My Math Forum Why is the value of s = 1/3?

 Differential Equations Ordinary and Partial Differential Equations Math Forum

October 23rd, 2014, 06:26 PM   #11
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Quote:
 Originally Posted by topsquark I am wondering about your system of units. Specifically if you are using feet for s then an extension of 4 inches would be 1/3 feet. -Dan
I am thinking that too, if you check the end of my post #7. I made a comment about that.

I emailed my teacher, hopefully I get some clarification.

Thanks, I'm glad someone is thinking the same thing I am.

Last edited by shreddinglicks; October 23rd, 2014 at 06:32 PM.

 October 23rd, 2014, 06:36 PM #12 Member   Joined: Oct 2014 From: NJ Posts: 46 Thanks: 2 Also, the answer key has initial conditions. x(0) = -1/4 , x'(0) = 0. Where does that come from?
 October 23rd, 2014, 06:54 PM #13 Member   Joined: Oct 2014 From: NJ Posts: 46 Thanks: 2 I think I figured it out, our assumption of the units must be correct. 3 inches is a (1/4) of a foot. That would satisfy the initial condition and our assumption of the units.
 October 23rd, 2014, 06:57 PM #14 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,671 Thanks: 2651 Math Focus: Mainly analysis and algebra I agree.
October 24th, 2014, 12:44 AM   #15
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Quote:
 Originally Posted by v8archie Isn't $k$ supposed to be the spring constant? The spring constant is the displacement per unit force on the spring. So that would be $\frac{4}{24}= \frac16$ in/lb. I apologise if I'm wrong about pounds. I've always equated pounds with kilograms (as do most converters).
As this thread shows, it's a lot better to use SI units for science.

Ironically, however, in everyday life, Kg is used as a measure of weight (force). You buy something by what it weighs.

When you see 2.2 lbs on some scales then that's right, but when you see 1 kg, then that's really 9.8N being measured.

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# a mass weighing 24 pounds attached to the end of a spring

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