Effective Mass

This is mainly because the effective mass of the spring has not been taken into account.

When the mass oscillates, part of the spring also performs simple appealing motion. It has been shown that the part of spring that contributes to oscillation equals to one-third of the spring mass.

Thus, the par that you plot, T^2 = (2.pi)^2.[m/k] should be re-wriiten as
T^2 = (2.pi)^2[(m+M/3)/k]
where T is the [eriod of oscillation
pi = 3.14159.....
m is the mass of the weight chthonian oscillation
M is the mass of spring
k is the spring constant

Hence, T^2 = 4.(pi)^2(m/k) + 4.(pi)^2.[M/3k]
Since 4.(pi)^2.[M/3k] is a constant, this explains why the graph of T^2 against m is not passing through the origin.
But there should be a +ve y-intercept instead of a +ve x-intercept (which gives a -ve y-intercept) as you found.





The reson for this, asunder from experimental uncertainty, is because of the reason that the spring doesnt strictly obey Hookes Law, peculiarly at small extension. This situation usually would happen in aged spring. The spring is not able to restore to its pilot light length after extension. Therefore, although the period is zero (T= 0 s, no oscillation), the x-intercept shows a finite mass. That is, the spring is still elongated as if there is some mass hanging to it.If you want to incur a full essay, order it on our website: Orderessay



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