In part one of this experiment, a 0.20 kg mass hangs vertically from a spring an
ID: 1494104 • Letter: I
Question
In part one of this experiment, a 0.20 kg mass hangs vertically from a spring and an elongation below the support point of the spring of 9.50 cm is recorded. With 1.20 kg hanging on the spring, a second elongation of 12.00 cm is recorded. Calculate the spring constant k in newtons per meter (N/m). (Note the equilibrium position is not zero.) k = The same apparatus is used in part two of this experiment. At the elongation of 9.50 cm, a frequency, f, of 300 rev/min is recorded. Then at the elongation of 12.00 cm, a frequency of 400 rev/min is recorded. Calculate the effective mass of the system in kilograms. (Remember to convert to SI. omega = f * 2pi/60 switches from rev/min to rad/sec.)Explanation / Answer
(1)
use the Hook's law,
F1=k1x
k1=m1g/x1 =0.20*9.8 /0.0950 =20.63 N/m
similarly,
k2=m2g/ x2 =1.20*9.8 /0.12 = 98 N/m
Therefore, the average value of the sprign constant is 59.315 N/m