Effective Mass

This is mainly because the effective portion of the arising has not been taken into account. When the mass oscillates, equatingt of the spring also performs simple openhearted 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 observational uncertainty, is because of the reason that the spring doesnt strictly obey Hookes Law, pecu liarly at small extension. This situation usually would happen in vulcanized spring. The spring is not able to restore to its pilot animated 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 baffle a full essay, order it on our website: BestEssayCheap.com

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