Spring-Mass Systems
Example 1.
A mass of 10 kg is attached to a spring stretched 20cm by a force of
10kg.m/s2 What is the frequency of oscillation of the system?
Solution: Need to find the frequency ω of oscillation of the system.
The system obeys
d2 y
m 2 + ky = 0
dt
or
k
d2 y
+ y=0
2
dt
m
or
d2 y
+ ω02 y = 0
dt2
where ω02 = k/m.
Mass m = 10kg If a force of 10kg.m/s2 stretches a spring 20 cm,
force = ky
10 kg.m/s2 = k(20cm)
50kg/s2 = k
So
50 1
ω 2 = k/m =
10 sec2
so
1
ω2 = 5 2
sec
or
√ 1
ω= 5
sec
Example 2: A mass on a spring is set in motion at x = −3 feet with initial
velocity
x′ (0) = 3 feet/sec. If the mass reaches a maximum displacement of
√
2 3 feet, at what speed does it pass through its equilibrium point?
Solution: x(x) = A cos ωt + B sin ωt is the solution to the equation
d2 x
m 2 + kx = 0
dt
(where ω 2 = k/m). Then
dx
= −Aω sin ωt + Bω cos ωt
dt
1
We have
x(0) = −3 so A = −3
and
q
x′ (0) = 3 so Bω = 3 or B = 3 m/k.
The equilibrium point is the point where x′ is maximum (in other words,
where x(t) = 0).
√
The speed at equilibrium is A2 + B 2 ω
=
s
q
q
9 + 9m/k k/m
k+m
=3
k
s
s
k
k+m
=3
.
m
m
√
√
Note since the maximum displacement is A2 + B 2 = 2 3 the equilibrium
speed is equal to
q
√
3 1 + k/m = 3 1 + 3 = 6 feet/sec
Example 3. A 10-kg mass is hanging from a spring whose spring constant
is 5 kg/s2 . The weight is set in motion at x = 3 m with an initial velocity of
−6 m/sec. Discuss the resulting motion if the surrounding medium offers a
resistance equal to 4 times the velocity.
2
+ kx = 0 where
Solution: m ddt2x + r dx
dt
2
k = 5 kg/s
m = 10 kg
2
k/m = 510kg/s
= 0.51/s2
kg
Resistance = 4× velocity:
r = 4, r/m = 4/10
Equation is
r dx
k
d2 x
+
+ x=0
2
dt
m dt
m
Solution is
where
are the roots of
x(t) = Ae−αt cos(βt − δ)
z = −α ± iβ
z2 +
r
k
z+
=0
m
m
2
in other words
1
1r
±
z=−
2m 2
or
r
,
α=
2m
So
q
β=
α = 2/5, β = 1/2 − 1/25 =
So the solution is
s
s
k
r2
−
4
m2
m
r
k
−
m
2m
2
q
23/25
q
x(t) = Ae−2t/5 cos( 23/25t − δ)
Impose initial conditions to determine δ:
x(0) = +3, so A cos(δ) = 3
dx
(0) = −6, so
dt
−αA cos(δ) − Aβ sin(−δ) = −6
αA cos(δ) − βA sin(δ) = 6
(1)
A cos δ = 3
(2)
Substitute (2) in (1):
s
sin(δ) = ± 1 −
so
s
3α ∓ βA 1 −
9
A2
9
=6
A2
√
2 ∓ β A2 − 9 = 6
√
∓β A2 − 9 = 4
A cos(δ) = 3
So
16 × 9
16
=
2
β
11
16 × 9 11 × 9
A2 =
+
11
11
A2 − 9 =
3
(3)
(4)
√
9 3
A= √
11
q
So cos(δ) = 3/A, sin(δ) = − 1 −
9
A2
or
tan(δ) = −
1
=−
3
s
1√ 2
A −9
3
14
16
=−
2
β
3β
4
4 3
= − √ = −√ .
3 11
11
4