Question

An unstretched vertical spring has length of L1 = 7.35 ± 0.05 cm. A mass is hung on the spring which then stretches to a length L2 = 12.50 ± 0.05 cm. If the mass added was 500.0 ± 0.1 g, calculate the spring constant k and its uncertainty. Show your workings.

Answer 1

The unstretched length is L₁ = 7.35cm ± 6.8% (0.05* 100%/7.35 = 0.68% )

Streched length L₂ = 12.50 ± 0.4 %

Mass m = 500 ± 0.02%

According to Hooke law

$k = \frac{mg}{\Delta L}$

$k = \frac{(500)(980)}{12.5-7.35} = 95145.6 dynes /cm$

to calculate the uncertainity of the demonimator we should find the relative uncertainity

(0.05+ 0.05)100% / (12.5 - 7.35) = 1.94%

Uncertainity in force constant

$k= 0.02\% + 1.94\% = 1.96\%$

The spring constant is

k = 95145.6 ± 1.96%

or

k = 95145.6 ± 1847 dynes /cm