mg and we just have to solve for x or something?

janiexo said:
So at the new point of equilibrium will the force of the spring equal the force of gravity? Would that mean that kx = mg and we just have to solve for x or something?
Right. To find the new equilibrium position, find the additional force that the ham adds to the spring (its weight, mg) and then how much further the spring must depress to cancel that force (using Hooke's law).

Ok so x=0.3*9.8/200 = 0.0147? It's just the additional mass (0.3kg) that I need to take into account, not the total mass (0.3+0.4)kg, right?

That's correct.

## 1. What is the formula for calculating the oscillation amplitude of a deli plate?

The formula for calculating the oscillation amplitude of a deli plate is: A = (D/2)*sin(θ), where A is the amplitude, D is the diameter of the plate, and θ is the angle of deflection.

## 2. How do I measure the diameter of the deli plate for this calculation?

To measure the diameter of the deli plate, simply use a ruler or measuring tape to measure the distance across the center of the plate. Make sure to measure in a straight line from one edge to the opposite edge.

## 3. What is the unit of measurement for the oscillation amplitude?

The unit of measurement for the oscillation amplitude is the same as the unit used to measure the diameter of the deli plate. For example, if the diameter is measured in inches, the amplitude will also be measured in inches.

## 4. Can the oscillation amplitude of a deli plate be negative?

Yes, the oscillation amplitude of a deli plate can be negative. This means that the plate is oscillating in the opposite direction from its starting point. In this case, the value of the amplitude will be represented with a negative sign.

## 5. How is the period of a deli plate's oscillation calculated?

The period of a deli plate's oscillation is calculated using the formula: T = 2π√(m/k), where T is the period, m is the mass of the plate, and k is the spring constant. This formula assumes that the plate is attached to a spring and is undergoing simple harmonic motion.

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