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Manipulating a formula 
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#1
Jun308, 02:06 PM

P: 8

1. The problem statement, all variables and given/known data
"An instructor sets up an oscillating vertical massspring 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 massspring 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. 


#2
Jun308, 02:20 PM

P: 36

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? 


#3
Jun308, 02:27 PM

P: 8

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. 


#4
Jun308, 02:38 PM

P: 677

Manipulating a formula



#5
Jun308, 02:48 PM

P: 8

Thanks for your patience, though, I realize I am very out of my league with physics. 


#6
Jun308, 02:51 PM

P: 36




#7
Jun308, 03:02 PM

P: 8

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 ***. 


#8
Jun308, 03:04 PM

P: 36




#9
Jun308, 03:08 PM

P: 8

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) 


#10
Jun308, 03:13 PM

P: 36




#11
Jun308, 03:20 PM

P: 8




#12
Jun308, 03:24 PM

P: 36




#13
Jun308, 03:26 PM

P: 70

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. 


#14
Jun308, 03:31 PM

P: 8




#15
Jun308, 03:36 PM

P: 70

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 


#16
Jun308, 03:42 PM

P: 8

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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