12 Vibrations: one degree of freedomn.ethz.ch/~mneumann/files/MechanicsIII/Solutions14... · 1 + k...

12 Vibrations: one degree of freedom

12.1 In-class problem A block of mass m is suspen-ded from two springs having a stiffness of k1 and k2,arranged

(a ) parallel to each other, and

(b ) as a series.

Determine the equivalent stiffness of a single spring withthe same oscillation characteristics and the period ofoscillation for each case.

Remark: Include gravity in your treatment, andassume that the unextended length of both springs is l.

Solution

(a)

I. Geometry and free-body diagram

We assume that the unextended length of both springs is l and indicate the forces bytheir magnitude:

We have

F1 = k1(x− l) (12.1)

F2 = k2(x− l) (12.2)

(12.3)

II. Apply the appropriate principle

The linear momentum principle reads

mx=mg − F1 − F2 (12.4)

=mg − (k1 + k2)x+ (k1 + k2)l. (12.5)

12-1

12 Vibrations: one degree of freedom 12-2

Introducing the equivalent stiffness

keq = k1 + k2 , (12.6)

we have

mx+ keqx = mg + keql. (12.7)

Introducing

y := x− l − mg

keq(⇒ y = x), (12.8)

we can shift the origin to the position of the equilibrium.1 We are now back to the usualequation of motion

y +keqmy = 0. (12.9)

With ω20 := keq

m, the period of oscillation T becomes

T =2π

ω0

= 2π

√m

k1 + k2

. (12.10)

(b)

The stretch x1 of the first spring should be equal to F ,

F = k1x1, (12.11)

and the same for the second one with stretch x2:

F = k2x2. (12.12)

1You can picture it by thinking of the equivalent spring that has stiffness keq: Its unextended lengthis l. Once we attach the mass m to its end, due to the gravitational force mg, the spring will extend tothe new length l + mg

keq. This is the term that we subtracted in (12.8).


The stretch of the equivalent spring, xeq = x1 + x2, multiplied by its stiffness keq shouldalso be equal to F :

F = keq(x− l1 − l2) = keqxeq = keq(x1 + x2). (12.13)

We have 3 equations to solve for keq. Equaling (12.11) and (12.12), we get

x2 =k1

k2

x1, (12.14)

and plugging into (12.13),

F = keq(x1 +k1

k2

x1). (12.15)

Comparing this to (12.11), we need

keq(x1 +k1

k2

x1) = k1x1. (12.16)

The result for keq is

keq =k1

1 + k1k2

=k1k2

k1 + k2

. (12.17)

The period of oscillation T is therefore going to be

T = 2π

√m(k1 + k2)

k1k2

. (12.18)


12.2 In-class problem

The seismograph consists of a mass m suspended from the ceiling via a spring of stiffnessk. Let the natural frequency of this system be ω0 = 2.31

s. Attached to the mass there is

a pen that plots the displacement between ceiling and mass m onto a rotating cylinder.Assume that the ceiling oscillates like

e(t) = e0 sinωt, (12.19)

with ω = 561s

being the frequency of the earthquake. Let the length of the unextendedspring be l.

(a ) Derive the equation of motion using the x-coordinate.

(b ) Find a particular solution xp(t).

(c ) Find initial conditions for this system so that the general solution is x(t) ≡ xp(t),i. e., the contribution of the homogeneous solution xh(t) is zero.

