Question

Just need help answering questions at end

Prelab for Simple Harmonic Motion Due: Beginning of lab Name: Fill in the missing data from the data table for Hooke's Law. The equilibrium position of the car is at Xetal 10cm Hanging Mass [kg] F-mg [N] Car Position [cm] x(cm 20-10, 10 0.03 0.03 x45 0.24920 30 20 0.ss 0.06 0.09 0.12 0.15 0. 8 40 1,176 150 り 160 46 니)

Answer 1

the equilibrium position of car is X initial = 10 cm

F = mg = 0.03*9.8 N = 0.294 N , 20-10=10 cm

for m = 0.06, F = mg = 0.06*9.8 N = 0.588 N , 30-10=20 cm

for m = 0.09, F = mg = 0.09*9.8 N = 0.882 N , 40-10=30 cm

for m = 0.12, F = mg = 0.12*9.8 N = 1.176 N , 50-10=40 cm

for m = 0.15, F = mg = 0.15*9.8 N = 1.47 N , 60-10=50 cm

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F= k*Dx

k = F/Dx

looks like a straight line equation passing through the rogin (y = mx)

F = k*Dx ==> k = F/Dx

from the graph F vs Dx , the slope is k = F/Dx

F is the force measured in newtons and DX is the stretch or compression in the spring measured in meters, (in S.I system)

the units of the spring constant is N/m, but from the given data (graph) the unit is N/cm

Let the points on the line are  

(x2,y2) = (0.50,1.4) , (x1,y1) = (0.30,0.8)

the slope is k = (y2-y1/x2-x1) = (1.4-0.8)/(0.50-0.30) = 3 N/m