Errors in measurementMedium
Question
The time period of a simple harmonic oscillator is $T = 2\pi\sqrt{\frac{k}{m}}$. The measured value of mass (m) of the object is 10 g with an accuracy of 10 mg , and time for 50 oscillations of the spring is found to be 60 s using a watch of 2 s resolution. Percentage error in determination of spring constant(k) is %.
Options
A.3.43
B.3.35
C.7.60
D.6.76
Solution
$\frac{\Delta K}{K} = \frac{2\Delta\text{ }T}{\text{ }T} + \frac{\Delta m}{m}$
$${T = \frac{60}{50} = 1.2sec }{\Delta T = \frac{2}{50} }{\therefore\frac{\Delta K}{K} = \frac{2 \times 2}{50 \times 1.2} + \frac{10 \times 10^{- 3}}{10} = 0.0676 }$$∴ % Error $= 6.76$ %
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