# Mecflu - oscila??o

(Parte 1 de 2)

CHAPTER 3 Oscillations

3-1.

gram cm

2210 gram2gram2k mππ π π

12 sec

or,

c) The maximum velocity is attained when the total energy of the oscillator is equal to the kinetic energy. Therefore,

3-2.

a) The statement that at a certain time 1tt= the maximum amplitude has decreased to onehalf the initial value means that

so that

b) According to Eq. (3.38), the angular frequency is

so that

which can be written as

That is, 1ν is only slightly different from 0ν.

OSCILLATIONS 81

3-3. The initial kinetic energy (equal to the total energy) of the oscillator is 201 2 mv, where m = 100 g and v. 01 cm/sec= a) Maximum displacement is achieved when the total energy is equal to the potential energy. Therefore, mvkx=

or,

b) The maximum potential energy is

3-4.

a) Time average: The position and velocity for a simple harmonic oscillator are given by

where 0kmω= The time average of the kinetic energy is

where

By inserting (2) into (3), we obtain

or, 2

In the same way, the time average of the potential energy is 2 t t

Uk x dt kA t dt kA

This equality is valid instantaneously, as well as in the average. On the other hand, when T and U are expressed by (1) and (2), we notice that they are described by exactly the same function, displaced by a time 2τ:

mA Tt

mA t Ut

Therefore, the time averages of T and U must be equal. Then, by taking time average of (9), we find

b) Space average: The space averages of the kinetic and potential energies are

OSCILLATIONS 83

and 2

A m Uk x dx x

(13) is readily integrated to give 2

To integrate (12), we notice that from (1) and (2) we can write ( )

Then, substituting (15) into (12), we find

Am TA x

m A A or,

From the comparison of (14) and (17), we see that

To see that this result is reasonable, we plot T = T(x) and U = U(x):

x Tm A

T = T(x) A

Energy mA ω

And the area between T(x) and the x-axis is just twice that between U(x) and the x-axis.

3-5. Differentiating the equation of motion for a simple harmonic oscillator,

0sin x t A

Therefore,

and substitution into (2) yields x t

Then, the fraction of a complete period that a simple harmonic oscillator spends within a small interval ∆x at position x is given by tx x

This result implies that the harmonic oscillator spends most of its time near x = ±A, which is obviously true. On the other hand, we obtain a singularity for tτ∆ at x = ±A. This occurs because at these points x = 0, and (2) is not valid.

3-6.

x x x m m

Suppose the coordinates of m and are and x and the length of the spring at equilibrium is . Then the equations of motion for m and are 1 2m 1x 2

OSCILLATIONS 85

From (2), we have

Substituting this expression into (1), we find

from which

Therefore, oscillates with the frequency 2x

We obtain the same result for . If we notice that the reduced mass of the system is defined as 1x

we can rewrite (6) as

k µ

This means the system oscillates in the same way as a system consisting of a single mass µ. Inserting the given values, we obtain µ 6.7 g and 12.74 radsω−⋅ .

3-7. A h h

Let A be the cross-sectional area of the floating body, its height, the height of its submerged part; and let ρ and bh sh

0ρ denote the mass densities of the body and the fluid, respectively.

The volume of displaced fluid is therefore VsAh=. The mass of the body is bMAhρ=.

There are two forces acting on the body: that due to gravity (Mg), and that due to the fluid, pushing the body up (00sgVghAρρ−−=).

