The time period of a simple harmonic oscillator is $T = 2\pi\sqrt{\dfrac{k}{m}}$. 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 ______ %.
Answer: (D) 6.76
(For a spring–mass system the period is really $T = 2\pi\sqrt{m/k}$; either way $k \propto m\,T^{\pm 2}$, so the percentage error is the same.)
$$\frac{\Delta k}{k}\times100 = \frac{\Delta m}{m}\times100 + 2\,\frac{\Delta T}{T}\times100$$
Mass: $\dfrac{\Delta m}{m} = \dfrac{10\ \text{mg}}{10\ \text{g}} = 0.001 \Rightarrow 0.1\%$.
Time: for 50 oscillations, $\dfrac{\Delta t}{t} = \dfrac{2}{60}$, and the same fractional error applies to $T = t/50$, i.e. $3.33\%$.
$$\frac{\Delta k}{k}\times 100 = 0.1 + 2(3.33) \approx 6.77\% \approx 6.76\%$$
Solution by Sreeraj P, M.Sc Physics