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Lab 9: Simple Harmonic Motion, Mass-Spring
Only 3 more to go!!
The force due to a spring is, F = -kx, where k is the spring constant and x is the
displacement from equilibrium.
+x
F = - kx
2
As you can see this is a cosine function.
Streched
Spring's Displacement
This type of force is called a “restoring force”
because it always acts to restore the system to
equilibrium. If we pull on the spring-mass system
and let it go it will oscillate back and forth. If we
make a plot of the spring’s position vs time we get:
Equilibrium
0
Compressed
-2
0
2
4
6
Time
8
10
It turns out we can write the displacement as a function of time, t:
x = xMAX cos(t)
The velocity can be written as: V = -  xMAX sin(t),
The acceleration: a = - 2 xMAX cos(t)
t – time in seconds
xMAX – maximum displacement, or amplitude
- angular frequency, tells us how many oscillations occur per second.
For a spring mass system:
where k, is the spring constant
k

and m, is the mass

m
Another useful quantity is the period, T. The period tells us how long it takes
to complete one oscillation, for a spring mass system:
2
m
T
 2

k
Let’s consider a spring-mass system, k= 30 N/m, m = 3 kg. The mass
oscillates with an amplitude of 0.1m
xMAX= 0.1 m
3 kg
What’s the
frequency of osc. ?
3 kg
k
30 N / m


 10 sec 2  3.2 sec s
m
3kg
What is the
displacement at t = 0?
What is the max. and min
acceleration?
x = xMAX cos(t) = 0.1 m cos (0) = 0.1 m
a = - 2 xMAX cos(t) , so the max accel. occurs
when the cos(t) term equals 1, when does this
happen?
This happens when (t) = 0,  (i.e. at the turning
points)
When is the velocity maximum? V = -  xMAX sin(t) velocity is max. when
sin(t) is max (i.e. equals 1), this happens when (t) = /2
What are we doing today? 2 experiments.
The first will allow us to measure the spring constant, k, of our spring.
You will hang the spring, measure the equilibrium length.
Next you will and mass, and measure how much it stretches.
Do this several times for different masses.
Make a plot of Weight (y-axis) vs Displacement from equilibrium (x-axis)
12
10
Weight (N)
Slope = k
8
6
4
2
1
2
3
4
5
Displacement from Equilibrium (x-m)
6
The next exp, will allow us to find the spring constant, k using, the oscillatory
motion of the spring
First, put 100g on the end of spring and pull down on the spring slightly. Release
and record the time it takes to make 25 oscillation. Calculate the period, T = time/25
Increase the mass and repeat the experiment 5 times
Normally, the period T:
m
T  2
k
But this doesn’t account for the fraction,f
of the spring that is also oscillating, to account for this we derive a new equation for
the period:
m  ( f  M S ) Where, f is the fraction of the spring that
T  2
is oscillating, and MS is the mass of the
k
spring
If I find the equation for T2:
2
2
4

f MS
4

2
T 
m
k
k
Now let’s plot T2 (y-axis) vs. mass, m (x-axis):
12
2
Slope = 4 / k
10
2
2
T (sec )
8
6
4
2
Y-int = 4 f MS/ k
2
1
2
3
4
mass (m - kg)
5
6