Simple Harmonic
Motion
7/2/2014
Ting Yeh
Yi Shen
Purpose:
In this experiment we means to find the spring constant in two different ways- static and dynamic.
In the first part of the experiment, we measured the displacement of the spring with different mass
2 to get the static constant. In part two we measured the period of simple harmonic motion of
releasing mass with consideration of the mass of spring in order to get dynamic constant.
Equipment:

Spring

Four weights (m1, m2, m3, and m4)

Ruler

Leveling instrument (000935)

Stand

Clamp
Data and analysis:
Table 1. Static case of Simple Harmonic motion
M2 (kg)
Xi (m)
0.20044
0.1
0.02
0.07077
0.22
0.12
0.19
0.27
0.5
Xf (m)
0.2
0.2
0.2
0.2
0.2
0.2
0.2
0.2
0.2
0.283
0.241
0.208
0.222
0.289
0.249
0.268
0.301
0.401
Table 2. Dynamic case of Simple Harmonic Motion
M (kg)
t (10 cycles) (s)
T=t/10 (s)
T^2 (s^2)
0.200
5.59
0.559
0.312
0.100
3.69
0.369
0.136
0.120
4.37
0.437
0.191
0.070
3.37
0.337
0.114
0.220
5.75
0.575
0.331
0.300
7.10
0.710
0.504
0.320
7.09
0.709
0.503
0.290
6.74
0.674
0.454
0.500
8.81
0.881
0.776
M is the mass of the load in the spring system
t is the time required for the mass spring system to complete 10 cycles
T is the period of the mass spring system
X is how much the spring stretched.
Standard curve between Mass (M2) and displacement (x).
Displacement X is y axis; M2 is x axis.
Standard curve between Mass and Period squared (T^2).
T^2 is y axis; m is x axis.
For simple Harmonic Motion
F   KX
2
d x
 KX  ma  m
dt

dt
2

K
x0
m
k
m

T
2
2
d x
2

Therefore: T  2
m
K
For Part I:
m g  kx  k  0.20044 * 9.76 / .083  23 .6n / m
2
g
x  m2
k
g
slope 
k
g
9.76
K


 24 .3n / m
static slope 0.4023
g  9.7946 m s 2
In Part II:
m
m  1 / 3m
4 2m
4 2m
eff
spring
spring
2
T  2
 2
T 

k
k
k
3k
assume :
T 2  am  b
slope  a 
e% 
4 2
4 2
K

 4 2 / 1.5836  24 .9n / m
dynamics
k
a
kd  k s
24 .9  24 .3
*100 % 
*100 %  2.4%
kd  ks
24 .3  24 .9
2
2
For the mass of the spring:
4 2 m
3k b
s  m  d  3 * 24 .9 * 0.004  0.007568 kg
s 4 2
3k
4 2
d
mmea su re  0.0281 kg
b
e% 
m
m
calculate
measure *100 %  0.007568  0.0281 *100 %  73 .1%
m
0.0281
measure
Note: that b is really small indicates that spring mass can be ignored
Discussion:
In this experiment, we got the static K=24.3 N/m and the dynamic K=24.9N/m respectively with an
error of 2.00%. It seems like the percentage error between the static and dynamic spring constant is fine.
However, once we consider the mass of the spring into the equation and redo the mass of it, the
percentage error comes up with 73.1%. As we noted, which can also be seen in the graph of T^2-m,
that, the b is relatively small. This fact indicates that the mass of the spring does not make significant
change in our calculation of dynamic k. After discussion in our group we concluded that:one of the
ways to avoid the error in the future would be to use a spring of negligible mass and also to not
consider the mass of the spring in the system.