The time period of a simple harmonic oscillator is $T=2\pi \sqrt{\frac{k}{m}}.$ 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 2s resolution. Percentage error in determination of spring constant (k) is _____ %.

[JEE Main 2026, 28 Jan (Shift 2)]

Question

The time period of a simple harmonic oscillator is $T=2\pi \sqrt{\frac{k}{m}}.$ 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 2s resolution. Percentage error in determination of spring constant (k) is _____ %.

[JEE Main 2026, 28 Jan (Shift 2)]

YESPYQ · Accepted Answer

Correct answer: d) 6.76. $\frac{\Delta K}{K}=\frac{2\Delta T}{T}+\frac{\Delta m}{m}$ $T=\frac{60}{50}=1.2\sec$ $\Delta T=\frac{2}{50}$ $∴\text{ }\frac{\Delta K}{K}=\frac{2\times 2}{50\times 1.2}+\frac{10\times {10}^{−3}}{10}=0.0676$ $∴$ %Error = 6.76%

The time period of a simple harmonic oscillator is \(T=2\pi \sqrt{\frac{k}{m}}.\) Measured value of mass ( m ) of the ob…

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