Physics Chapter 15 Oscillations

Oscillations

CONCEPTS IN CONTEXT

Cone

The body-mass measurement device shown is used aboard the International Space Station for th e daily measureme nt of th e masses of th e astron aut s. The device consists of a spring coupled to a chair into which the astron aut is strapped. Pu shed by the spring, the chair with the astronaut oscillates back and forth. W e will see in thi s chapter that the frequ ency of oscillation of the mass-spring system dep end s on th e mass, and th erefore the frequency can serve as an indi cator of the ma ss of th e astrona ut. W hile learn ing abo ut oscillating systems , we will consider such question s as:

? W hen the spring pushes and pulls the astronaut, wh at is the position of the astronaut as a functio n of time ? The velocity of the astronaut? (Example 4, page 478)

'

..

;n Conte

15.1

469

Simple Harmonic Motion

? What is th e total mechani cal ener gy of the astronaut- spring system? What are the kineti c and potential energies as th e spring begins to push? At later times? (E xampl e 5, page 482)

? Good oscillato rs have low friction. H ow do we measure the quality of an oscillator? (Ex ample 10, page 490)

T

he motion

ofa p art icle or ofa system ofp articles is periodic, or cyclic , if it repeats again

an d agai n at regular interv als oftime . The orbital motion of a planet aro und the Sun, th e uniform rotational motion of a carousel or of a circular saw blade, th e back rn d- forth mot ion of a pisto n in an auto mobile engine or in a wat er pump, th e swing:ng motion of a pendulum bob in a grandfather clock, and the vibration of a guitar string are example s of periodic motion s. Ifthe periodic motion is a back-and-jOrth motion .dong a straight or curv ed lin e, it is called an oscillation. Thus, the motion of th e piston :5 an oscillatio n, and so are th e mo tio n of the pendulum and th e moti on of th e indi vidual particles of the gu itar string. In this chapter we will examine in some detail the motion of a mass oscillating back and forth under the push and pull exerted by an ideal, massless spring. The equa rions that we will develop for the description of thi s mass-spring system are of great im portance because analogou s equations also occur in the description of all other oscillating systems. We will also examine some of these other oscillating systems, such as :he pendulum .

15.1 SIMPLE HARMONIC MOTION

Online

Concep t Simple harm onic mot ion is a special kind of one-dimensional periodic motion . In any of one-dimensional periodic motion, th e particle moves back and forth alon g J straigh t line, rep eating th e same moti on again and again . In the special case of simple harmonic motion, thep art icle'sposition can be exp ressed as a cosine or a sinef unc.lon ofti me. As we will see later, the motion of a mass oscillating back and forth under :he push and pull of a spring is simple harm onic (Fig. l5.l a), and so is the moti on of J pendulum bob swinging back and forth (provided the amplitude of swing is small; see Fi g. l5 .lb), and so is th e up -and-d own moti on of th e blad e of a sa be r saw Fig. 15 .Ic), However, in th is first section we will merely deal with th e mathematical description of simple harm onic mo tion, and we will postpone until th e next section (he question ofw hat causes the moti on. As a num erical example of simple harmonic motion, suppose th at the tip of th e blade in Fig. l5.lc moves up and down between x = -0.8 cm and x = +0.8 cm (where

Tutorial



.a)

-

FIGURE 15 .1 (a) T he motio n of a particle oscilla ting back an d for th in response to the push and pu ll of a spring is sim ple harmonic. (b) The mot ion of a pendulum bob is approximately simple ha rmo nic. (e) T he mo tion of a saber saw blade is simple harmonic.

(bl

(el

W hen motor rum, wheel. . .

...blade moves up and down.

