Question

For a linear spring the force is proportional to the stretch. F =kx k is the elastic constant of the spring in N/m

Masses are placed on a spring and the stretch is measured (no bouncing). Then the spring was allowed to bounce and the period was measured.

mass Stretch(cm) Period (s)

0.15 kg 9 0.6

0.25 15 0.76

0.35 20.5 0.9

0.45 26.5 1.0

Calculate k for the spring. Be careful mass must be converted to weight in Newtons and the stretch must be in meters.

Plot a graph of period squared (y) vs. mass. Calculate the slope. The slope is 4 pi² / k. Calculate k from the slope.

Get the % difference between the two k's.

Answer 1

Mass	Stretch (cm) =X	Force = massX g(9.8) Force = Kx	K N/m
.15	9		16.33
.25	15		16.33
.35	20.5		16.73

.45	26.5		16.64
		Average	16.50

Hence from the above data K = 16.50 Newton/ meter = K(Theoretical)

From the above graph slope comes out to be : 17.085 = K(Observed)

Hence K = 17.08Newton/metre

% error = [K(Theo) -K(Observed)]/K(Observed) * 100

= [16.50-17.08]/17.08   *100

=3.39 %

{Note : We have neglected -negative sign}