The time period of a simple harmonic oscillator is T=2 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 2 s resolution. Percentage error in determination of spring constant (k) is _ _ _ _ %.

Step-by-step solution

The given formula for time period is T = 2 k m . Squaring both sides, we get T^2 = 4 ^2 k m , which gives k = 4 ^2 m T^2 . The relative error in k is given by k k = m m + 2 T T . Given: m = 10 g, m = 10 mg = 0.01 g. Total time for n=50 oscillations is t = 60 s with resolution t = 2 s. Since T = t/n , the relative error in T is the same as in t : T T = t t . Substituting the values: k k = 0.01 10 + 2 ( 2 60 ) . k k = 0.001 + 2 30 = 0.001 + 0.0666... Percentage error = (0.001 + 1 15 ) 100 = 0.1 + 100 15 = 0.1 + 6.666