SME
University Physics I: Mechanics
Oscillations
Use this for “A 2 kg block on an 80 N/m spring starts 0.1 m from its rest position, moving at −0.6 m/s. Find its amplitude, phase and position 1 s later.”
Diagram
A
=
√(
x₀
2
+
(v₀/ω)
2
)
=
0.05
m
φ = atan2(−v₀/ω, x₀) = −0.927 rad
x(0) = A cos φ = 0.03 m = x₀
This is an example. Type your own values to start a new problem.
x(
t (s)
)
=
A (m)
cos(
ω
×
t
+
(
φ
)
)
=
x (m)
m
Mass
example
k
Spring constant
example · N/m
x₀
Start position
example
v₀
Start velocity
example
Formulas
ω
=
√(
k
m
)
10
=
√(
50
0.5
)
A
=
√(
x₀
2
+
(
v₀
ω
)
2
)
0.05
=
√(
0.03
2
+
(
0.4
10
)
2
)
φ
=
atan2(−
v₀
ω
,
x₀
)
−0.9273
=
atan2(−
0.4
10,
0.03)
x
=
A
×
cos(
ω
×
t
+
φ
)
0.02389 = 0.05 × cos(10 × 0.2 + (−0.9273))
Assumptions
•
No friction: x = A cos(ωt + φ) solves mẍ = −kx with ω = √(k/m).
•
At t = 0, x₀ = A cos φ and v₀ = −Aω sin φ; squaring and adding gives A, and their signs pick φ’s quarter.
•
The phase φ is in radians, from −π to π; the angle ωt + φ is in radians too.
Step-by-step
WE KNOW
m = 0.5 kg, k = 50 N/m, x₀ = 0.03 m, v₀ = 0.4 m/s, t = 0.2 s
FIND
angular frequency (ω), amplitude (A), phase (φ), position at t (x)
Step 1 · Find angular frequency (ω)
Use ω = √(k ÷ m)
The
natural
angular
frequency:
the
square
root
of
k
over
m
.
ω
=
√(
50
0.5
)
ω
=
√
100
ω
=
10
rad/s
Step 2 · Find amplitude (A)
Use A = √(x₀² + (v₀ ÷ ω)²)
x₀
=
A
cos
φ
and
v₀
ω
=
−A
sin
φ
,
so
x₀
2
+
(
v₀
ω
)
2
=
A
2
(cos²
φ
+
sin²
φ
)
=
A
2
.
A
=
√(
0.03
2
+
(
0.4
10
)
2
)
A
=
√
0.03
2
+
0.04
2
A
=
√
0.0009 + 0.0016
A
=
√
0.0025
A
=
0.05
m
Step 3 · Find phase (φ)
Use φ = atan2(−v₀ ÷ ω, x₀)
cos
φ
has
x₀
’s
sign
and
sin
φ
has
−
v₀
’s
sign:
atan2
finds
the
angle
in
that
quarter.
φ
=
atan2(−
0.4
10,
0.03)
φ
=
−0.9273
rad
Step 4 · Find position at t (x)
Use x = A × cos(ω × t + φ)
The
position
at
time
t
,
the
angle
ωt
+
φ
in
radians.
x
=
0.05
×
cos(10
×
0.2
+
(−0.9273))
x
=
0.05
×
cos(2
+
(−0.9273))
x
=
0.05
×
cos(1.0727)
x
=
0.05
×
0.4778
x
=
0.02389
m
Check
Put the numbers back in. Both sides should match.
10
=
√(
50
0.5
)
✓
0.05
=
√(
0.03
2
+
(
0.4
10
)
2
)
✓
−0.9273
=
atan2(−
0.4
10,
0.03)
✓
0.02389 = 0.05 × cos(10 × 0.2 + (−0.9273)) ✓
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A damped oscillator: its frequency, Q and fading amplitude
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