The equilibrium situation occurs when the total force vanishes:

Mg gV gAh gh Asρ ρ ρ

which gives the relation between and : shbh

For a small displacement about the equilibrium position (), (1) becomes sh→+x

Thus, the motion is oscillatory, with an angular frequency g A g

where use has been made of (2), and in the last step we have multiplied and divided by A. The period of the oscillations is, therefore,

3-8. y ms 2a

The force responsible for the motion of the pendulum bob is the component of the gravitational force on m that acts perpendicular to the straight portion of the suspension string. This component is seen, from the figure (a) below, to be

OSCILLATIONS 87 where α is the angle between the vertical and the tangent to the cycloidal path at the position of m. The cosine of α is expressed in terms of the differentials shown in the figure (b) as

α α m

F dxdy S

(a)(b)

The differentials, dx and dy, can be computed from the defining equations for x(φ) and y(φ) above:

sin dx a d dy a d φ φφφ

Therefore,

2 2 1c os sin 2 1c os ds dx dy ad aad 2d4s

= (5) so that

Thus, sin

2s in 2 cos cos 2 dy a d ds ad φ φ φ φ

2s in 2

4c os 2 ds d va dt dt d a from which d va

Letting cos2 zφ≡ be the new variable, and substituting (7) and (9) into (1), we have

which is the standard equation for simple harmonic motion,

where we have used the fact that . 4a=A

Thus, the motion is exactly isochronous, independent of the amplitude of the oscillations. This fact was discovered by Christian Huygene (1673).

OSCILLATIONS 89

The homogeneous solutions for both (3) and (4) are of familiar form ()itittAeBeωωξ−=+, where kmω=. A particular solution for (3) is Fkξ=. Then the general solutions for (3) and (4) are it i tF

Then

and, from (5), we have

1c os sin it i t it it

F t t Ce De

F t t i Ce De

(10)

The equations in (10) can be rewritten as: ()

1c os sin it i t it i t

F Ce De t iF Ce De t

Then, by adding and subtracting one from the other, we obtain it it it i t

F Ce e

F De e

2 cos cos i t it i t it it t i t tit it

F e e e

F e e e

F t t

()00coscos; F

3-10. The amplitude of a damped oscillator is expressed by

Substituting this relation into the equation connecting 1ω and 0ω (the frequency of undamped

24nn ωωω ωππ

Therefore,

so that

3-1. The total energy of a damped oscillator is

OSCILLATIONS 91

Substituting (2) and (3) into (1), we have

cos sin 2s in cos tA Et e m k t m t mt t β2 β ωδ ω ω δ

(4)

Rewriting (4), we find the expression for E(t):

Taking the derivative of (5), we find the expression for dE dt :

tdE mA et t β βω β ω δ

The above formulas for E and dE reproduce the curves shown in Figure 3-7 of the text. To find the average rate of energy loss for a lightly damped oscillator, let us take dt 0βω . This means that the oscillator has time to complete some number of periods before its amplitude decreases considerably, i.e. the term 2teβ− does not change much in the time it takes to complete one period. The cosine and sine terms will average to nearly zero compared to the constant term in dEdt, and we obtain in this limit

3-12.

mg sin θ θ

singθθ=− A (2) If θ is sufficiently small, we can approximate sinθθ≅, and (2) becomes

where 0gω=A and where 0θ is the amplitude. If there is the retarding force 2mgθ A, the equation of motion becomes

we identify 0ωβ=. This is just the case of critical damping, so the solution for θ(t) is [see Eq. (3.43)]

If we assume a solution of the form

2t xy e ye xy e ye yeββββ βββ−− −− −

OSCILLATIONS 93 and

3-14. For the case of overdamped oscillations, x(t) and ()xt are expressed by

cosh 2 yye ey −+=, sinh or, cosh sinh cosh sinhyyey ey−

Using (4) to rewrite (1) and (2), we have

cosh sinh cosh sinh cosh sinh xt t t A A t t

A At ββ ω β ω ω

(6)

with the values A = 1 cm,101 radsω−=⋅, 101 sβ−=., and δ = π rad. The position goes through x = 0 a total of 15 times before dropping to 0.01 of its initial amplitude. An exploded (or zoomed) view of figure (b), shown here as figure (B), is the best for determining this number, as is easily shown.

x (cm) v (cm/s) t (s)

(b)

(c)

t (s)

3-16. If the damping resistance b is negative, the equation of motion is

From this equation, we see that the motion is not bounded, irrespective of the relative values of 2β and 20ω.

The three cases distinguished in Section 3.5 now become:

OSCILLATIONS 95

where

This solution also increases continuously with time.