CHAPTER 15

470

Oscillations

the x axis is assumed to be vertical); furth er suppose that the blade c.=. pletes 50 up-and-down cycles each second. Figure 15.2 gives a plot :--posi tion of the tip of the blade as a fun ction of tim e. The plot in 15.2 has th e mathematical form of a cosin e function of the time t.

x

x = 0.8COS(1007Tt)

wh ere it is assumed th at dist ance is measured in centimeters and -in seconds, and it is assumed th at th e "angle" 1007Tt in the cosine r - tion is reckoned in radi ans. [The factor 1007T multiplying t in Eq. (": • -0.8 has been selected so as to obtain exactly 50 complete cycles each sec - M otion is simple harmonic which is typical for saber saws; we will see below in Eq. (15 .5) ho w u; if position is a cosine (or factor multiplying tin E q. (15.1) is related to the period of the mo ues sine) function of time. Cosines and sines are called harmonic functions, which is why we _ FIGURE 15 .2 P lot of positionvs. time the motion harmonic. For the harm onic motion plotted in Fig. 15.2, ar . = for a case of sim ple harmonic m otion up 0, the blade tip is at its maximum upward displacement [evaluating Eq. (15.1) at t = and down along th e x axis. we have cos 0 = 1, so x = 0.8 ern] and is just starting to move; at t = 0.005 S, it P" through the midpoint [since coS(1007T X 0.005) = cos (7T/ 2) = 0, Eq . (15.1) gives x = 0]; at t = 0.010 S, it reaches maximum downward displacement [COS(7T) = -1, so x = em]; at t = 0.015 S, it again passes through midpoint . Finally,at t = 0.020 s, the tip rercrr, to its maximum upward displacement, exactly as at t = O-it has completed one the mot ion and is ready to begin the next cycle.Thus, the period T, or the repeat ti .• the moti on (th e number of seconds for one complete cycle of the motion), is

a

T = 0.020 s and the frequencyf of the motion, or the rate of repetiti on of the motion (the nu ru: of cycles per second), is 1 T

1 0.020 s

f = - = - - = 50/s The points x = 0.8 em and x = -0.8 em, at whi ch the x coordinate attains its mac mum and minimum valu es, are the turning points of the motion ; and the po.r.' x = 0 is the midpoint. Equation (15.1) is a special example of simple harm onic motion. More genera.' the mot ion of a particle is simple harmonic if the dependence of position on tim e 1-._ th e form of a cosine or a sine fun ction, such as simple harmonic motion

x = A cos(wt + 0)

(15.-

The quantities A, co, and 0 are consta nts. The quantity A is called the amplitude .the motion; it is simply th e distance between the midpoint (x = 0) and either of '._ turning points (x = +A or x = -A) .The quantity w is called the angular its value is related to the period T To establish the relation ship between wand T, northat if we increase the time by T (from t to t + T), the argument of the cosine in Eq. increases by w T. For this to be one cycle of the cosine function, we must require w T = 2-;T hus, the repetition tim e of th e motion, that is, the peri od Tofthe motion, is relate i to the angular frequen cy by

period and angular frequency

T= 27T W

or

W =

27T T

The repetition rate, or th e frequency of the motion, is l iT, so we may write

(15.5

15.1

or

w

Simple Harmonic Motion

= 27rf

471

frequency and angular frequency

(15 .6)

N ote th at th e angu lar frequency w and th e frequency f differ by a facto r of 27r, -.vhich corresponds to 27r rad ians = 1 cycle. The units of angu lar frequ ency are radi m s per second (radia ns/s) ,The units of frequency are cycles per secon d (cycles/s) . Like the label revoluti on th at we used in revl s in rotational mot ion , th e label cycle in cyclels . an be o mitted in th e course of a calculation, and so can th e label radian in radian/so 3 ut it is useful to ret ain the se lab els wh erever th ere is a chance of confus ion. The 51 urut offrequency is called th e hert z (Hz): 1 hert z

= 1 H z = 1 cyclels = l i s

(15 .7)

hertz (Hz)

For ins tance, in the example of the motion of th e saber saw blade , the per iod of th e 'lotion is T = 0.020 s, th e frequ en cy isf = l / T = 1/ (0.020 s) = SOls = 50 Hz, and .he angular frequency is w

= 27rf = 27r X

SOl s = 314 radiansls

ere, in th e last step of the calculati on, th e label radians has been insert ed , so as to ':i5tingu ish th e angul ar frequ ency w fro m the ordin ary frequ encyf T he argu ment (wt + 8 ) of the cosine funct ion is called th e phase of the oscilla. n, and th e quantity 8 is called th e phase constant. This cons tan t determi nes at what mes th e part icle reaches th e point of maxim um displacement, wh en cos(wt + 8) = 1. ne such inst ant is when

