Lab Report
Physics 215
October 30, 2013
Experiment 9
Simple Harmonic Motion
Sosena Aseffa
Introduction
Harmonic motion is a periodic back and forth
motion that obeys Hooke’s Law. We will take
a look at a pendulum and a mass-spring
system to determine what effects their period.
A period is the time (in seconds) it takes to
complete one cycle.
Theory
Harmonic motion must be periodic and follow
Hook’s Law.
F=Kx
Equation 1
F is the magnitude of the restoring force
K is a constant
X is the amplitude
Period of a harmonic motion:
T=1/f
Equation 2
T is the period in seconds and f is frequency in
hertz.
The period of a pendulum that is undergoing
harmonic motion:
T=2pi square root of L/g Equation 3
L is the length of the pendulum and g is
acceleration due to gravity. Pi =3.14
The period of a mass on a spring:
T=2pi Square root of m/k Equation 4
K is the spring/force constant (obtained using
Hooke’s law) and m is the mass being
suspended by the spring.
The variable for the pendulum include:
 Length of string
 Amplitude
 Mass of the pendulum bob
The variables for the mass-spring system
include:
 Amplitude
 Mass suspended
 Spring
Experimental procedure
Collect the following equipment:
 Support rod and clamp
 One brass spring and one steel
balance
 Timer Meter stick
 One wooden ball and one brass ball
 Set of masses
Part A
Start of by determining the mass of each
ball to be used as a pendulum bobs. Create
a pendulum with a 100cm string with the
brass ball attached. Set the amplitude at
10cm apart using a meter stick. After you
release the ball divide the number of cycle
by one minute to determine the frequency.
Use equation 2 to figure out the period.
Repeat the same procedure with 20 cm
amplitude. Than change the brass ball to
the wooden ball and repeat the procedure
using 10cm amplitude. Calculate the
acceleration due to gravity using equation
3. Below is a model of a pendulum.
Part B
Measure the length of the brass spring before and after attaching a 500 gram mass. Calculate
the displacement distance by subtracting the initial length from the final, and the restoring
force using the mass value, g (9.80 m/s square), and the spring constant. Repeat the same
procedure for the steel spring.
Part C
Hang 500 grams mass on the brass spring and pull it down to 5cm amplitude. Count the
number of cycles in one minute and repeat with 10 cm amplitude. Calculate the frequency and
period. Change the mass to 1000 grams, and repeat this experiment using only 5 cm
amplitude. Next use the steel spring, and 5cm amplitude and count the number of frequency.
Determine the frequency using the k value found in part B, and equation 4.
Data/ Analysis
Part A
Pendulum
bob
Length
(cm)
Amplitude
(cm)
Frequency
(Hz)
Period
(s)
Brass Ball
Brass Ball
Brass Ball
Wooden Ball
100.
100.
50.
50.
10.
20.
10.
10.
0.5
0.48
0.68
0.68
60
60
60
60
Brass ball: 66.5 grams
Wooden ball: 6.4 grams
f=# of cycle/60 sec
f= 30/60
f=0.5
g=L(2pi)/T square
g=100(2pi)/(1/0.5) square
g= 157.08
f: frequency
g: acceleration due to gravity
T; period (1/f)
L: length
G
(cm/s
square)
157.08
144.76
145.4
145.4
Part B
Spring
Brass
Steel
Initial
length Li
(m)
Final
length Lf
(m)
Displacement Mass m
x=(Lf-Li)
(kg)
(m)
Restoring
force F=mg
(N)
0.43
0.23
1.06
0.45
0.63
0.22
4.9
4.9
0.5
0.5
Displacement:
x=(Lf-Li)
x=(1.06-0.43)
x=0.63
x: displacement
Lf: Final length
Li: Initial length
Restoring force:
F=mg
F=0.5 x9.8
F=4.9
F: restoring force
m: mass
g: acceleration due to gravity
Spring constant:
k=F/x
k=4.9/0.63
k=7.78
k: spring constant
Part c
Spring
Brass
Brass
Brass
Steel
Mass
(g)
500.
500.
1000.
500.
Amplitude
(cm)
5.0
10.0
5.0
5.0
Frequency
(Hz)
0.63
0.62
0.47
1.2
Period
(s)
60
60
60
60
Spring
constant
k=F/x
(N/m)
7.78
22.27
Frequency
(Equ. 4)
1.59
1.59
2.25
0.99
Frequency:
T=2pi square root of m/k
T=2pi square root of .500 kg/7.78
T=1.59
Discussion of results
The period does not depended on the mass of the pendulum bob, but it depends on the length of
the pendulum. The shorter the pendulum the higher the frequency becomes. The amplitude of
the pendulum makes very little change in the period. This answer is consistent with equation 3.
The steel spring is stronger than the brass spring. The period of the spring does not depend on
the amplitude. It depends on the mass and spring constant. Period increases as the mass
increases, and increase in spring constant decrease in period. The measurement
Conclusion
A lot of errors in this experiment come from measurement, counting the cycles, and calculation.
Repeating the experiment three times and getting an average might be time consuming but it
will definitely decrease the percent error. The method is accurate because my values are
consistence with my equation. We need to be extra careful on calculation, and we had to redo
some. In the future, the device needs to have less friction.
H