(d ) In the situation of part (c), estimate the precision of the device. To that end, computethe relative error f between the real signal e(t) and the measured signal e(t):

f =|e| − |e||e|

. (12.20)

Solution

(a)

I. Geometry and free-body diagram

We indicate the forces by their magnitude:


The magnitude of the restoring force is

F = k(x− e(t)− l). (12.21)

II. Apply the appropriate principle

From the linear momentum principle we get

mx = mg − k(x− e(t)− l), (12.22)

which we rearrange to

mx+ kx = ke(t) + (mg + kl) , e(t) = e0 sinωt. (12.23)

(b)

We introduce the following Ansatz for the particular solution xp(t) of (12.23):

xp(t) = A sinωt+B, (12.24)

which implies

xp(t) = −Aω2 sinωt. (12.25)

Plugging the above two expressions into the ODE (12.23), we find

−mAω2 sinωt+ kA sinωt+ kB = ke0 sinωt+ (mg + kl), (12.26)

and after regrouping terms:[A(−mω2 + k)− ke0

]sinωt+ [kB − (mg + kl)] = 0. (12.27)

The two terms inside the square brackets have to be zero each, therefore we need

A = e0k

k −mω2= e0

ω20

ω20 − ω2

, ω20 :=

k

m, (12.28)

and

B = l +mg

k. (12.29)


Plugging these expressions for A, B into the Ansatz (12.24), we obtain the particularsolution

xp(t) = e0ω2

0

ω20 − ω2

sinωt+(l +

mg

k

). (12.30)

(c)

With the homogeneous solution labeled as xh(t), the general solution is x(t) = xh(t) +xp(t). In order to have x(t) ≡ xp(t) for all time, we need the two initial conditions for theODE (12.23) to be picked according to this requirement:

1. We need

x(0) = xp(0), (12.31)

which means we should take

x(0) = l +mg

k. (12.32)

2. We have xp(t) = e0ωω2

0

ω20−ω2 cosωt. In order to get xp(t) = xh(t) + xp(t)

!= xp(t) for all

time, we need to pick

x(0) = xp(0) = e0ωω2

0

ω20 − ω2

. (12.33)

(d)

After somehow initializing the device according to the initial conditions (12.32), (12.33),the trajectory of the tip is given by

x(t) = e0ω2

0

ω20 − ω2

sinωt+(l +

mg

k

). (12.34)

The second term should be considered a geometric constant of the device and thereforesubtracted. Also subtracting the displacements of the cylinder due to the earthquake, e(t),we finde the measured signal to be

e(t) = e0ω2

0

ω20 − ω2

sinωt− e0 sinωt ≡(

ω20

ω20 − ω2

− 1

)e(t). (12.35)

With the last expression at hand, we evaluate the relative error:

f =|e| − |e||e|

(12.36)

=

∣∣∣∣ ω20

ω20 − ω2

− 1

∣∣∣∣− 1 (12.37)

ω0<ω=

(1− ω2

0

ω20 − ω2

)− 1 (12.38)

=ω2

0

ω2 − ω20

. (12.39)

For the numerical values ω0 = 2.31s

and ω = 561s, we obtain f ≈ 0.17 %. Since f > 0, we

are slightly overestimating the magnitude of the earthquake.


12.3 Homework The three diagrams A, B, C shown below are graphs of dampedvibrations. The vertical axis measures the oscillator’s displacement y, the horizontal axisis time t. We later use M for the oscillator’s mass, δ for the characteristic damping, andk for the spring constant.

(a ) Which curve corresponds to free linear vibrations? Discuss each diagram.

(b ) In this setting, we introduce the logarithmic decrement as

Λ = logy(t)

y(t+ T ), (12.40)

with T being the oscillation period, and log the natural logarithm. Use the definitionsfrom the lecture notes to show that Lehr’s damping D can be computed as

D =

√Λ2

Λ2 + 4π2. (12.41)

This formula can be used to obtain D directly from the graph of y.

(c ) Use Eq. (12.41) to compute Lehr’s damping D from the graph that you chose in part(a).

(d ) Considering the mass M to be given, compute the characteristic damping δ and thespring constant k.