· at is, whe n phose constant and time of maximum displacement

(15.8) H ence th e particle reach es th e po in t of maximum di splacem ent at a time

0/w before

· = 0 (see Fig. 15.3) . O f course, the particl e also passes th rough thi s point at periodic

ntervals before and after this tim e. If the pha se constant is zero (8 = 0), th en th e max:::1Um displacem ent occurs at t = O. N ote that th e precedi ng equ ations con necting angularfrequ ency, peri od , and frecu ency are formally th e same as the equa tions connecting angular velocity, period, and :iequency of u niform rot ati onal motion [see E qs. (12.4) and (12.5)]. This coincidence (b) x

(c) x

0= - rr/ 4

Positive phase constant advances cosine peak to before ! = O.

Negative phase constant delays cosine peak to after! = O.

• GURE 15.3 Examples of cosin e fu nction s cos(wt + 8) for sim ple harm on ic moti on wi th d ifferent -hase cons tants. (a) 8 = O. The par ticl e reac hes maximu m displacem ent at t = O. (b) 8 = 71'/4 (or 45°). Th e parti cle reach es maximu m displaceme nt befor e t = O. (c) 8 = -71'/4 (or -45°). T he particle -eaches maxi mu m dis placem en t af ter t = O.

472

CHAPTER 15

O scillation s

arises from a special geometrical relation ship between simple harm onic and un iform circular mot ion . Sup pose that a particle moves with si monic motion according to Eg . (15.4), with amplitude A and angular \ and consider a "satellite" particle th at is constrained to move in un if :lar motion with angular velocity w along a circle of radius A, centere midp oint of the harm onic motion, that is, centered on x = O. F igure 1:,thi s circle, called th e reference circle. At time t = 0, both th e par ri cie _ satellite are on the x axis at x = A . Mter this time, the particle move th e x axis, so its position is x = A cos(wt)

M eanwhil e, th e satellite moves around the circle, and its angular posi

e=

FIGURE 15.4 Par ticle oscilla ting along th e x axis and satelli te p article mov in g around referen ce circle. T he particle and th e satellite are always aligned verti cally; that is, th ey h ave th e sam e x coordinat e.

wt

Now note that the x coordinate of the satellite is the adjacent side of the tria ngl: in Fi g. 15.4: x sat

= A co s e = A cos(wt)

Co m paring thi s with E g. (15.9), we see th at th e x coordinate of the satelli e coincides with the x coordi nate of th e particle; th at is, the particle and th e : always have exactly the same x motion. This mean s that in Fi g. 15.4 the S;1: _ always on that point of th e referen ce circle directly above or directly below the r _This geomet rical relation ship between simple harmonic motion and uni fo:cular mo tion can be used to generate simple harm onic moti on from uniform _ _ moti on . Figur e 15.5 shows a simple mech anism for accomplishing this by mea slotted arm placed over a peg that is attached to a wheel in uniform circular rn The slot is vert ical, and the arm is constrained to move horizontally.The pc the role of "satellite," and th e midp oint of the slot in th e arm plays th e role or"; _cle." The peg drags the ar m left and right and makes it move with simple harrr moti on. A mechanism of thi s kind is used in electr ic saber saws and other dcvi convert the rotational motion of an electric motor int o the up-and-down motion :' saw blade or other moving component. Finally, let us calculate th e instanta neous velocity and instant ane ous acceler in simp le harm onic motion. If the displ acement is (F _

x = Acos(wt + o )

th en differentiation of thi s displacement gives th e velocity v

dx

=- = dt

- w/l sin( wt

+ 0)

Circular motion is converted into linear motion.

FIGURE 15.5 Rotating w heel with a peg driving a slotted arm back and forth .

Slotted arm is constrained to move horizontally.

