Effective Mass

This is mainly because the effective troop of the bounciness has not been interpreted into account. When the mass oscillates, tallyt of the spring also performs simple harmonic 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 elderly spring. The spring is not able to restore to its pilot thinly 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 come up a full essay, order it on our website: BestEssayCheap.com

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