Solution

(a)

We see oscillations with decaying amplitudes. We are therefore looking for free, under-damped vibrations as discussed in Sec. VII.1.1, Part C. of the lecture notes. The generalsolution can then be written as

y(t) = e−δt (C1 cosωδt+ C2 sinωδt) , (12.42)

where

ωδ =√ω2

0 − δ2, (12.43)

with ω0 =√k/M being the natural frequency of the undamped system. As you know

from the lecture, Eq. (12.42) may also be written as

y(t) = C0 e−δt sin(ωδt+ ϕ0), (12.44)

where ϕ0 is a constant. This makes explicit that we have an oscillation with constantperiod Tδ = 2π

ωδand decaying amplitude C0 e−δt.

• Curve A decays linearly, and not like C0 e−δt.

• Curve C shows oscillations with variable period.

• Curve B is the only choice consistent with (12.44).

(b)

We list the following definitions and identities:

δ=ω0D, (12.45)

Tδ =2π

ωδ, (12.46)

ωδ =√ω2

0 − δ2 = |ω0|

√1−

(δ

ω0

)2

= ω0

√1−D2. (12.47)

Here we used that ω0 is not an angular velocity component, but a circular frequency:It cannot be negative, i.e., |ω0| = ω0. Now derive the expression for the logarithmicdecrement Λ:

Λ = logy(t)

y(t+ Tδ)(12.48)

= loge−δt sin(ωδt+ ϕ0)

e−δ(t+Tδ) sin(ωδ(t+ Tδ) + ϕ0)(12.49)

= log eδTδ (12.50)

= δTδ (12.51)

(12.45)−(12.47)=

2πD√1−D2

(12.52)


Solving the last equation for (positive) D, we get

D =

√Λ2

Λ2 + 4π2. (12.53)

(c)

In order to obtain Λ, we need to measure y(t) and y(t + Tδ) from the graph of y (Fig. Bin the assignment text):

On my printout,

y(t) ≈ 2 cm, y(t+ Tδ) ≈ 1.65 cm,

and therefore

Λ = logy(t)

y(t+ Tδ)≈ 0.19. (12.54)

Using Eq. (12.53), we get

D ≈ 0.03. (12.55)

(d)

We need to obtain the numerical value for δ. From Eq. (12.51), we see that

δ =Λ

Tδ≈ 0.15 s−1. (12.56)

From the graph, we measure Tδ ≈ 1.25 s, and therefore δ = ΛTδ≈ 0.15 s−1.

In order to relate the known values to k and M , recall the following definitions:

D =δ

ω0

, and ω0 =

√k

M. (12.57)

Putting these expressions together by eliminating ω0, we find

k

M=

(δ

D

)2

. (12.58)

Plugging in δ ≈ 0.15 s−1 and D ≈ 0.03 yields

k

M≈ 25

1

s2. (12.59)


12.4 Homework Consider the following pendulum consisting of a homogeneous thinrod with center of mass at C, and of a point mass m at the rod’s endpoint B. The roditself is of mass M and length l. It is suspended at point O, and may rotate withoutfriction. We label its angular displacement with respect to the vertical axis by ϕ. Betweenpoints B and A, there is a massless spring of stiffness k. The displacement of point A withrespect to O is given by a horizontally, and b = l/

√2 vertically. Let the restoring force of

the spring be zero when |rBA| = 0.

(a ) Consider M = m. Derive the equation of motion for ϕ.Hint: The result is

ϕ =3

4

ka

mlcosϕ− 3

8

(√2k

m+ 3

g

l

)sinϕ. (12.60)

You may use it to solve the remaining parts of this exercise.

(b ) Determine a such that the equilibrium is at ϕ0 = π4.

(c ) Introducing the new variable x(t) = ϕ(t)−ϕ0, expand the equation of motion (12.60)around ϕ0 = π

4, to obtain the linear ordinary differential equation that describes free

vibrations about the equilibrium. Find an expression for the natural frequency ω0.

