Stopping an Underdamped Oscillator
Problem Statement. Suppose the positive constants m, c, and k make the differential operator involved below underdamped:
mÿ(t) + cẏ(t) + ky(t) = f (t),
y(0) = y0 , ẏ(0) = v0 .
(∗)
Choose the constants a and τ > 0 for which the input
f (t) = aδ(t − τ )
leads to a solution of (∗) obeying y(t) = 0 for all t > τ .
Superposition Approach. Split the solution of (∗) into y(t) = yc (t) + yi (t), with
subscripts c for “complementary” and i for “impulsive”:
mÿc (t) + cẏc (t) + kyc (t) = 0,
yc (0) = y0 , ẏc (0) = v0 ,
mÿi (t) + cẏi (t) + kyi (t) = a δ(t − τ ),
yi (0) = 0,
ẏi (0) = 0.
It’s easy to check that this splitting respects the full ODE and IC’s in (∗). Further,
the complementary solution is well known:
2mv0 + cy0
sin(ωt) ,
y0 cos(ωt) +
yc (t) = e
2mω
2mcv0 + (4m2 ω 2 + c2 )y0
−ct/(2m)
v0 cos(ωt) −
ẏc (t) = e
sin(ωt) .
4m2 ω
−ct/(2m)
with
These expressions use the standard notation
ω = ωd =
√
4mk − c2
.
2m
(Writing ω instead of ωd eliminates a few subscripts.)
To find yi (t), use the Laplace Transform:
(ms2 + ks + c)Yi (s) = ae−sτ ⇐⇒ Yi (s) =
ae−sτ
.
ms2 + ks + c
To recover yi (t), we complete the square and remember the definition of ω:
k
c
2
s+
ms + cs + k = m s + 2
2m
m
"
2 #
c 2 4mk
c 2
c
2
s+
=m
+
=m s+
+ω .
−
2m
4m2
2m
2m
2
File “stop-osc”, version of 05 Dec 2013, page 1.
Typeset at 10:20 December 5, 2014.
2
PHILIP D. LOEWEN
Rearranging this, with the “un-cancelling” trick 1 = ω/ω, gives
Yi (s) =
m
h
ae−sτ
a −sτ
i =
e
2
c
mω
s+
s + 2m
+ ω2
ω
c 2
2m
+ ω2
.
Now Shift Theorems #1 and #2 lead to the inverse transform
(
)
a −1 −sτ
ω
yi (t) =
e
L
c 2
mω
s + 2m
+ ω2
(
)
a
ω
=
u(t − τ )L−1
c 2
mω
+ ω 2 t→(t−τ )
s + 2m
ω
a
−ct/(2m) −1
u(t − τ ) e
L
=
mω
s2 + ω 2 t→(t−τ )
h
i
a
−ct/(2m)
=
u(t − τ ) e
sin(ωt)
mω
t→(t−τ )
a
−c(t−τ )/(2m)
=
u(t − τ )e
sin(ω(t − τ ))
mω
Combining this yi (t) with yc (t) above will work best if all the trig functions involved
have the same phase. This can be arranged in various ways. The all-real approach
uses standard trig identities:
yi (t) =
ae−ct/(2m)
u(t − τ )ecτ /(2m) [ sin(ωt) cos(ωτ ) − cos(ωt) sin(ωτ )] .
mω
Now we need to arrange y(t) = 0 for t > τ . This is equivalent to requiring, for t > τ ,
0 = 2mωect/(2m) [yc (t) + yi (t)]
= 2mωy0 cos(ωt) + (2mv0 + cy0 ) sin(ωt) + 2aecτ /(2m) [ sin(ωt) cos(ωτ ) − cos(ωt) sin(ωτ )]
h
i
h
i
= 2mωy0 − 2aecτ /(2m) sin(ωτ ) cos(ωt) + 2mv0 + cy0 + 2aecτ /(2m) cos(ωτ ) sin(ωt).
Take the limit as t → τ + to get an equation that determines τ :
(2mv0 + cy0 ) sin(ωτ ) + 2mωy0 cos(ωτ ) = 0.
Outside cases where division by 0 occurs, this gives
2mωy0
,
tan(ωτ ) = −
2mv0 + cy0
−1
i.e., ωτ = nπ − tan
2mωy0
2mv0 + cy0
, n ∈ Z.
Now to get y(t) = 0 for all t > τ , both sinusoidal terms above must have zero
coefficients. Work with the coefficient of cos(ωt) because it’s slightly simpler:
mωy0 = aecτ /(2m) sin(ωτ ).
File “stop-osc”, version of 05 Dec 2013, page 2.
Typeset at 10:20 December 5, 2014.
3
tan2 θ
reveals that
1 + tan2 θ
The identity sin2 θ =
(2mωy0 )2
(ωy0 )2
=
(2mv0 + cy0 )2 + (2mωy0 )2
(ωy0 )2 + v0 +
sin2 (ωτ ) =
So our condition defining a becomes
2
m
ω 2 y02
(ωy0 )2
2 cτ /m
=a e
(ωy0 )2 + v0 +
Rearranging, with a little cancellation, gives
2
2 −cτ /m
a =m e
cy0 2
2m
!
.
cy0 2
2m
.
cy0 2
2
+ (ωy0 ) .
v0 +
2m
Report. It is only possible to stop the oscillator with an impulse at an instant when
it is at the equilibrium location. Here is a list of those instants:
nπ
1
τ=
− tan−1
ω
ω
2mωy0
2mv0 + cy0
, n ∈ Z.
Choose τ from this list, and then apply the impulse with scalar factor
a = ±me−cτ /(2m)
r
v0 +
cy0 2
+ (ωy0 )2 .
2m
Choose the sign here to match the sign of sin(ωτ ) (remember that τ is known).
Physics Review. With τ as above, the momentum in the mass at time τ − is
2mcv0 + (4m2 ω 2 + c2 )y0
−
mẏc (τ ) = me
v0 −
sin(ωτ )
4m2 ω
me−cτ /(2m) =−
(2mv0 + cy0 )2 + 4m2 ω 2 y02 sin(ωτ ).
2
4m ωy0
−cτ /(2m)
−
2mv0 + cy0
2mωy0
Cancelling this momentum requires an impulse of the opposite sign. The expression
for sin(ωτ ) above provides more detail about the momentum:
me−cτ /(2m) 2mωy0 sgn(sin(ωτ ))
2
2 2 2
p
(2mv
+
cy
)
+
4m
ω
y
0
0
0
4m2 ωy0
(2mv0 + cy0 )2 + (2mωy0 )2
r
cy0 2
) + (ωy0 )2 sgn(sin(ωτ )).
= −me−cτ /(2m) (v0 +
2m
mẏc (τ − ) = −
This exactly matches the findings above.
File “stop-osc”, version of 05 Dec 2013, page 3.
Typeset at 10:20 December 5, 2014.