_ I (a)

(b)

15.1

MATH HELP

Simple Harmonic Motion

473

DERIVATIVES OF TRIGONOME TRIC FUNCTIONS

Under the assump tion tha t the argumen t of each trigo nometri c function is expressed in rad ians, the derivatives or the sine, cosine, and tan gent are d - cos bu = - b sin bu du

d .

- srn bu = b cos bu du

d 2 b - tan bu = b sec bu = - -2du cos bu

and differentiation of th is velocity gives the accelerat ion a

=

d 2x

-

dt

2

dv

=-

dt

=

2

- w A cos(wt

+ 0)

(15.13)

Here we have used the standa rd form ulas for the derivatives of the sine function and the cosine func tion (see M ath H elp: D erivatives of Tr igonometric Function s). Bear in mind that the argu ments of the sine and cosine func tions in this chapter (and also the next) are always expressed in radians, as required for the validity of the standard for:nulas for derivatives. As expected, the instantaneou s velocity calculated from Eq. (15.12) is zero for wt + 0 = 0, when the particle is at the turn ing point. Furthermore, the ins tantaneous velocity attains a maximum magnitude of (15 .14)

maximum velocity

.or co: + 0 = 7T/ 2, when the particle passes thr ough th e midpoi nt (note th at the maxunum magn itude of sinwt is 1). F igure 15.6 shows a multiple-exposure photograph of the oscillations of a particle .n simple harm onic motion .The picture illustrates the variationso f speed in simple har:nonic motio n: the particle moves at low speed (smaller displacements between snapshots) :-.ear the turn ing points, and at high speed (larger displacements) near the midpoint . The velocity (15.12) is a sine function, whereas the displacemen t (15.11) is a cosine .u nction , When t he co sine is at its max im um (say, cos O = 1), th e sine is small sin 0 = 0); whe n the cosine is smal l (say, cos 7T/ 2 = 0) , the sine is at it s maxim um

fiGURE 15 .6 Seq uence of snap sho ts at uniform tim e in tervals of an oscillating mass on a spring (a- h). N ote t hat the mass moves slowly at the extreme s of its moti on. (a)

(b)

(c)

(d)

(e)

(f)

(g)

(h)

474

CHAPTER 15

Oscillati ons

(sin 7T/ 2 = 1). Hence the displacement and the velocity are out of step- whehas a large magnitude, th e other h as a small magnitude, and vice versa. Figure, and b compare the velocity and the displacem ent for simple harmonic moti on a :.... ferent tim es. Graphically, th e velocity is the slope of the position vs. time curve. 1 , th e position goes through a maximum or minimum, the slope is zero; when the tion goes through zero , th e magnitude of th e slope is a maximum . Comparison of Eqs. (15.11) and (15.1 3) shows that

acceleration in simple harmonic motion

(a)

Thus, the acceleration is always prop ortional to the displacement x, but is in th e o:r site direction; see Fi g. 15.7 c. This proportio nality is a characteristic feature of si harmonic motio n, a fact th at will be useful in the next section. Even when a nomenon does not involve motion along a line (for example, rotational motion c behavior of electric circuits), harmonic beh avior occurs wh enever the second tive of a qu antity is proportional to the negative of that quantity, as in Eq. (15. The sine and cosine functions (or a combination of them) are the only fun ctio ns . . have thi s property.

x

(b)

v

When displacement has a large magnit ude, velocity has a small magnitude.

Consider the blade of a saber saw m oving up and dow:". simple harmonic moti on with a frequ enc y of 50.0 H z, or angular frequen cy of 314 radians/so Suppose that the amplitu de of the morioc 1.20 em and th at at time t = 0, the tip of the blade is at x = 0 and its velocin positive. What is the equation describing the position of the tip of the blade function of time? How long does the blade take to travel from x = 0 to x = 0.60 To 1.20 em?

