Experiment # 7
Simple Harmonic Motion
OBJECTIVE/ INTRODUCTION:
In this experiment, the objective is to understand the natural equilibrium position of
a spring. We will then attach a mass M to the spring leading to its new equilibrium position,
which we will call the hanging equilibrium. We can also observe these forces with a free
body diagram of the suspended mass indicating how these three forces are acting on the
spring. With the suspended mass in equilibrium we will be able to use Newton’s second law,
other given equations along with the data to formulate the average mass density and to obtain
a constant of proportionality. In the lab the main concept is to visualize the horizontal spring
being attached to a mass. As we manipulate the masses affecting its equilibrium we can draw
conclusions based on the effect it has along with concluded calculations.
Calculations & Analysis:
Data sheet
Measuring the Spring Elongation
M
𝛥𝛥𝛥𝛥
(kg)
(m)
0.050
0.100
0.150
0.200
0.250
0.300
0.350
1𝑚𝑚
� = 0.019𝑚𝑚
1.9𝑐𝑐𝑐𝑐 = (1.9𝑚𝑚) �
100𝑐𝑐𝑐𝑐
1𝑚𝑚
� = 0.038𝑚𝑚
3.8𝑐𝑐𝑐𝑐 = (3.8𝑐𝑐𝑐𝑐) �
100𝑐𝑐𝑐𝑐
1𝑚𝑚
� = 0.053𝑚𝑚
5.3𝑐𝑐𝑐𝑐 = (5.3𝑐𝑐𝑐𝑐) �
100𝑐𝑐𝑐𝑐
1𝑚𝑚
� = 0.067𝑚𝑚
6.7𝑐𝑐𝑐𝑐 = (6.7𝑐𝑐𝑐𝑐) �
100𝑐𝑐𝑐𝑐
1𝑚𝑚
� = 0.085𝑚𝑚
8.5𝑐𝑐𝑐𝑐 = (8.5𝑐𝑐𝑐𝑐) �
100𝑐𝑐𝑐𝑐
1𝑚𝑚
� = 0.099𝑚𝑚
9.9𝑐𝑐𝑐𝑐 = (9.9𝑐𝑐𝑐𝑐) �
100𝑐𝑐𝑐𝑐
1𝑚𝑚
� = 0.115𝑚𝑚
11.5𝑐𝑐𝑐𝑐 = (11.5𝑐𝑐𝑐𝑐) �
100𝑐𝑐𝑐𝑐
M(kg)
𝛥𝛥𝛥𝛥(𝑚𝑚)
0.050
0.019
0.100
0.038
0.150
0.053
0.200
0.067
0.250
0.085
0.300
0.099
0.350
0.115
