The time period of a simple harmonic oscillator is T = 2 mk. 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 %.
- 3.43
- 7.60
- 3.35
- 6.76
Answer
(4) 6.76
Solution
T = 2 mk k = 4 2 mT2
kk 100 = mm 100 + 2 TT 100
mm 100 = 10 mg10 g 100 = 0.0110 100 = 0.1\%
TT = tt = 260 (t = time for 50 oscillations) TT 100 = 3.33\%
kk 100 = 0.1\% + 2 3.33\% = 6.76\%
Answer: (4)