JEE Main 2026 — Jan 28, Shift 2 — Physics solution

JEE Main2026Jan 28, Shift 2Physics

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

A
3.43
B
3.35
C
7.60
D
6.76

Solution

From $T = 2 π m / k$ , $k = \frac{4 π ^{2} m}{T ^{2}}$ , so $\frac{Δ k}{k} = \frac{Δ m}{m} + 2 \frac{Δ T}{T}$ .

$T = \frac{60}{50} = 1.2$ s; $Δ T = \frac{2}{50} = 0.04$ s. So $2 \frac{Δ T}{T} = \frac{2 \times 0.04}{1.2} = \frac{0.08}{1.2} \approx 0.0667$ .

$\frac{Δ m}{m} = \frac{10 \times 1 0 ^{- 3}}{10} = 0.001$ .

$\frac{Δ k}{k} \approx 0.0667 + 0.001 = 0.0676 \Rightarrow 6.76%$ .

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