The tree cases describe motions in which the particle is either always moving away from its initial position, as in cases b) or c), or it is oscillating around its initial position, but with an amplitude that grows with the time, as in a).

Because b < 0, the medium in which the particle moves continually gives energy to the particle and the motion grows without bound.

3-17. For a damped, driven oscillator, the equation of motion is

mAT ω

Let the frequency n octaves above 0ω be labeled 1ω and let the frequency n octaves below 0ω be labeled 2ω; that is

The average kinetic energy for each case is

Multiplying the numerator and denominator of (5) by , we have 42n

Hence, we find

3-18. Since we are near resonance and there is only light damping, we have 0Rωωω , where ω is the driving frequency. This gives 02Qωβ . To obtain the total energy, we use the solution to the driven oscillator, neglecting the transients:

The energy lost over one period is

0 energy lost over one period42 which proves the assertion.

3-19. The amplitude of a damped oscillator is [Eq. (3.59)]

Let us find the frequency, ω = ω′, at which the amplitude is 1

2RD:

(3)

Solving this equation for ω′, we find

OSCILLATIONS 97

For a lightly damped oscillator, β is small and the terms in 2β can be neglected. Therefore,

which gives

3-20. From Eq. (3.6),

Ax )tω ω δ

Therfore, the absolute value of the velocity amplitude v is given by

A v

The value of ω for v a maximum, which is labeled 0vω, is obtained from

and the value is 0vωω=.

Since the Q of the oscillator is equal to 6, we can use Eqs. (3.63) and (3.64) to express β in terms

v A A 2ω

Substituting for β in terms of 0ω from (4), and by squaring and rearranging terms in (5), we obtain

from which

Solving for 1ω, 2ω we obtain

It is sufficient for our purposes to consider 1ω, 2ω positive: then

so that

A graph of vs. ω for Q = 6 is shown. 0v

OSCILLATIONS 9

magnified view of figure (a). The dashed line is the path that all paths go to asymptotically as t → ∞. This can be found by taking the limits.

so that in this limit, v = –βx, as required.

x (cm) v (cm/s) x (cm) v (cm/s)

(a) (b)

3-2. For overdamped motion, the position is given by Equation (3.4)

The time derivative of the above equation is, of course, the velocity: ( )

The initial conditions and v can now be used to solve for the integration constants 0x01A and 2A.

3-23. Firstly, we note that all the δ = π solutions are just the negative of the δ = 0 solutions. The 2δπ= solutions don’t make it all the way up to the initial “amplitude,” A, due to the retarding force. Higher β means more damping, as one might expect. When damping is high, less oscillation is observable. In particular, 209β=. would be much better for a kitchen door than a smaller β, e.g. the door closing (δ = 0), or the closed door being bumped by someone who then changes his/her mind and does not go through the door (2δπ=).

OSCILLATIONS 101

1 β = 0.1, δ = π/2 β = 0.9, δ = 0 β = 0.9, δ = π/2 β = 0.5, δ = 0 β = 0.5, δ = π/2

1 β = 0.1, δ = π

1 β = 0.1, δ = 0

3-24. As requested, we use Equations (3.40), (3.57), and (3.60) with the given values to evaluate the complementary and particular solutions to the driven oscillator. The amplitude of the complementary function is constant as we vary ω, but the amplitude of the particular is increased further. The plot closest to resonance here has 1ωω=., which shows the least distortion due to transients. These figures are shown in figure (a). In figure (b), the 16ωω=

ω/ω = 6

xc xp x

Legend:

(b) A = 1A = 20

3-25. This problem is nearly identical to the previous problem, with the exception that now Equation (3.43) is used instead of (3.40) as the complementary solution. The distortion due to the transient increases as ω increases, mostly because the complementary solution has a fixed amplitude whereas the amplitude due to the particular solution only decreases as ω increases. The latter fact is because there is no resonance in this case.

