SME
Differential Equations · Problem type
Forced swings and resonance
Use this for “For y″ + 0.5y′ + 4y = cos(ωt), how large is the steady swing at ω = 2, and at ω = 1?”
Table
Forcing frequency (ω)
Steady amplitude (R)
R against ω, worked out for every ω from 1 to 3
The most R ≈ 1.008 comes at ω ≈ 1.969
This is an example. Type your own values to start a new problem.
y″
+
b
y′
+
c
y
=
A
cos(
ω
t)
R
=
,
φ
=
ωᵣ
Frequency of the largest swing
worked out (not typed)
Rₘₐₓ
Largest steady amplitude
worked out (not typed)
Formulas
R
=
A
÷
√(
(
c
−
ω
2
)
2
+
(
b
×
ω
)
2
)
1
=
1
÷
√(
(4 −
2
2
)
2
+
(0.5 × 2)
2
)
φ
=
π
2
1.5708
=
π
2
ωᵣ
=
√(
c
−
b
2
2
)
1.9685
=
√(4
−
0.5
2
2
)
Rₘₐₓ
=
A
b
×
√
c
−
b
2
÷ 4
1.0079
=
1
0.5 ×
√
4 −
0.5
2
÷ 4
Assumptions
•
With damping (b > 0) the free part dies away, leaving the steady swing y(t) = R cos(ωt − φ) at the forcing frequency.
•
Putting it into the equation gives R = A ÷ √((c − ω²)² + (bω)²), with the lag φ between 0 and π.
•
The swing is largest at ωᵣ = √(c − b² ÷ 2), close to the natural frequency √c when the damping is light (b² < 2c).
Step-by-step
WE KNOW
b = 0.5, c = 4, A = 1, ω = 2
FIND
steady amplitude (R), phase lag (φ), frequency of the largest swing (ωᵣ), largest steady amplitude (Rₘₐₓ)
Step 1 · Find steady amplitude (R)
Use R = A ÷ √((c − ω²)² + (b × ω)²)
Put
y
=
R
cos(ωt
−
φ
)
into
the
equation:
the
cos
and
sin
parts
give
R
(
c
−
ω
2
)
=
A
cos
φ
and
Rbω
=
A
sin
φ
.
Square
and
add.
R
=
1
÷
√(
(4 −
2
2
)
2
+
(0.5 × 2)
2
)
R
=
1
÷
√(
(4 − 4)
2
+
1
2
)
R
=
1
÷
√
0
2
+
1
2
R
=
1
÷
√
1
R
=
1
Step 2 · Find phase lag (φ)
Use φ = π ÷ 2
At
ω
2
=
c
the
spring
and
the
mass
cancel:
only
the
damper
resists,
and
the
swing
lags
by
a
quarter
turn.
2
2
=
4:
at
resonance
y″ + 0.5y′ + 4y = cos(2t)
y(t) = cos(2t − 1.5708)
φ
=
1.5708
rad
Step 3 · Find frequency of the largest swing (ωᵣ)
Use ωᵣ = √(c − b² ÷ 2)
R
is
largest
where
(
c
−
ω
2
)
2
+
(
bω
)
2
is
smallest.
Its
derivative
in
ω
2
is
−2(
c
−
ω
2
)
+
b
2
,
which
is
0
at
ω
2
=
c
−
b
2
2
.
0.5
2
<
2
×
4:
a
peak
ωᵣ
=
√(4
−
0.5
2
2
)
ωᵣ
=
√(4
−
0.25
2
)
ωᵣ
=
√
4 − 0.125
ωᵣ
=
√
3.875
ωᵣ = 1.9685
Step 4 · Find largest steady amplitude (Rₘₐₓ)
Use Rₘₐₓ = A ÷ (b × √(c − b² ÷ 4))
Put
ω
2
=
c
−
b
2
2
into
R
:
the
bottom
becomes
b
2
(
c
−
b
2
4
).
Less
damping
makes
this
peak
taller
and
sharper.
Rₘₐₓ
=
1
0.5 ×
√
4 −
0.5
2
÷ 4
Rₘₐₓ
=
1
0.5 ×
√
4 − 0.25 ÷ 4
Rₘₐₓ
=
1
0.5 ×
√
4 − 0.0625
Rₘₐₓ
=
1
0.5 ×
√
3.9375
Rₘₐₓ
=
1
0.5 × 1.984
Rₘₐₓ
=
1
0.992
Rₘₐₓ = 1.0079
Check
Put the numbers back in. Both sides should match.
1
=
1
÷
√(
(4 −
2
2
)
2
+
(0.5 × 2)
2
)
✓
1.5708
=
π
2
✓
1.9685
=
√(4
−
0.5
2
2
)
✓
1.0079
=
1
0.5 ×
√
4 −
0.5
2
÷ 4
✓
Topic check
See what you know: 5 questions on this topic
New questions each time. Pass to mark it complete.