Question

How would I calculate the spring constant without knowing the 'x'?

Answer 1

what do you know?

Answer 2

if you know period:
\[T=2\pi\sqrt{m/k}\]
or frequency f=1/T, you can get k

Answer 3

I should have specified that I am doing this as an experiment (dropping a soccer ball) from a certain height and I know the velocity at any time but I can't really see any compression of the ball looking in my video that I've taken of the experiment.

Answer 4

sorry I don't know about deformation of ball

Answer 5

Thank you anyway.

Answer 6

When looking at the video, are there two frames where in which you can definitely tell the ball is going downwards in the first and upwards in the second?

If so, you might be able to finagle some **rough info out of it. If you halve the time in that instance, you can use a kinetic eq to coax out an acceleration

\[ \cancel{v^2}^0 = v_i^2 + 2ay\]
\[a=-\frac{v_i^2}{2y}\]
then
\[ma = F_{avg \ spring}\]
\[m \left(-\frac{v_i^2}{2y} \right) = -\frac{1}{2}mky\]
Then from conservation of energy, you know that all of its kinetic energy comes from potential, and at the point of max compression
\[K=U_{spring}\]
so
\[U_{grav}=U_{spring}\]
\[mgh = \frac{1}{2}ky^2\]


There's a huge amount of error in both of those, but if you're looking for ballpark figures that's a decent solution I think ^_^

Answer 7

nope, that actually doen't work at all. You just eliminate both variables. :P Hmmm...

Answer 8

wait I have acceleration.. I..again forgot to mention that I was using logger pro to calculate the velocity so I copied the data to excel and found the acceleration from the slope of the velocity when then ball was bouncing up. It was about \[81\frac{ m }{ s ^{2} }\]

Answer 9

I think mechanical energy is not conserved here

Answer 10

maybe you can assume it is conserved

Answer 11

what would be mechanical energy in this case?

Answer 12

if you know a, and v before hit, using the first equation of AllTehMaffs you can calculate x, then use it to find k

Answer 13

\[y=v_it + \frac{1}{2}at^2\]
then from above
\[a=-\frac{v_i^2}{2y}\]
\[y^2=v_it y - \frac{1}{4}v_i^2t^2\]
here we go
\[y^2-v_it y + \frac{1}{4}v_i^2t^2=0\]
Now you can quadractic it
\[y_{1,2}=\frac{v_i t ± \cancel{\sqrt{(-v_i t )^2 - v_i^2t^2}}^0}{2}\]
\[y = \frac{v_i t}{2}\]
?

That seems exceptionally wrong!

@bourne13 It's definitely not conserved, you're right, I was just playin' around with em.

Answer 14

There's so much energy lost to sound and to the act of deforming the ball I can't image it'd work out well...

Answer 15

here is the data

Answer 16

yeah, it eats up almost half of the energy each bounce

Answer 17

*please ignore my bad spelling

Answer 18

so is there anything that I could do to find the spring constant? with this data?

Answer 19

If you're using Logger pro you have the velocity too, right?

Answer 20

it's sampling like 35 times a second-ish, yah?

Answer 21

yes

Answer 22

I have velocity and position at any point of time from my video analysis.

Answer 23

can you subtract (the distance closeest :) the ball gets to earth) and radius of the ball

Answer 24

Anyways, right before the first bounce you should definitely have the velocity of the ball, and at h=0 (the point right after .5s), the velocity should absolutely be zero, so if you use the velocity of of the ball right before it hits, and say that your time is about half a half of the sample time (prolly around 1/60 s) then you can use
\[y=\frac{v_i t}{2}\]
and that's how far the ball compressed, then find a spring constant using
\[k=\frac{v_i^2}{y^2}\]
shown above.

You can verify the "y" using the acceleration and
\[a=-\frac{v_i^2}{2y}\]

Answer 25

Remember that's the acceleration of the ball from vi to 0, not from downward velocity to upwards velocity.

Answer 26

in your above derivation why do you have mass for spring force

Answer 27

thanks

Answer 28

because I'm a dumb! ^_^ You're right, it shouldn't be there.

so that makes it
\[k=\frac{mv_i^2}{y^2}\]
right? Then the units are actually correct :P

Answer 29

@AIITehMaffs, you were right the mechanical energy will be conserved