The time period of a simple harmonic oscillator is T = 2π√(m/k)… | JEE Main 2026 (28 Jan, Shift 2)

Q35JEE Main 2026 · 28 JAN · SHIFT 2Difficult

Errors in Measurement

The time period of a simple harmonic oscillator is $T = 2\pi\sqrt{\frac{m}{k}}$ . 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 ________%.

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Answer: (d)

$T = 2\pi\sqrt{m/k} \Rightarrow k = \dfrac{4\pi^2 m}{T^2}$ , so $\dfrac{\Delta k}{k} = \dfrac{\Delta m}{m} + 2\dfrac{\Delta T}{T}$ .
$\dfrac{\Delta m}{m} = \dfrac{10\text{ mg}}{10\text{ g}} = 0.1\%$
$\dfrac{\Delta T}{T} = \dfrac{\Delta t}{t} = \dfrac{2}{60} = 3.33\%$ , so $2\dfrac{\Delta T}{T} = 6.67\%$
Total $\approx 0.1\% + 6.67\% \approx 6.76\%$ , answer (d).

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