EXAMPLE 1

SOLUTION : The pos ition as function of time is given by Eq. (15.4): (c)

x =A cos(wt

:

1:""""'i

t

•

!

t

+ 0)

with to = 314 radians/s and A = 0.0120 m. Since x = 0 at t= 0, we must adop; • value of 0 such that cos 0 = O. The smallest values of 0 th at satisfy thi s cond itio , _ 0= 7T/2 and 0 = -7T / 2 (other possible values of 0 differ from the se by :±:27T, = ,:etc.). From Eq . (15.12), we see th at to obtain a positive value of vat t = 0, we nee; a negat iv e value of 0; that is, 0 = - 7T/ 2. So the equation describing the motion "

If W hen displacemen t is large and positive, acceleration is large and negative.

x

= (0.0120 m) cos [ (314/s)t -

The tip of the blade reache s x

;]

= 0.00 60 m when

FIGURE 15 .7 (a) Position , (b) veloci ty, and (c) acceler ation of a particle in simple harmo nic m ot ion as fun cti on s of tim e.

0.0060 m

=

(0.0120 m) cos [ (314/s) t - ; ]

m.

that is, when cos[(314/ s)t - 7T/2] = With our calculator we obtain cos - 1 = - 1.05 radians (here, we have to select a negative sign, since th e argu me nt of t1:: cosine is initially negative, and remains negative until the motion reache s the fu:.: amplitu de, x = 0.0120 m). So 7T

(314/ s)t - -2

=

-105 .

15.1

Simple Harmonic Motion

475

from which t =

+ (1T/2)

-1.05

314 /s

=

0.0017 s

To find when the tip of the blade reaches x = 0.0120 m, we can use Eq. (15.8), which gives

8

t = - -

w

= -

(-1T /2) 314 / s

=

0.0050 s

COMMENT: Note that the time taken to reach a distance of one-half of the ampli-

tude is not one-half of the time taken to reach the full amplitude, because the motion does not proceed at constant speed.

In an atomic-force microscope (AFM), a cantilever beam with a sharp tip (Fig . 15.8a) oscillates near a surface. We can map the topog raphy of a surface (see Fig . 15.8b) by slowly moving the tip laterally as it oscillates vertically, much like a blind person tapping a cane on the ground. The AFM tip shown in Fig . 15.8 a oscillates with a period of3.0 X 10- 6 s. The tip moves up and down with amplitude 9.0 X 10- 8 m. What is the maximum vertical acceleration of the tip? It s maximum vertical velocity?

EXAMPLE 2

(a)

(b)

SOLUTION : As discu ssed above, the largest acceleration occurs at the point of

FIGURE 15 .8 (a) Atom ic-force micro-

maximum displacem ent. From Eq . (15.13) thi s maximum acceleration is [since the maximum value of cos(wt + 8) is 1]

scope (AFM) cantilever and tip . (b) AFM image of the surface of a crystal, obt ained by scanning the vibrating tip across the surface. The area shown is 2 fLm X 2 us«. The ragged terr aces are single atomic "steps."

amax -- w 2A

From Eq . (15.5) and the period T w

Thus, with A

21T T

=-

= 9.0 X

=

(15.16)

= 3.0 X 10- 6 s, we obtain the angular frequency

21T 6 3.0 X 10- s

= 2.1

X

6

10 radians/s

10- 8 m, the maximum acceleration is

This is more than 40000 standard g's, an enormous acceleration. The maximum velocity is, from Eq. (15.12), v max = wA = 2.1 X 10 6 radians/s X 9.0 X 10- 8 m = 0.19 m/s

476

CHAPTER 15

O scillation s

m

Checkup 15.1

QUESTION 1: I s th e rotation al motion of the E arth abou t its axis periodi c m o ti O scillator y moti on? QUESTION 2 : For a particle with sim ple harmoni c motion, at what point of the m does th e velocity attain maximum magnitude? Minimum magnitude? QUESTION 3 : For a particle with simple harmonic moti on, at what point of th e m doe s the acceler ati on attain maximum magnitude? M inimu m magnitude ? QUESTION 4 : Two particles execute simple harmoni c moti on with the same an:,.

tude. One particle ha s twi ce the frequenc y of the oth er. Comp are th eir maxim velocities and accelerations . QUESTION 5 : Are the x coordin at es of the particle and the satellite particle in

t:

-

15.4 alwa ys th e same? The y coo rd ina tes? The veloc ities ? The x comp onents 0:' :: velocities? The accelerations? The x co mpo nents of the acceleratio ns? QUEST ION 6 : Suppose that a particle with sim ple harm onic moti on passes th r

the equ ilibrium point (x = 0) at t = O. In thi s case, wh ich of the followin g is a ble valu e of th e phase constant 8 in x = A cos(wt + 8)?

