Question

A string with a total length of 270.8 0.05 cm has a total mass of 12.0 +0.1g. It is fixed at one end and stretched between that end and a pulley, two points that are 159.6 0.05 cm apart, by hanging a mass of 200.0 +0.1 g. When the string is removed from the hanging mass, that portion of the string that was between the two fixed points is 145.0 ± 0.05 cm. From the above information show the following calculations 1. The linear mass density of the stretched string: 2. The percent uncertainty in the linear mass density of the stretched string

Answer 1

(145.0 Plusminus 0.05) Cm of unstretched string corresponds to (159.6 plusminus 0.05)Cm of Stretched string Therefore 270.8 cm of unstretched string = 270.8 times 159.6/145.0 cm of stretched string. Linear mass density of the stretched string Mu = 12.0 times 145.0/270.8 times 159.6 g/cm Mu = 4.06 times 10^-2 g/cm Delta Mu/Mu = 0.1/12.0 + 0.05/145.0 + 0.05/270.8 + 0.05/159.6 Delta Mu/Mu = 0.009 percent uncertainty = 0.9%