# Mechanical vibration spring/mass help

1. Jan 19, 2014

Hey guys,

I'm working on solving a problem of figuring out what an impact velocity is going to be roughly. The numbers seem very small to me though so I was wondering if you guys could check out my work. This isn't a homework problem, but rather a real problem. I"m trying to build a test device that will generate an impact shock. Basically I'm taking two plates, using bungee cord, and slamming them together. I've modeled this after the classic spring and mass problem (mechanical vibrations, free and undamped) using it's standard second order differential equation. I realize some of the real world problems (friction etc...) that will inhibit my performance, but I'm just trying to get a rough idea here so I'm assume all ideal conditions.

The idea here is that I've got a plate attached to bungee, or a spring, stretched 1 meter past it's equilibrium point and it is locked here with the bungee/spring providing 100,000N of force. The initial position x(0) = 1 meter and the initial velocity x'(0) is zero. I then pull the release pin, it shoots down and impacts the bottom plate.

I'm trying to find the velocity just before impact. I feel like this velocity should be pretty high since my force is extremely large (100kN), but I'm getting just under 7 meters per second. That seems slow. Anyways, I've attached my work so if someone could check it out and let me know if I'm correct or I've messed up and where I'm going wrong. Any general recommendations for building an impact shock table are helpful, but I'm looking mainly for guidance because I'm worried about there being no way for me to possibly create enough force to accelerate a plate to create an impact of 10's of thousands of g's (i'd like 20 or 30kg's) which is the end goal for me.

Here is a link to an example problem thats similar.
http://www.stewartcalculus.com/data/CALCULUS%20Concepts%20and%20Contexts/upfiles/3c3-AppsOf2ndOrders_Stu.pdf

Attached is a pdf of my work and a picture of my system as a spring mass problem. Thanks again for any help I really appreciate it.

Rob

#### Attached Files:

File size:
1.1 MB
Views:
91
• ###### Velocity work.pdf
File size:
1.5 MB
Views:
88
2. Jan 19, 2014

### voko

There is no need to solve differential equations to find the max velocity of a mass on a spring. You can use conservation of energy for that.

Your solution is also correct - in the method, but not in details. I agree that ${\pi \over 2} \ \text{rad} = 100 t$ (which you wrote as $1.57 \ \text{rad} = 100 t$). But I do not understand what you do next that gets you $4 = 100 t$. That is certainly wrong.

Note, also, that you do not really need to compute $t$. It is enough that $100 t = {\pi \over 2}$, this you can plug in the expression for velocity directly.

3. Jan 19, 2014

Thank you so much for the reply. My thought process was taking 1.57rad = 100t and dividing (2*pi)/1.57 to get a time in seconds because I thought you had to plug a seconds time in for t into the velocity equation and I thought that the conversion would give me seconds.

recalculating with a t in radians (1.57 or pi/2) gives me a velocity of 39m/s or 89mph. This would make a lot more sense to than 8 m/s. Do you think I am now correct in my calculations after this? Using the radians value for t? Thanks again for your time!

Rob

4. Jan 19, 2014

### voko

I am not sure what you mean by "t in radians". As I said, I would not even bother calculating the time. Since $\omega t = {\pi \over 2}$, then $v = A \omega \sin \omega t = A \omega \sin {\pi \over 2} = A \omega = 100 \ \text {m/s}$. This, again, could have been obtained from conservation of energy.

5. Jan 19, 2014

I see now. For some reason I was thinking you had to convert radians to something else and I thought that was something else was seconds. After talking to you it makes sense you just stay with radians. I think I was just having a brain malfunction.

So one more thing while I have your attention. Could you point me in the right direction to figure out what the acceleration (or negative acceleration) level would be when this moving plate impacts a stationary plate that will not move, not even on impact?

6. Jan 19, 2014

### voko

This is not straightforward and depends on the properties of the colliding materials. A ballpark estimate could model both bodies as elastically deforming bars, then you would need their geometries and Young's modulus.