SME
University Physics I: Mechanics · Problem type
A damped oscillator: its frequency, Q and fading amplitude
Use this for “A 2 kg block on a 200 N/m spring has damping b = 1 kg/s and starts with a 0.1 m amplitude. Find ω′, Q and the amplitude after 4 s.”
Diagram
ωₙ
=
√(
k
m
)
=
√(
50
0.5
)
=
10
rad/s
ζ
=
b
2
√
km
ζ
=
0.4
2
√
50 × 0.5
=
0.04
ζ
<
1,
underdamped:
ω′
=
ωₙ
√
1 −
ζ
2
=
9.99
rad/s,
less
than
ωₙ;
the
crests
shrink
inside
±A₀
exp(−ζωₙt).
This is an example. Type your own values to start a new problem.
A (m)
=
A₀ (m)
×
e
−
×
÷
(2
×
)
A₀ is how far the block swings at the start and A how far at time t: the height of the shrinking crests on the graph then.
k
Spring constant
example · N/m
ω′
Damped angular frequency
worked out (not typed) · rad/s
Q
Quality factor
worked out (not typed)
Formulas
ω′
=
√(
k
m
−
(
b
2 ×
m
)²)
9.992
=
√(
50
0.5
−
(
0.4
2 × 0.5
)²)
Q
=
√
m
×
k
b
12.5
=
√
0.5 × 50
0.4
A
=
A₀
×
e
−
b
×
t
÷ (2 ×
m
)
0.006767
=
0.05
×
e
−0.4 × 5 ÷ (2 × 0.5)
Assumptions
•
A drag force −bv acts on the block, so mẍ = −kx − bẋ, solved by x = A₀e−bt/2m cos(ω′t) when the damping is light.
•
Underdamped only: b < 2√(mk). At b = 2√(mk) the block is critically damped and returns without swinging, so ω′ has no value.
•
Q = mω₀ ÷ b with ω₀ = √(k/m), which is √(mk) ÷ b; Q above 0.5 means it swings.
Step-by-step
WE KNOW
m = 0.5 kg, k = 50 N/m, b = 0.4 kg/s, A₀ = 0.05 m, t = 5 s
FIND
damped angular frequency (ω′), quality factor (Q), amplitude at t (A)
Step 1 · Find damped angular frequency (ω′)
Use ω′ = √(k ÷ m − (b ÷ (2 × m))²)
Damping
slows
the
swing
a
little
below
ω₀
=
√
k
m
;
there
is
no
swing
once
b
2
m
reaches
ω₀.
ω′
=
√(
50
0.5
−
(
0.4
2 × 0.5
)²)
ω′
=
√(
50
0.5
−
0.4
2
)
ω′
=
√(
50
0.5
−
0.16)
ω′
=
√
100 − 0.16
ω′
=
√
99.84
ω′ = 9.992 rad/s
Step 2 · Find quality factor (Q)
Use Q = √(m × k) ÷ b
Q
=
mω₀
b
,
and
mω₀
=
m
√
k
m
=
√
mk
:
a
high
Q
rings
for
many
cycles.
Q
=
√
0.5 × 50
0.4
Q
=
√
25
0.4
Q
=
5
0.4
Q
=
12.5
Step 3 · Find amplitude at t (A)
Use A = A₀ × e−b × t ÷ (2 × m)
The
envelope
of
the
swing:
the
amplitude
falls
by
a
factor
e
every
2
m
b
seconds.
A
=
0.05
×
e
−0.4 × 5 ÷ (2 × 0.5)
A
=
0.05
×
e
−0.4 × 5
A
=
0.05
×
e
-2
A
=
0.05
×
0.13534
A
=
0.006767
m
Check
Put the numbers back in. Both sides should match.
9.992
=
√(
50
0.5
−
(
0.4
2 × 0.5
)²)
✓
12.5
=
√
0.5 × 50
0.4
✓
0.006767
=
0.05
×
e
−0.4 × 5 ÷ (2 × 0.5)
✓
Topic check
See what you know: 5 questions on this topic
New questions each time. Pass to mark it complete.