Solution

(a)

I. Geometry, reference system and coordinates

Please refer to the assignment text. We will need the following vectors:

rOC =l

2cosϕi+

l

2sinϕj ≡ rCB, (12.61)

rOB = l cosϕi+ l sinϕj, (12.62)

rOA = bi+ aj, (12.63)

rBA = rOA − rOB. (12.64)


II. Free-body diagram

We cut twice, introducing g = gi, and reaction forces R = Rxi+Ryj and T = Txi+ Tyj:

III. Apply the appropriate principles to eliminate the internal forces and derivethe equation of motion for ϕ

• Linear momentum principle for mass m:

mrOB = −T +mg + krBA. (12.65)

Differentiating (12.62) twice, using the other expressions from Step I to write outkrBA, and then solving for T , we get

Tx = kb+mg − kl cosϕ+ml cos(ϕ)ϕ2 +ml sin(ϕ)ϕ, (12.66)

Ty = ka− kl sinϕ+ml sin(ϕ)ϕ2 −ml cos(ϕ)ϕ. (12.67)

• Linear momentum principle for the rod:

MrOC = R +Mg + T , (12.68)

which yields

Rx =−1

2

[2Mg + 2Tx +Ml cos(ϕ)ϕ2 +Ml sin(ϕ)ϕ

], (12.69)

Ry =1

2

[−2Ty −Ml sin(ϕ)ϕ2 +Ml cos(ϕ)ϕ

]. (12.70)

• Angular momentum principle for the rod, with respect to its center of mass C:

HC = M extC . (12.71)

Using the centroidal moment of inertia IC , the angular momentum HC can bewritten as

HC = IC ϕ k. (12.72)


From the lecture notes p. 55, we read off the expression for the moment of inertiaIC :

IC =1

12Ml2. (12.73)

We evaluate the external torque M extC using the points O and B shown in the free-

body diagram:

M extC = rCO ×R + rCB × T (12.74)

=l

2[(−Ry + Ty) cosϕ+ (Rx − Tx) sinϕ] k. (12.75)

Putting everything together, and replacing the internal forces R, T using the ex-pressions we derived from the linear momentum principle, the k-component of theangular momentum principle (12.71) reads:

1

12Ml2ϕ = −1

4l (−4ka cosϕ+ 2(2kb+ 2mg +Mg) sinϕ+ l(4m+M)ϕ) .

(12.76)

Solving the last equation for ϕ yields

ϕ =3

2· 2ka cosϕ− (2kb+ (2m+M)g) sinϕ

(3m+M)l. (12.77)

Using b = l/√

2 and M = m, this simplifies to

ϕ =3

4

ka

mlcosϕ− 3

8

(√2k

m+ 3

g

l

)sinϕ. (12.78)

(b)

At equilibrium, we have ϕ!

= 0 and ϕ!

= 0. The second condition means that we needto plug the desired equilibrium value ϕ = ϕ0 into Eq. (12.78) and set the left hand sideto zero:

0!

=3

4

ka

mlcosϕ0 −

1

8

(3√

2k

m+ 9

g

l

)sinϕ0. (12.79)

With ϕ0 = π4, and thus sin π

4= cos π

4= 1√

2, we get

a =l√2

+3

2

mg

k. (12.80)

With this choice of parameter, the ODE (12.78) becomes

ϕ+3

8

[√2k

m+ 3

g

l

](sinϕ− cosϕ) = 0. (12.81)


(c)

Now we linearize about ϕ0 = π4

by introducing the new variable x = ϕ − ϕ0. The ex-pressions we need to plug into (12.81) are therefore

ϕ = ϕ0 + x =π

4+ x ⇒ ϕ = x ⇒ ϕ = x. (12.82)

We also need the following Taylor expansions, valid for small x (make sure you know howto derive these):

sin(π

4+ x)

=1√2

+1√2x+O(x2), (12.83)

cos(π

4+ x)

=1√2− 1√

2x+O(x2). (12.84)

With these equations, Eq. (12.81) can be expanded as

x+3

8

[√2k

m+ 3

g

l

](1√2

+1√2x− 1√

2+

1√2x+O(x2)

)= 0. (12.85)

The linear ODE is therefore

x+3

4

(k

m+

3√2

g

l

)x = 0, (12.86)

describing free undamped vibrations with natural frequency

ω0 =

√3

4

(k

m+

3√2

g

l

). (12.87)


12.5 Homework A wheel of mass m is suspendedfrom three equal-length cords. When we give it a smallangular displacement of θ about the z-axis and thenrelease, we observe that the period of oscillation is τ .

