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[SOLVED] Manipulating a formula |
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| Jun3-08, 02:06 PM | #1 |
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[SOLVED] Manipulating a formula
1. The problem statement, all variables and given/known data
"An instructor sets up an oscillating vertical mass-spring system (k=6.05 N/m). The maximum displacement is 81.7 cm and the maximum speed is 2.05 m/s. What is the mass of the oscillator? Answer: 0.961kg 2. Relevant equations vmax=A*the square root of k/m (maximum speed of a mass-spring system) 3. The attempt at a solution Well, I probably could solve this, if only I could manipulate the formula for m. I've tried m = the square root of V*A divide by k squared (sorry, I don't know how to use symbols) and other similar manipulations, but it just doesn't work for me. Note: I am a complete "n00b" at physics and am doing practice problems to improve. |
| Jun3-08, 02:20 PM | #2 |
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Two things i have to point out:
a) Make sure all your metrics are in the right SI form (so km to meters etc etc...) b) In the first formula, you say that m is under the root, compared to vmax. There is no other root or square in the formula. When you would transform the equation to m =... , would it be possible to still have a root there? with something that is not squared underneath? |
| Jun3-08, 02:27 PM | #3 |
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b) I don't really understand what you are saying... I can try to explain the formula I expect I need to use better: Vmax = A√k/m That symbol being the square root, of course. The reason I mentioned "squaring" some things is because I assume the way to get things out of a square root (especially in this context) is to square them. |
| Jun3-08, 02:38 PM | #4 |
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[SOLVED] Manipulating a formula |
| Jun3-08, 02:48 PM | #5 |
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Thanks for your patience, though, I realize I am very out of my league with physics. |
| Jun3-08, 02:51 PM | #6 |
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| Jun3-08, 03:02 PM | #7 |
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m = vmax2/A(k)2 m = SQUR(vmax2/A(k)2) m = vmax2/Ak m = vmax/A(k)2 etc. But I'm still not getting it. ![]() I have a physics exam tomorrow and am practicing so that I can actually pass, and this question for whatever reason or other is kicking my ***. |
| Jun3-08, 03:04 PM | #8 |
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| Jun3-08, 03:08 PM | #9 |
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a. vmax = A*SQUR(k/m) b. vmax2 = A * (k/m)2 (now out of the square root) c. vmax2/A = k/m2 (divided by A) d. vmax2/(Ak2) = m2 (divided by k, unsure if it should be squared or not) e. SQUR(vmax2/(AK2)) = m (not sure if this is necessary) |
| Jun3-08, 03:13 PM | #10 |
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| Jun3-08, 03:20 PM | #11 |
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| Jun3-08, 03:24 PM | #12 |
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| Jun3-08, 03:26 PM | #13 |
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Also, to add...
First fix your step from a-->b Also, from c-->d if you divide by k, how did m end up on the top and nothing changed on the left hand side. |
| Jun3-08, 03:31 PM | #14 |
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| Jun3-08, 03:36 PM | #15 |
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Well, let me try something like this for an example, just some random variables that have nothing to do with anything and lets solve for p
v*r/p = T/m 1/p = T / (m*r*v) p = (m*r*v) / T <--- flipped the equation because the step before, it was 1/p EDIT: What you're really doing from step 2-->3 is... 1/p *p = T / (m*r*v) *p (mult both sides by p) 1 = T*p / (m*r*v) (mult both sides by (m*r*v)) (m*r*v) = T*p (divide both sides by T) (m*r*v) / T = p |
| Jun3-08, 03:42 PM | #16 |
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A2*K/vmax2 = m = 0.9609 = 0.961kg Thank you Swatje, dirk_mec1, and lukas86, thank you very much! Off to more studying, but I will most likely be back soon :) Thanks again! |
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