What is the mass of the oscillator in an oscillating mass-spring system?

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In summary: So take a good look at that again.You didnt only forget to square A, but you also squared k/m when not necessary, because of the root.... So take a good look at that again.In summary, the problem is asking to find the mass of an oscillator in an oscillating vertical mass-spring system with a spring constant of 6.05 N/m and a maximum displacement of 81.7 cm and a maximum speed of 2.05 m/s. The equation used to solve this is vmax = A * the square root of k/m. After manipulating the formula by squaring both sides and rearranging the variables, the final answer is 0.4147 kg.
  • #1
Dangelo
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[SOLVED] Manipulating a formula

Homework Statement


"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

Homework Equations


vmax=A*the square root of k/m
(maximum speed of a mass-spring system)

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.
 
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  • #2
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
Swatje said:
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?

a) Yes, when attempting I made sure that A was 0.817kg.

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
Dangelo said:
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.

First of all what do you mean with get the things out of the root? You have to square both sides so that the root drops out. Then bring m to one side using multiplication and division, you think you can do this with this info I gave you?
 
  • #5
dirk_mec1 said:
First of all what do you mean with get the things out of the root? You have to square both sides so that the root drops out. Then bring m to one side using multiplication and division, you think you can do this with this info I gave you?

Okay, while this does help and I feel I am on the verge of getting it right, i still didn't get it. I am left with equations such as m=vmax squared divide by A*K and variations of those variables being squared.

Thanks for your patience, though, I realize I am very out of my league with physics.
 
  • #6
Dangelo said:
Okay, while this does help and I feel I am on the verge of getting it right, i still didn't get it. I am left with equations such as m=vmax squared divide by A*K and variations of those variables being squared.

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

Hehe, this isn't that hard. You just made a small mistake in derivation of your formula for m out of the formula for vmax. Redo it, try and find your error. ;).
 
  • #7
Swatje said:
Hehe, this isn't that hard. You just made a small mistake in derivation of your formula for m out of the formula for vmax. Redo it, try and find your error. ;).

Tried numerous times, getting things such as:

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. :frown:
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 ass.
 
  • #8
Dangelo said:
Tried numerous times, getting things such as:

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. :frown:
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 ass.

try writing down step by step on here. ill try and help you along.
 
  • #9
Swatje said:
try writing down step by step on here. ill try and help you along.

Ok

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
Dangelo said:
Ok

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)

ok. Your first step. From a to b. If you square both sides, keep in mind that a square cancels out a root, and variables that arent under a root get a square...
 
  • #11
Swatje said:
ok. Your first step. From a to b. If you square both sides, keep in mind that a square cancels out a root, and variables that arent under a root get a square...

Okay, so I see that I didn't square A... but when I do, I am left with the answer 0.4147...
 
  • #12
Dangelo said:
Okay, so I see that I didn't square A... but when I do, I am left with the answer 0.4147...

You didnt only forget to square A, but you also squared k/m when not necessary, because of the root. :)
 
  • #13
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
Swatje said:
You didnt only forget to square A, but you also squared k/m when not necessary, because of the root. :)

D'oh. Okay, well I've taken that into consideration and am now getting 0.922~.

lukas86 said:
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.

I thought there was something wrong with that, but I don't see what I should have done... i should multiply by m? :confused:
 
  • #15
Well, let me try something like this for an example, just some random variables that have nothing to do with anything and let's 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
 
Last edited:
  • #16
lukas86 said:
Well, let me try something like this for an example, just some random variables that have nothing to do with anything and let's solve for pv*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

Okay, I finally got it!

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!
 

Related to What is the mass of the oscillator in an oscillating mass-spring system?

1. How do I manipulate a formula?

To manipulate a formula, you can use basic algebraic operations such as addition, subtraction, multiplication, division, and exponentiation. You can also rearrange the variables or use inverse operations to isolate a specific variable.

2. Why would I need to manipulate a formula?

Manipulating a formula allows you to solve for a specific variable, simplify complex equations, or convert the formula to a different form that is more useful for your particular problem.

3. Can I manipulate a formula in any way I want?

While there are certain rules and principles that govern how you can manipulate a formula, you have some flexibility in choosing the operations and rearrangements you use. However, it is important to ensure that your manipulations follow the rules of algebra and do not change the meaning of the original formula.

4. Are there any common mistakes to avoid when manipulating a formula?

Yes, some common mistakes to avoid include forgetting to perform the same operation on both sides of the equation, not distributing correctly, and making arithmetic errors. It is also important to double check your work and make sure you have not changed the meaning of the original formula.

5. Can I use a calculator to manipulate a formula?

Yes, a calculator can be a helpful tool when manipulating a formula. However, it is important to understand the steps and principles behind the manipulation rather than solely relying on the calculator's results. Additionally, make sure to use the correct order of operations when entering the formula into the calculator.

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