Hookes’s Law: PART I determine the spring constant using Hookes’s law: F = - kx
ID: 1502911 • Letter: H
Question
Hookes’s Law:
PART I
determine the spring constant using Hookes’s law:
F = - kx
Where x is the extension that is caused by a mass m hanging at the end of the spring. So the force is just the weight hanging.
This is the data that a student has from an experiment he performs.
Mass (kg)
Spring Length (m)
Force, F = mg (N)
Extension, x (m)
0
0.52
0.1
0.54
0.2
0.57
0.3
0.595
0.4
0.625
0.5
0.655
0.6
0.68
0.7
0.71
0.8
0.74
Complete the data in the table and plot a graph in Excel between Force, F and extension, x.
What does the slope equal?
Determine the spring constant.
PART II
In a second part to the experiment, masses are hung at the bottom of the spring and time taken to complete 10 complete cycles is determined.
The data collected is as follows:
Mass (kg)
Time for 10 cycles (s)
Time period T (s)
T2 (s2 )
0.35
6.59
0.45
7.34
0.55
8.00
0.65
8.71
0.75
9.32
0.85
10.2
0.95
10.6
Complete the table.
It you plot a graph of T2 vs m what is the slope of the graph?
Determine the slope of the graph by plotting it in Excel.
From the slope determine the spring constant.
Based on possible sources of error, which experiment would give a more accurate value for the spring constant?
Calculate the percent difference error between the two values obtained.
Percent difference error = | A – B | * 100 %
(A + B)/2
Mass (kg)
Spring Length (m)
Force, F = mg (N)
Extension, x (m)
0
0.52
0.1
0.54
0.2
0.57
0.3
0.595
0.4
0.625
0.5
0.655
0.6
0.68
0.7
0.71
0.8
0.74
Explanation / Answer
PART I ]
when you plot the graph between Force and extension , it will linear and slope gives (spring constant)
So, spring constant = 32.7 N/m
for PART 2) I don't understand meaning of T2(S2)
Mass(Kg) spring length(m) Force = mg(N) Extension(m) 0 0.52 0 0 0.1 0.54 0.981 0.02 0.2 0.57 1.962 0.05 0.3 0.595 2.943 0.075 0.4 0.625 3.924 0.105 0.5 0.655 4.905 0.135 0.6 0.68 5.886 0.160 0.7 0.71 6.867 0.190 0.8 0.74 7.848 0.220Related Questions
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