The time period of a simple harmonic oscillator is
T = 2π√(m/k).
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 i $s _______$ %.

Question

The time period of a simple harmonic oscillator is

T = 2π√(m/k).

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 i $s _______$ %.

G1 Physics · Accepted Answer

From T = 2π√(m/k), we get k = 4π²m/T². Therefore the maximum fractional error is Δk/k = Δm/m + 2ΔT/T. Given, m = 10 g, Δm = 10 mg = 0.010 g. Thus, Δm/m = 0.010/10 = 0.001. Time for 50 oscillations is 60 s, so T = 60/50 = 1.2 s. The watch resolution is 2 s for the total measured time. Therefore the error in one period is ΔT = 2/50 = 0.04 s. Hence, Δk/k = 0.001 +2(0.04/1.2) = 0.06767. Therefore, percentage error ≈ 6.76%. Correct option: D.