The wheel’s center of mass is on the z-axis. Itscentroidal moment of inertia about that axis, Iz, is un-known. We can however write it in the form

Iz = mρ2z, (12.88)

where we introduced the so-called radius of gyration ρz,with 0 < ρz < r.

Find an expression for the period of oscillation τ ,and from that, determine ρz.

Some remarks: Assume that the cords remain taut,and that L is at least on the order of r. You can therefore treat the cords as straight lineswhich for small nonzero θ enclose a small angle with the vertical axis. The force that eachof them exerts on the wheel always remains aligned with the respective cord.

Solution

I. Geometry

This is a problem with one degree of freedom, and we want to use θ as the generali-zed coordinate. The following picture shows the relevant geometric quantities, with φbeing the angle between each rope and the vertical axis:

We could have drawn this three times, but by symmetry, the situation is the same foreach cord. (This reminds of the planetary gear problem in Ex. 8.2.)

We already work out a relation that allows us to express φ in terms of θ: Since weare only deriving a model that holds for small angles θ and φ, from the picture we getthat

L tanφ ≈ r tan θ. (12.89)


The Taylor expansion of tan x about x = 0 is

tanx = x+O(x2), (12.90)

which is a good approximation (only) for small x with |x| � 1. We use this to approximateEq. (12.89) as

φ =r

Lθ. (12.91)

II. Free-body diagram

You can explicitly introduce the tension for each string as a force vector, e.g. writeT = Txi+ Tyj + Tzk with |T | = T .

III. Apply the appropriate principles

First we apply the linear momentum principle for the wheel to determine the stringtension T . Taking the z-component:

mz = 3T cosφ−mg. (12.92)

With z = const., i.e. z = 0, we get that

T =mg

3 cosφ. (12.93)

Next we apply the angular momentum principle with respect to the wheel’s centerat x = y = z = 0, which is its center of mass. In the z-component:

Iz θ = 3M strO,z, (12.94)

where M strO,z is the z-component of the torque that each cord exerts on the wheel, with

respect to the origin O at x = y = z = 0. Considering the free-body diagram, we obtain2

M strO,z = −rT sinφ. (12.95)

2E.g. use the vector indicated on the left of the free-body diagram: MstrO,z = (ri × T ) · k = rTy =

r·(−T cos(90◦ − φ)), or use the formula (a×b)z = (|a||b| sin∠(a, b))z for two vectors a, b in the x−y−plane.


Together with Iz = mρ2z, the angular momentum principle yields

mρ2z θ = −3rT sinφ. (12.96)

We eliminate T using (12.93):

mρ2z θ = −rmg sinφ

cosφ. (12.97)

Now we replace φ by θ using (12.91), to obtain a non-linear equation of motion for θ:

θ +gr

ρ2z

tan( rLθ)

= 0. (12.98)

Within the small-angle approximation, we may use tan x = x +O(x2) again and get thelinear ODE

θ +gr2

ρ2zLθ = 0. (12.99)

This describes free vibrations with natural circular frequency

ω0 =r

ρz

√g

L, (12.100)

and period of oscillation

τ =2π

ω0

= 2π

(ρzr

√L

g

). (12.101)

This means that the radius of gyration ρz can be determined from τ via

ρz =τr

2π

√g

L. (12.102)

12 Vibrations: one degree of freedomn.ethz.ch/~mneumann/files/MechanicsIII/Solutions14... · 1 + k...

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