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JEE Mains · Physics · STD 11 - 1. units,dimensions and measurement

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 ___________ %.

  1. A 3.43
  2. B 3.35
  3. C 7.60
  4. D 6.76
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Answer & Solution

Correct Answer

(D) 6.76

Step-by-step Solution

Detailed explanation

\(\frac{\Delta K}{K}=\frac{2\Delta T}{T}+\frac{\Delta m}{m}\) \(T=\frac{60}{50}=1.2~sec\) \(\Delta T=\frac{2}{50}\) \(\therefore \frac{\Delta K}{K}=\frac{2\times2}{50\times1.2}+\frac{10\times10^{-3}}{10}=0.0676\) \(\therefore \%Error=6.76\%\)
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