OSCILLATIONS 103 ω/ω = 1.1

0 ω/ω = 6, A = 6 xcxpxLegend: 3-26. The equations of motion of this system are cosmx kx b x x F t

The electrical analog of this system can be constructed if we substitute in (1) the following equivalent quantities:

Lq R q q q t C

Using the mathematical device of writing exp(iωt) instead of cos ωt in (2), with the understanding that in the results only the real part is to be considered, and differentiating with respect to time, we have itI LI R I I i e

Then, the equivalent electrical circuit is as shown in the figure:

ε cos ωt L

1 2C The impedance of the system Z is

where is given by 1Z

Then,

R R R L i L R Z

and substituting (6) into (4), we obtain

1 R R R L i R L L R R L

3-27. From Eq. (3.89),

cos sin2 n n a a n t b n t)Ft ω ω

We write

cos2 n Ft a c n t )nω φ which can also be written using trigonometric relations as

cos cos sin sin2 n t a c nt nt nF ω φω

Comparing (3) with (2), we notice that if there exists a set of coefficients c such that n

OSCILLATIONS 105 cos

sin n n

tan n n n n ca b ba φ with and b as given by Eqs. (3.91). nan

3-28. Since F(t) is an odd function, F(–t) = –F(t), according to Eq. (3.91) all the coefficients vanish identically, and the b are given by na sin sin sin

1 cos cos

4 for odd

0 for even nbF t n t dt nt dt n t dt nt n t n n n n n

Then, we have

Terms 1 + 2 t

Terms 1 + 2 + 3 t

Terms 1 + 2 + 3 + 4 t

3-29. In order to Fourier analyze a function of arbitrary period, say 2Pτω= instead of 2πω, proportional change of scale is necessary. Analytically, such a change of scale can be represented by the substitution t x

for when t = 0, then x = 0, and when 2Ptτω==, then 2xπω=.

Thus, when the substitution tPxπ= is made in a function F(t) of period 2Pω′, we obtain the function

and this, as a function of x, has a period of 2πω. Now, f(x) can, of course, be expanded according to the standard formula, Eq. (3.91):

cos sin2 n n a a n x b n x)fx ω ω where

OSCILLATIONS 107 cos

sinnn af x n x d bf x n x d ω ωπ = x

If, in the above expressions, we make the inverse substitutions t x

the expansion becomes t t nt ntaP F t a b

and the coefficients in (4) become

cos sin

P n

P n nt aF t nt bF t d

For the case corresponding to this problem, the period of F(t) is 4π ω, so that P = 2π. Then,

obtain

sin nt aF t nt bF t and substituting into (6), the expansion for F(t) is nt nta a b

Substituting F(t) into (8) yields 2 sin cos 2

sin sin nt at nt bt

Evaluation of the integrals gives

0 eve b n

a a n nπ and the resulting Fourier expansion is t t Ft t

3-30. The output of a full-wave rectifier is a periodic function F(t) of the form t Ft

The coefficients in the Fourier representation are given by sin cos sin cos

sin sin sin sinnn at n t dt t n bt n t dt t n t dt

(2)

Performing the integrations, we obtain

4 ;if even or 0

0;if odd

0for all n n n na

The expansion for F(t) is

The exact function and the sum of the first three terms of (4) are shown below.

OSCILLATIONS 109

Sum of first three termsF(t)sin ωt

3-31. We can rewrite the forcing function so that it consists of two forcing functions for t > τ:

Ft at t

at at t

During the interval 0 < t < τ, the differential equation which describes the motion is

The particular solution is , and substituting this into (2), we find pxCt=+D

from which

Therefore, we have

which gives

2p a x t 4β

Thus, the general solution for 0 < t < τ is

2 cos sint a

and then,

Therefore, the response function is cos 1 sinttae e t t tββ 2xt

two equations to obtain the total response function:

2 cos cos 1

sin sin t t ae xt e t e t te t βββ β

When τ → 0, we can approximate eβτ as 1 + βτ, and also 11sinωτωτ≅, co1s1ωτ≅. Then, () () ()

(Parte 1 de 2)