Finding spring constant of mass

[Smartguy94]

Homework Statement

The left side of the figure shows a light (`massless') spring of length 0.300 m in its relaxed position. It is compressed to 67.0 percent of its relaxed length, and a mass M= 0.190 kg is placed on top and released from rest (shown on the right).

[URL]http://loncapa.gwu.edu/res/msu/physicslib/msuphysicslib/13_EnergyConservation/graphics/prob24_CompSpring.gif[/URL]

The mass then travels vertically and it takes 1.30 s for the mass to reach the top of its trajectory. Calculate the spring constant, in N/m. (Use g=9.81 m/s2). Assume that the time required for the spring to reach its full extension is negligible.

Homework Equations

Fs=-kx
Us=(1/2)kx^2

The Attempt at a Solution

first off this is the data that i use

l(relax)=.3
l(tensed)=.201
so the X for spring = .3-.201 = .099
x=.099
m=.190
V(final)=0
t=1.3
a=-9.81
k=?

i found V(initial)

vFinal=vInitial+at
vInitial = 12.753

next i found the distance

x = v(initial) + (1/2)at^2
x= 8.2979

then i put

K+U(gravity)+U(spring)=Total energy
15.45+15.466+.0049K=0
K=6308.74

and it is wrong... I'm really confused

 

[SammyS]

Yes, it's wrong.

x = v(initial)t + (1/2)at² is for constant acceleration.

Use y = -Acos(2πt/T) , where A is the amplitude of the motion and T is the period.

You should be able to get A and T from the given information.

 

[Smartguy94]

Yes, it's wrong.

x = v(initial)t + (1/2)at² is for constant acceleration.

Use y = -Acos(2πt/T) , where A is the amplitude of the motion and T is the period.

You should be able to get A and T from the given information.

I don't think i have learned how to get A and T yet man..

i know T=2∏√(m/k)

but how can I get the K if that's what the question is asking?

beside T=2∏√(m/k) i believe i haven't learned any other equation yet that is relating to T, and I have no idea what A is..

 

[SammyS]

T=2∏√(m/k) will work just fine. Solve it for k.

 

[Smartguy94]

T=2∏√(m/k) will work just fine. Solve it for k.

I'm honestly quite confused on how to solve for k when I don't know the value of T

 

[SammyS]

How would you describe on period of the motion of this spring & mass combination?

(You are given information in this problem from which you can find T.)

 

[NascentOxygen]

i found V(initial)

vFinal=vInitial+at
vInitial = 12.753

next i found the distance

x = v(initial) + (1/2)at^2
x= 8.2979

What is this x, and why do you need it? If you don't explain what you are doing, it's difficult to say whether you are on the right track or not.

K=6308.74

I used the same basic method as you, but I found k a bit under 1800 N/m

Work done by spring = Integral from 0.099 to 0 of k.x

Equate this to the (K.E. of the mass with your vInitial) + (mgh for 0.099m rise)

 

[NascentOxygen]

next i found the distance

x = v(initial) + (1/2)at^2
x= 8.2979

then i put

K+U(gravity)+U(spring)=Total energy
15.45+15.466+.0049K=0

It seems to me that you might be counting the same energy twice here.
K.E. + U(gravity) are constant, once the mass has lost contact with the spring. You don't sum their maximum values. You either include K.E. when U is zero, or include U at the peak of the trajectory where speed is zero.

 

[NascentOxygen]

After having pondered this puzzle some more, I have are a few comments.

First, what I outlined above is not a complete answer. I should subtract from the 1.3 secs for flight that length of time during which the spring acts on the mass. It will be a small time, but that is no basis on which to overlook it. Unless we are being instructed to do so, viz., "Assume that the time required for the spring to reach its full extension is negligible."

Other respondents seem to view the arrangement as a spring bob, with the mass attached to the spring and giving SHM. I read it as a spring flinging the mass into the air, and the two becoming separated early in the piece. It would have helped had Smartguy94 included the diagram he was given which undoubtedly would clear this up.