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

[Dangelo]

[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.

 

[Swatje]

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?

 

[Dangelo]

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.

 

[dirk_mec1]

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?

 

[Dangelo]

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.

 

[Swatje]

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. ;).

 

[Dangelo]

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.
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.

 

[Swatje]

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.
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.

 

[Dangelo]

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)

 

[Swatje]

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...

 

[Dangelo]

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...

 

[Swatje]

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. :)

 

[lukas86]

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.

 

[Dangelo]

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~.

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?

 

[lukas86]

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

 

[Dangelo]

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!