Using excel we get the graph of Δy vs M, and the equation of the line that best fit(regression).
dy(m)
dy(m) vs M(kg)
0.2
0.1
0
0.019
0
0.038
0.1
0.053
0.067
0.085
0.2
0.099
0.115
0.3
0.4
M(kg)
𝑦𝑦 = 0.005 + 0.316𝑥𝑥
This graph has 99.94% correlation, and slope m = 0.316
𝑺𝑺𝑺𝑺𝑺𝑺𝑺𝑺𝑺𝑺 = 𝟎𝟎. 𝟑𝟑𝟑𝟑𝟑𝟑
𝒌𝒌𝒔𝒔𝒔𝒔,𝟏𝟏
𝒎𝒎
𝒌𝒌𝒌𝒌
𝒎𝒎
𝟗𝟗. 𝟖𝟖𝟖𝟖
𝒈𝒈
𝒔𝒔² = 31.074 𝒌𝒌𝒌𝒌(𝒎𝒎) � 1 � = 𝟑𝟑𝟑𝟑. 𝟎𝟎𝟎𝟎𝟎𝟎 𝑵𝑵
=
=
𝒔𝒔𝒔𝒔𝒔𝒔𝒔𝒔𝒔𝒔 𝟎𝟎. 𝟑𝟑𝟑𝟑𝟑𝟑 𝒎𝒎
𝒔𝒔𝟐𝟐
𝑚𝑚
𝒎𝒎
𝒌𝒌𝒌𝒌
𝑴𝑴𝑻𝑻𝑻𝑻𝑻𝑻
Measuring the period 𝝉𝝉
Trial 1 Trial 2
Trial 3 𝝉𝝉𝒂𝒂𝒂𝒂𝒂𝒂 = 𝝉𝝉𝟏𝟏 +𝝉𝝉𝟐𝟐 +𝝉𝝉𝟑𝟑 = (s)
𝟑𝟑
(𝝉𝝉𝒂𝒂𝒂𝒂𝒂𝒂 )²
(kg)
𝝉𝝉𝟏𝟏 (𝒔𝒔)
𝝉𝝉𝟐𝟐 (𝒔𝒔)
(s)²
𝝉𝝉𝟑𝟑 (𝒔𝒔)
0.250 0.625
0.616
0.616
0.300 0.662
0.658
0.664
0.350 0.708
0.708
0.708
0.400 0.757
0.753
0.752
0.450 0.793
0.798
0.791
0.500 0.835
0.826
0.829
𝝉𝝉𝒂𝒂𝒂𝒂𝒂𝒂,𝟏𝟏 =
𝟎𝟎.𝟔𝟔𝟔𝟔𝟔𝟔+𝟎𝟎.𝟔𝟔𝟔𝟔𝟔𝟔+𝟎𝟎.𝟔𝟔𝟔𝟔𝟔𝟔
𝝉𝝉𝒂𝒂𝒂𝒂𝒂𝒂,𝟑𝟑 =
𝟎𝟎.𝟕𝟕𝟕𝟕𝟕𝟕+𝟎𝟎.𝟕𝟕𝟕𝟕𝟕𝟕+𝟎𝟎.𝟕𝟕𝟕𝟕𝟕𝟕
𝝉𝝉𝒂𝒂𝒂𝒂𝒂𝒂,𝟓𝟓 =
𝟎𝟎.𝟕𝟕𝟕𝟕𝟕𝟕+𝟎𝟎.𝟕𝟕𝟕𝟕𝟕𝟕+𝟎𝟎.𝟕𝟕𝟕𝟕𝟕𝟕
𝟑𝟑
𝝉𝝉𝒂𝒂𝒂𝒂𝒂𝒂,𝟐𝟐 =
𝟎𝟎.𝟔𝟔𝟔𝟔𝟔𝟔+𝟎𝟎.𝟔𝟔𝟔𝟔𝟔𝟔+𝟎𝟎.𝟔𝟔𝟔𝟔𝟔𝟔
𝝉𝝉𝒂𝒂𝒂𝒂𝒂𝒂,𝟒𝟒 =
𝟎𝟎.𝟕𝟕𝟕𝟕𝟕𝟕+𝟎𝟎.𝟕𝟕𝟕𝟕𝟕𝟕+𝟎𝟎.𝟕𝟕𝟕𝟕𝟕𝟕
𝝉𝝉𝒂𝒂𝒂𝒂𝒂𝒂,𝟔𝟔 =
𝟎𝟎.𝟖𝟖𝟖𝟖𝟖𝟖+𝟎𝟎.𝟖𝟖𝟖𝟖𝟖𝟖+𝟎𝟎.𝟖𝟖𝟖𝟖𝟖𝟖
𝟑𝟑
𝟑𝟑
𝟑𝟑
𝟑𝟑
𝟑𝟑
=0.619 s
=0.661 s
=0.708 s
=0.754 s
=0.794 s
=0.830 s
(𝝉𝝉𝒂𝒂𝒂𝒂𝒂𝒂𝒂𝒂 )² = (𝟎𝟎. 𝟔𝟔𝟔𝟔𝟔𝟔 𝒔𝒔)² = 𝟎𝟎. 𝟑𝟑𝟑𝟑𝟑𝟑 𝒔𝒔²
(𝝉𝝉𝒂𝒂𝒂𝒂𝒂𝒂𝒂𝒂 )² = (𝟎𝟎. 𝟕𝟕𝟕𝟕𝟕𝟕 𝒔𝒔)² = 𝟎𝟎. 𝟓𝟓𝟓𝟓𝟓𝟓 𝒔𝒔²
(𝝉𝝉𝒂𝒂𝒂𝒂𝒂𝒂𝒂𝒂 )² = (𝟎𝟎. 𝟕𝟕𝟕𝟕𝟕𝟕 𝒔𝒔)² = 𝟎𝟎. 𝟓𝟓𝟓𝟓𝟓𝟓 𝒔𝒔²
(𝝉𝝉𝒂𝒂𝒂𝒂𝒂𝒂𝒂𝒂 )² = (𝟎𝟎. 𝟖𝟖𝟖𝟖𝟖𝟖 𝒔𝒔)² = 𝟎𝟎. 𝟔𝟔𝟔𝟔𝟔𝟔 𝒔𝒔²
t(s^2)
(𝝉𝝉𝒂𝒂𝒂𝒂𝒂𝒂𝒂𝒂 )² = (𝟎𝟎. 𝟔𝟔𝟔𝟔𝟔𝟔 𝒔𝒔)² = 𝟎𝟎. 𝟒𝟒𝟒𝟒𝟒𝟒 𝒔𝒔²
0.8
0.383
0.4
0.437
0.501
0.567
0.63
0
Using
0.1
excel
0.2
0.3
we
𝟎𝟎. 𝟓𝟓𝟓𝟓𝟓𝟓 𝒔𝒔²
𝟎𝟎. 𝟓𝟓𝟓𝟓𝟓𝟓 𝒔𝒔²
𝟎𝟎. 𝟔𝟔𝟔𝟔𝟔𝟔 𝒔𝒔²
𝟎𝟎. 𝟔𝟔𝟔𝟔𝟔𝟔 𝒔𝒔²
0.4
get
0.5
