# Help with understanding an experiment with masses on a spring

1. Dec 14, 2013

We did an experiment not so long ago that involved suspending masses on a spring and not knowing much about the background to what was found is hindering my ability to fully grasp the results. The experiment was not part of any assessment so I thought this forum would be best but feel free to move if necessary.

I knew of Hooke's law before hand, F=-kx, but that was it, kind of lacking knowledge with the rest of theory that is involved.

For the first part we simply added masses onto the spring, one by one (0.05kg incremenets) and measured the extension. Plotted a graph of the weight of the masses versus the extension and then drew a line of best fit. Found the spring constant by taking the reciprocal of the gradient.

For the second part, we did the same thing by adding masses incrementally (in 0.05kg increments) but this time we lifted the masses slightly and let them bounce and we timed 30 oscillations (to get better accuracy of one period). We then calculated what ω^2 is by (2π/T)^2 and plotted a graph of the reciprocal of the masses (1/m) against ω^2 and drew a best fit line. And found the gradient to get a different value for k the spring constant to compare with the previous value.

I just dont understand how the reciprocal of the gradient in part one gets you the spring constant and in part two, which I especially dont understand how doing what we did provided us with a another value for the spring constant.

Any help with a general understanding of the theory behind this experiement would be very much appreciated. Merry Christmas :)

2. Dec 14, 2013

### sophiecentaur

The first experiment just shows how metals follow Hooke's Law. k = gΔm/Δx
The second shows how the period of oscillation of a simple harmonic oscillator depends upon mass and spring constant.
T (period) = 2∏√(m/k). This is a more advanced bit of the course and involves some harder Maths.
You say there was some disagreement between the two experiments but you don't say by how much. The mass of the spring will affect the natural oscillation frequency (the spring's own mass also has to be pulled about by its stiffness.). Each coil of the spring moves a different amount as the spring flexes and each piece will have a mass. The total effect of all the pieces (and the stiffnesses of the various lengths of spring where each coil sits) will be like another added mass. - an 'equivalent extra mass'. The end portion of the spring will have more effect than the higher portions.

You could calculate this equivalent mass by assuming that k is the same in both cases. If it goes as I'm suggesting, this equivalent mass would be something less than the actual mass of the spring (can you get a spring and measure its mass?)

3. Dec 14, 2013

The two values do technically agree if you take the error in k into account but the error in k for the first part was like 1.5, calculated using a 'line of worst fit' which wasn't the most accurate thing ever. If the two absolute values were looked at there was a difference of 1.12 between them.

I know it is quite a big difference, it was not the most accurate experiment all together but it was interesting and fun t do nonetheless. Even more so now I am beginning to get an idea of what actually was going on.

Thanks.

4. Dec 15, 2013

Right ok I think I get that bit. Since F=kx, k=F/x. And in this case the force is the weight, mg, of the masses? I can see now why the gradient can produce the spring constant, thanks.

In this equation, is the m the mass of the masses attached to the spring, or the mass of the spring itself?

Also, I have seen the equation $f=\frac{1}{2\pi}\sqrt{\frac{m}{k}}$ on the web relating to harmonic oscillators, is this just a different permeation of the one you stated but using the frequency instead of the period?

Last edited: Dec 15, 2013
5. Dec 15, 2013

### sophiecentaur

It is always a good idea to read around a subject as well as asking questions on a Forum. The formula for frequency of a mass on a spring refers to an ideal, massless spring with a mass one end and clamped firmly, at the other.
As Period = 1/Frequency (by definition), either way round is OK and the one you use will depend upon the circumstances.
My comments about the spring mass were applied to a real spring (the one you used) and, if you take the spring mass into account, can sometimes explain the difference between the two measurements of k. If the error is as high as yours, the spring mass is not enough to account for the difference, IMO. Find out the mass of the spring and see for yourself how close you can get by including the spring mass, as I described.
Perhaps my comments have confused you rather than enlightened you -sorry. Later, they may make more sense to you.

6. Dec 15, 2013