Sam Student 00000P
Acoustics and the Physics of Sound 2016 Demo #1
Problem 1
Problem 1
Transient response of a driven oscillator. A mass m is attached to a moving platform via a string of length
l. A force is applied to the platform to force a horizontal displacement of µ = µ0 cos(ωt) on the system as
shown in the figure below. Find the transient response of the system when the conditions x = 0, ẋ = 0 are
imposed at t = 0.
μ = μ0 cos(ωt)
μ
θ
l
l
T
T
m
m
x-μ
mg
mg
x
Figure 1: Driven pendulum
The transient response is the superposition of what is caused by the initial conditions (free-response) and
what is caused only by the force (steady-state-response). To solve this exercise, you have to find the freeand steady-state-responses and combine them. Due to linearity, the system as a whole is simply described
as the sum of the steady-state and the free-state solutions.
x = f ree + steady = Ae−γt/2 cos(ωc t + φ) + |H|cos(ωt + ∠H)
(1)
Start by writing the forces for a snapshot of the pendulum in oscillation. In the figure above, T is tension
and mg is force due to gravity. Assuming that θ is small, we have cos(θ) ≈ 1 and the tension in the string
is thus T cos(θ) ≈ T ≈ mg.
Write equation from Newton’s 2nd
F = ma = mẍ
(2)
F = −damping − tension
(3)
mẍ = −bẋ − T sinθ = −bẋ − mgsin(θ)
(4)
Since T ≈ mg
From the geometry of the pendulum snapshot, we see that the driving force term µ is embedded into the
sinusoid of theta
x−µ
l
(5)
x−µ
opposite
=
hypotenuse
l
(6)
sin(θ) =
If this isn’t clear, remember that
sin(θ) =
Insert sin(θ) into Newton’s
mẍ = −bẋ − mgsinθ = −bẋ − mg
Problem 1 continued on next page. . .
1
x−µ
mgx mg
= −bẋ −
+
µ
l
l
l
(7)
Sam Student 00000P
Acoustics and the Physics of Sound 2016 Demo #1
Problem 1 (continued)
Insert the definition of µ into the above
mẍ = −bẋ −
mgx mg
+
µ0 cos(ωt)
l
l
(8)
We can clean up the equation by using the damping coefficient γ = b/m and natural frequency ω02 = g/l.
To do this, we divide everything by m and insert γ and ω02 where appropriate.
g
g
b
ẋ − x + µ0 cos(ωt)
m
l
l
(9)
ẍ = −γ ẋ − ω02 x + ω02 µ0 cos(ωt)
(10)
ẍ + γ ẋ + ω02 x = ω02 µ0 cos(ωt)
(11)
ẍ = −
Rearrange to fit eq. of harmonic osc.
Lets now derive the transfer function H.
H=
X
”out”
=
”in”
F (ω)
(12)
Assuming our force is a complex exponential, we can write
x = HF = Hejω ,
ẋ = jωHejωt ,
ẍ = −ω 2 Hejωt
(13)
Insert the above into previous eq.
− ω 2 Hejωt + jγωHejωt + ω02 Hejωt = ω02 µ0 ejωt
(14)
− ω 2 H + jγωH + ω02 H = ω02 µ0
(15)
H −ω 2 + jγω + ω02 = ω02 µ0
(16)
Exponentials cancel
Solve for H
H=
ω02 µ0
−ω 2 + jγω + ω02
(17)
Solve the angle from the above and plug into the steady state solution to obtain the full expression for the
steady state.
x = |H|cos(ωt + ∠H)
(18)
The equation describing the system as a whole is the superposition of the steady-state and the free-state
solutions. We now move on to solving the free-response.
x = Ae−γt/2 cos(ωc t + φ) + |H|cos(ωt + ∠H)
(19)
The free response of the system is determined by A and φ. In order to solve these, we impose the conditions
t = 0, x = 0, ẋ = 0 and form a system of equations from the resultant expressions for x and ẋ at t = 0.
Inserting t = 0 and x = 0 simplifies the above expression for x into
0 = Acos(φ) + |H|cos(∠H)
(20)
Differentiate the expression for x to obtain an expression for ẋ. Below the two terms in x are differentiated
separately for readability. Use the product rule (f ∗ g)0 = f 0 ∗ g + f ∗ g 0 .
Problem 1 continued on next page. . .
2
Sam Student 00000P
Acoustics and the Physics of Sound 2016 Demo #1
Problem 1 (continued)
First part
d −γt/2 d
d −γt/2
(cos(ωc t + φ)) + Ae−γt/2 (cos(ωc t + φ))
Ae
cos(ωc t + φ) =
Ae
dt
dt
dt
d −γt/2
γ
Ae
cos(ωc t + φ) = − Ae−γt/2 cos(ωc t + φ) − ωc Ae−γt/2 sin(ωt + φ)
dt
2
Second part
d
(|H|cos(ωt + ∠H)) = −ω|H|sin(ωt + ∠H)
dt
The first derivative is then
γ
ẋ = − Ae−γt/2 cos(ωc t + φ) − ωc Ae−γt/2 sin(ωt + φ) − ω|H|sin(ωt + ∠H)
2
γ
cos(ωc t + φ) + ωc sin(ωt + φ) − ω|H|sin(ωt + ∠H)
ẋ = −Ae−γt/2
2
Impose ẋ = 0 at t = 0. Exponential disappears and trig. functions lose time-dependence.
γ
ẋ = −A
cos(φ) + ωc sin(φ) − ω|H|sin(∠H) = 0
2
Write system of equations based on imposed conditions.