the
M(kg)
(𝝉𝝉𝒂𝒂𝒂𝒂𝒂𝒂 )²
0.250
0.383
0.300
0.437
0.350
0.501
0.400
0.567
M(kg)
0.450
0.630
0.6
0.500
0.689
0.689
0.2
0
𝟎𝟎. 𝟒𝟒𝟒𝟒𝟒𝟒 𝒔𝒔²
(𝝉𝝉𝒂𝒂𝒂𝒂𝒂𝒂𝒂𝒂 )² = (𝟎𝟎. 𝟕𝟕𝟕𝟕𝟕𝟕 𝒔𝒔)² = 𝟎𝟎. 𝟔𝟔𝟔𝟔𝟔𝟔 𝒔𝒔²
ĩ² vs Mass
0.6
𝟎𝟎. 𝟑𝟑𝟑𝟑𝟑𝟑 𝒔𝒔²
graph
of
Using excel we got the graph and the best fit for the equation of the line of (𝝉𝝉𝒂𝒂𝒂𝒂𝒂𝒂 )² vs M.
𝒔𝒔𝟐𝟐
𝑦𝑦 = 0.068 + 1.243𝑥𝑥 where the slope 𝒎𝒎 = 𝟏𝟏. 𝟐𝟐𝟐𝟐𝟐𝟐 (𝒌𝒌𝒌𝒌)
𝒌𝒌𝒔𝒔𝒔𝒔,𝟐𝟐
𝟒𝟒𝝅𝝅𝟐𝟐
𝟒𝟒𝝅𝝅𝟐𝟐
𝑵𝑵
=
=
= 𝟑𝟑𝟑𝟑. 𝟕𝟕𝟕𝟕𝟕𝟕
(𝒎𝒎)
𝒔𝒔𝒔𝒔𝒔𝒔𝒔𝒔𝒔𝒔 𝟏𝟏. 𝟐𝟐𝟐𝟐𝟐𝟐
%𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸 = �
CONCLUSION:
k sp,2 − k sp,1
(31.761 − 31.074)N
� x 100% = �
� x 100% = 𝟐𝟐. 𝟏𝟏𝟏𝟏𝟏𝟏%
k sp,2
31.761𝑁𝑁
In this experiment, we calculated the spring constant 𝒌𝒌𝒔𝒔𝒔𝒔,𝟏𝟏 using the mass M(kg) and
the change in the vertical component Δy. We used Microsoft Excel to approximate the slope
of the line the best fit the graph of Δy vs M(kg), from there it was easy to calculate the spring
𝒈𝒈
constant 𝒌𝒌𝒔𝒔𝒔𝒔,𝟏𝟏 = 𝒔𝒔𝒔𝒔𝒔𝒔𝒔𝒔𝒔𝒔. For the second part we had to calculate the spring constant 𝒌𝒌𝒔𝒔𝒔𝒔,𝟐𝟐 using
the average period square vs the mass(M). Same process was use, Microsoft Excel to
𝟒𝟒𝝅𝝅𝟐𝟐
calculate the slope, and then using 𝒌𝒌𝒔𝒔𝒔𝒔,𝟐𝟐 = 𝒔𝒔𝒔𝒔𝒔𝒔𝒔𝒔𝒔𝒔 to calculate the spring constant 𝒌𝒌𝒔𝒔𝒔𝒔,𝟐𝟐 . At last
we calculate the %Error between the two calculated spring constant which was a small error
of plus or minus 2.164%. Since the error is not significant, we can conclude that we have a
good approximation of the spring constant.