1. Calculating the damping constant of a spring + hook
system
By Ariel Garcia, Andrew Huynh and N’Vida Yotcho
Math 238 –Differential equations

2. Introduction:
A spring of 0.094 kg with initial length of 0.052 m has a stretch of
0.118 m when a hook of 0.512 kg is attached to it (Figure 1).
The spring + hook system is stretched from equilibrium position to - 0.116
m and released at t= 0.6745 s (Figure 2). With a motion detector and the
Vernier software LoggerPro, a position-time graph of the system had been
plotted (Figure 3). Find the damping constant c of the damping force
applying by the air on the spring + hook system.

3. FIGURE 1.
The spring without the hook (left)
The spring + hook (right)

4. FIGURE 2.
The
experimental
dispositive

5. FIGURE 3. Position-time graph of the system after release.

6. FIGURE 4.
The amplitudes An at time tn
n An tn
0 0.1373 1.05
1 0.1131 1.9
2 0.1035 2.75
3 0.08678 3.6
4 0.07114 4.4
5 0.06867 5.3
6 0.06510 6.1
7 0.05138 6.95

7. FIGURE 5. Amplitude vs. time graph (exponential envelope)

8. Initial Value Problem:
m*y’’ + c*y’ + k*y = 0,
y(0.6745) = -0.116, y’(0.6745)=0
==> y’’ +
𝒄
"
*y’ +
#
"
*y = 0
m: effective mass of the system (kg)
c: damping coefficient (N*s/m)
k: spring constant (N/m)

9. ____Effective mass of the system m:
m =
$%&&
()
*+,
&-./01
2
+ 𝑚𝑎𝑠𝑠
𝑜 𝑓
𝑡ℎ𝑒
ℎ 𝑜𝑜𝑘
=
=.=?@
2
+ 0.512
m = 0.543 kg
____Spring constant k:
Tension of the spring = k *stretch of the spring
Tension of the spring = m * g
è k = A,0&/(0
()
*+,
&-./01
&*.,*B+
()
*+,
&*./01
k =
$∗1
&*.,*B+
()
*+,
&*./01
=
=.D@2∗?.E
=.FFE
k = 45.097 N/m

10. _____ The Damping coefficient c: is unknown.
But c, k and m are related. Indeed, the damping ratio Ỷ is equal to:
Ỷ =
B
G #∗$
è c = Ỷ * (2 𝑘 ∗ 𝑚)
è
H
$
= Ỷ *2* 𝑘 ∗
$
$
è
H
$
= Ỷ *2*
#
$
, and ω0 =
#
$
B
$
= Ỷ * 2 * ω0
Rewriting the IVP in function of ω0 (undamped natural frequency), and Ỷ:
y’’ +
𝒄
"
*y’ +
#
"
*y = 0
è y’’ + Ỷ * 2 * ω0*y’ + ω02 *y = 0

11. Solving the IVP:
y’’ + Ỷ * 2 * ω0*y’ + ω02 *y = 0, y(0.6745) = -0.116, y’(0.6745)=0
r2+ Ỷ 2 ω0 r+ ω02 = 0
r=
IG
J
K=
G
± i
@
K=L
(FIJL)
G
è y(t) = A e -Ỷ ω0t *cos[µμ ∗ t −
δ]
μ: quasi-frequency --- μ = ω0 1 − ζG
δ : The phase

12. _____Amplitude decay envelope (Figure 5):
A(t) = A0 e -Ỷ ω0t
For tn : An= A0 e -Ỷ ω0tn
For tn+1: An+1=A0 e -Ỷ ω0(tn+1)
An
An+1
=
A0
e
−ζ
ω0t
A0
e
−ζ
ω0(t+1) è
An
An+1
= e -Ỷ ω0tn +Ỷ ω0(tn+1)
An
An+1
= e Ỷ ω0(tn+1- tn)

13. T = tn+1-tn
T is the period between each peak
But we know that T =
G]
^0
è T =
G]
ω0 1−ζ2
So that :
An
An+1
=
𝑒
JK=(
L`
ω0 1−ζ2 )
è
An
An+1
= 𝑒
(
La`
1−ζ2 )
Logarithmic decrement or the logarithm of the amplitudes is for any two
successive peaks: ln
[
An
An+1
] =
GJ]
1−ζ2
In general : ln
[
c=
cd
] =
GJd]
FIJL

14. FIGURE 6. Ln (A0/An) vs. n
0
0.2
0.4
0.6
0.8
1
1.2
0 1 2 3 4 5 6 7 8
ln(A0/An)
n

15. Our data follows a linear trend. So, it is valid that our model is a linear
damping model, so that, the damping ratio can be estimated by:
Ỷ =
e0
@]fe0L
with 𝛽𝑛 =
F
0
[ln
i=
i0
]
n
𝛽𝑛 4𝜋 + 𝛽𝑛G Ỷ Ỷ
average
is:
0.006
2 0.019965224 3.55 0.00563
3 0.023387802 3.55 0.00659
4 0.027020665 3.55 0.00761
5 0.019201971 3.55 0.00541
6 0.016754876 3.55 0.00472
7 0.019716947 3.55 0.00556

16. Conclusion:
Ỷ = 0.006 è Ỷ > 0.01 and that is what was
expected for a metal system.
c = Ỷ * (2 𝑘 ∗ 𝑚)
= 0.006 *(2 45.097
∗ 0.543
)
c = 0.0593820227
𝒄 ≃ 0.059