From x = 0:
Acos(φ) = −|H|cos(∠H) = −a
From ẋ = 0:
−A
γ
cos(φ) + ωc sin(φ) = ω|H|sin(∠H) = b
2
Where we introduce auxiliary variables a and b for clarity.
Rearrange the above eqs. and solve for φ by exploiting the trig. identity
Solve for Asin(φ) using b
Replace Acos(φ) with −a
(22)
(23)
(24)
(25)
(26)
(27)
(28)
sin(θ)
= tan(θ)
cos(θ)
(29)
γ
− A cos(φ) − Aωc sin(φ) = b
2
(30)
γ
− Aωc sin(φ) = b
2
γ
− Aωc sin(φ) = b + a
2
b + a γ2
Asin(φ) = −
ωc
a
Divide and solve tangent
(21)
b + a γ2
Asin(φ)
=
= tan(φ)
Acos(φ)
aωc
b + a γ2
−1
φ = tan
aωc
(31)
(32)
(33)
(34)
(35)
Solve for A using φ
−a
cos(φ)
Insert a, φ and ∠H back into the expression for x to obtain the solution
A=
x = Ae−γt/2 cos(ωc t + φ) + |H|cos(ωt + ∠H)
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(36)
(37)
Sam Student 00000P
Acoustics and the Physics of Sound 2016 Demo #1
Problem 2
Problem 2
A string of length L is attached to a rigid wall on one end, and to a vertical pole via a frictionless ring on the
other. The string has two masses of mass m at equidistant positions along the string. What are the normal
modes of vibration?
l
l
L
l
m
m
θ1
m
m
l
θ2
y2 - y1
y1
L
Figure 2: String attached to a wall and a frictionless vertical pole.
Draw freebody diagram of a snapshot in vibration. Write forces acting on masses in the y-dimension. The
magnitudes of the vertical components of the string tension are
”opposite”
y1
T sin(α1 ) = T sin
(38)
= T sin( )
hypotenuse
l
”opposite”
y2 − y1
T sin(α2 ) = T sin
= T sin
(39)
hypotenuse
l
assume that the angles α1 , α2 are small and hence sin(α) ≈ α.
y y 1
1
=T
T sin(α1 ) = T sin
l
l
y2 − y1
y2 − y1
T sin(α2 ) = T sin
=T
l
l
(40)
(41)
Write equations of motion. Mind the sign (±) of the forces. The first mass experiences two opposing tension
forces.
ma = F
(42)
y y2 − y1
1
mÿ1 = −T
+T
(43)
l
l
For the second mass we have just one component
mÿ2 = −T
Rearrange and clean up equations using
y¨1 =
T
lm
y2 − y1
l
(44)
= ω02 .
T
(y2 − 2y1 ) = ω02 (y2 − 2y1 )
ml
(45)
T
(y2 − y1) = −ω02 (y2 − y1)
(46)
ml
In normal mode, the masses oscillate at the same frequency. Assume that the displacements are described
by complex exponentials:
y1 = A1 ejωt ,
y2 = A2 ejωt
(47)
y¨2 = −
Problem 2 continued on next page. . .
4
Sam Student 00000P
Acoustics and the Physics of Sound 2016 Demo #1
Problem 2 (continued)
Where A1 and A2 are the amplitudes of oscillation and ω is the same for both masses. Write second
derivatives of solutions
y¨1 = −ω 2 A1 ejωt ,
y¨2 = −ω 2 A2 ejωt
(48)
Insert y1 , y¨1 , y2 and y¨2 into previous eqs. to arrive at a system of equations from where we solve the unknowns
(−ω 2 A1 ejωt ) = ω02 (A2 ejωt − 2A1 ejωt )
(49)
(−ω 2 A2 ejωt ) = −ω02 (A2 ejωt − A1 ejωt )
(50)
Cancel out the exponentials as they are common to all terms
(−ω 2 A1 ) = ω02 (A2 − 2A1 )
(51)
(−ω 2 A2 ) = −ω02 (A2 − A1 )
(52)
A1 (2ω02 − ω 2 ) = A2 ω02
(53)
A1 ω02
(54)
Rearrange to get A-terms.
=
A2 (ω02
2
−ω )
A1 and A2 depend on the initial conditions of the system so we don’t care about their absolute values but
rather their ratio A1 /A2 . From this ratio we can solve for ω. Form the ratios for both eqs. separately
ω02
A1
=
2
A2
2ω0 − ω 2
(55)
ω2 − ω2
A1
= 0 2
A2
ω0
(56)
Equate the two equations above and solve for ω
ω02
ω2 − ω2
= 0 2
2
ω0
−ω
(57)
2ω02
Multiply out and rearrange into quadratic form
ω04 = 2ω04 − ω02 ω 2 − 2ω02 ω + ω 4
(58)
0 = ω 4 + ω 2 (−3ω02 ) + ω04
(59)
Lets define x = ω 2 and rewrite the above quartic as a quadratic in x
0 = x2 + x(−3ω02 ) + ω04
(60)
Solve the resulting quadratic for x = w2
2
ax + bx + c = 0,
x=
−b ±
√
b2 − 4ac
2a
b = −3ω02 , c = ω04
p
3ω02 ± (−3ω02 )2 − 4 ∗ 1 ∗ ω04
2
x=ω =
2∗1
a = 1,
(61)
(62)
(63)
The solutions for ω 2 can then be inserted into A1 /A2 to solve for the amplitudes. In one mode the masses
vibrate in phase and in the other in anti-phase.
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