Simple Harmonic Motion and Springs

[MRMooneyham]

An object of mass m is supended from a vertical spring of force constant 1269 N/m. When the object is pulled down 0.068 m from equilibrium and released from rest, the object oscillates at 10 Hz. The mass of the object is ____ kg.



τ= 2∏*sqrt(M/K)
M = mass
K = spring constant
1/f = τ

The Attempt at a Solution

1/10 = .1
.1/2∏ = .15708
.15708² = M/K
.024674*K = M
.024674*1269 = 31.3113

and I keep getting told i have the wrong answer...help?

 

[hypnosam]

Well, the energy in a spring is 1/2Kx^2. This then relates to the kinetic energy, as at the top the potential energy in the spring is all converted into gravitational potential energy. Now I think, i could be wrong on this, that the amplitude above the equilibrium point is the same as the amplitude below, ie how far below the spring is initially below the equilibrium point.

You can set these two equal to each other, and use h in mgh as 2x. Rearrange and you can calculate the force. This, if its correct gives an answer of 4.39kg

 

[MRMooneyham]

umm i understand what your trying to say, but 4.39kg is wrong aswell

 

[hypnosam]

Better way, your way works, mine is more complicated, though since T=2*pi*Sqrt(m/K) 0.01*1269/(4*pi^2) which gives 0.321441kg

 

[truesearch]

you have made some calculator errors.
0.1/2pi = 0.0159 not 0.157
You have calculated (0.1/2) x pi

 

[MRMooneyham]

ahh i see, I have a ti-89 calculator, and i did my grouping wrong so it was doing things a little out of what i wanted it to lol, bad grouping on my part, but thanks a whole bunch!

 

[MRMooneyham]

An object is hung on the end of a vertical spring and is released from rest with the spring unstressed. If the object falls 0.17 m before coming to rest, the period of the resulting oscillatory motion will be _____ s.

im am uttlerly lost on this...

 

[hypnosam]

Are you given anything else? Like the mass?

 

[hypnosam]

Not convinced by this, but try it neverthe less. At equilibrium F=kx =mg, for it to be in equilibrium, by that meaning, m/k = x/g. so putting that into the period equation. T=2pi*sqrt(m/k) is the same as T=2pi*sqrt(x/g) which is 0.827 seconds.

 

[MRMooneyham]

nope...
but I've been looking around and i think the equation y_o = mg/k
the one to find the difference in the equilibrium points of when there is no mass and when there is a mass.
but i don't know if my algebra is bad but i rearrange and get y*g = k/m which is ω
but i also have seen some do g/y = k/m
but i think it could also be y*g = m/k
and with the right algebraic rearrangement you could use T = 2∏/ω

 

[hypnosam]

Well its g/y=k/m which is the same thing as i said, as I am taking x to be y, as there is no other info supplied. By the fact that omega is 2pif, then by the t equation you used in the first part, omega squared =m/k, from that you can get the time period.

 

[MRMooneyham]

tried your solution, still saying incorrect answer

 

[MRMooneyham]

odd, using what you said to work it out, i got T=.108994, so something is wrong if we got different answers doing the same thing...

nvm i forgot to sqrt the g/y, i did get the same value as you

 

[Doc Al]

The Attempt at a Solution

1/10 = .1
.1/2∏ = .15708

Do this over. You're off by a factor of 10. (And don't round off pi when doing the arithmetic.)

Edit: Oops... I see that truesearch already pointed this out in post #5. And I see that you are piling several problems into the same thread.

 

[hypnosam]

Good :) glad that it helped

 

[Doc Al]

An object is hung on the end of a vertical spring and is released from rest with the spring unstressed. If the object falls 0.17 m before coming to rest, the period of the resulting oscillatory motion will be _____ s.

im am uttlerly lost on this...

As the mass falls, what's conserved?

You can also use the equilibrium position to determine the ratio m/k, but where is that point?

 

[MRMooneyham]

the potential energy of the spring?

 

[Doc Al]

the potential energy of the spring?

The potential energy by itself is not conserved--it changes as the mass drops. But what is conserved?

 

[MRMooneyham]

if your referring to what does not change...g,m,k,T,f

 

[Doc Al]

if your referring to what does not change...g,m,k,T,f

Stick to energy. What other forms of energy are involved as the mass falls?

 

[MRMooneyham]

the total mechanical energy

 

[Doc Al]

the total mechanical energy

Right! So compare the total mechanical energy at the top and bottom of the motion.

 

[MRMooneyham]

how? to do that i need to know k, or omega, or something that I am not given

 

[Doc Al]

how? to do that i need to know k, or omega, or something that I am not given

In order to calculate the period, what do you need? (Look at the formula.) That's what you're going to solve for.

 

[MRMooneyham]

i could get the period from either 1/f or 2pi/omega...

 

[Doc Al]

i could get the period from either 1/f or 2pi/omega...

Well, yeah. But I meant the other formula. (From post #1.)

 

[MRMooneyham]

T=2pi/sqrt(m/k) ?

 

[Doc Al]

T=2pi/sqrt(m/k) ?

Yes, that's the one. If you knew m/k, then you can calculate the period. Use energy conservation to figure out m/k.

 

[MRMooneyham]

thats what I've been trying to do, but I am not sure what way to find m/k
one way is to use the equilibrium displacement equation, y = mg/k, but i didnt get the right answer from there

 

[Doc Al]

one way is to use the equilibrium displacement equation, y = mg/k, but i didnt get the right answer from there

That's certainly one way to solve for m/k. But what would you use for y? Where is the equilibrium position.

Conservation of mechanical energy is another way to solve for m/k. Both methods give the same answer, of course.