(A) 0

(B) 7T/4

(C) 7T/2

(D) 37T/4

-r -

(E -

15.2 THE SIMPLE HARMONIC OSCILLATOR I

«

When displaced and released, the mass will oscillate about equilibrium.

• x

FIGURE 15 .9 A mass atta ched to a spring slides back and forth on a frict ionless surface . W e reg ard the m ass as a par ticle, whose position coincides with th e cen ter of th e ma ss.

The simple harmonic oscillator consists ofaparticle coupled to an ideal, masslessspri ,obeys Hooke's Law, that is, a spring th at provide s a forc e prop ortional to the elongazi or compression of the spring . One end of the spring is attached to th e part icle. - _ the other is held fixed (see F ig. 15.9). W e will ign ore gravity and friction , so th e SFi - _ force is th e only force acting on the particle. The system has an equilibrium posi - . corresponding to the relaxed len gth of the spring. If th e particle is initially at s distance from this equilibrium position (see Fig. 15.10 ), then the stretc he d sori _ supplies a restoring force that pull s th e particle toward th e equilibrium positio n. -=--particle speeds up as it moves tow ard th e equ ilib riu m p ositi on , and it overs the equilibrium positi on. Then, th e particle begins to com press the spring and ,,: down, coming to rest at the other side of the equilibriu m p ositi on, at a distan ce ec to its initial distanc e. The compressed spring then pushes the particle back towar " equilibrium position .The particle again speeds up, overshoot s the equilibrium posio and so on. The result is that the particle oscillates back and forth about the equili b ' position-forever if th ere is no frict ion . The great importance of the simple harm onic oscillato r is that many physical ,tem s are mathe matically equivalent to simple harmonic oscillators; that is, th ese ' . . ' tems have an equation of motion of the same mathematical form as the simple harmo oscillator. A pendulum, the balance wh eel of a watch, a tuning fork, the air in an or", pipe, and the at om s in a diatomic molecule are systems of thi s kind ; the restor ing fo r and the inerti a are of the same math em atical form in these systems as in the simple moni c oscillato r, and we can tran scribe th e ge neral mathem atic al results dir ectly fr the latter to th e fo rmer. To obtain the equation of motion of the simple harmonic oscillator, we begi n \,.i' · Hooke's Law for th e restoring force exerted by the spring on the particle [co mpare Eq. (6.11)]:

F= -kx

15.2

The Simple Harmonic Oscillator

477

Here the displacement x is measured from the equilibrium position, which corresponds to x = O. The con stant k is the spring constant. Note th at the force is negative if x is positive (stretched spring; see Fig . 15. LOa) ; and the force is positive if the displacement is negative (compressed spring; see Fig. 15. lOb). With the force as given by Eq. (15.16), the equ ation of motion of the particle is

(15.18)

equation of motion for simple harmonic oscillator

This equation says that the acceleration of the particle is always proportional to the distance x, but is in the opposite direction. We now recall, from Eq. (15.15), that such a proportionality of acceleration and distance is characteri stic of simple harmonic motion, and we therefore can immediately conclude that the motion of a particle CQU?led to a spring must be sim ple harmonic motion. By comparing Eqs. (15 .18) and 15.15), we see that these equations become identical if k

2

w=m

and we therefore see that the angular frequency w of th e oscillation of the particle on J spnng IS



(15 .19)

Con sequently, the frequency and the period are

(15.20)

angular frequency, frequency, and period for simple harmonic oscillator

"-TId 1

T=

f

=

{;z 27T\j1;

(15.21)

(a)

With the value (15.19) for the angular frequency, the expression (15.4) for the posi20n as a function of time becomes

-A

+A

x

Spring force always acts toward equilibrium position.

(15.22) (b)

According to Eq. (15.20) the frequency of the 'motion of the simple harm onic osciltor depends only on the spring con stant and on the mass. The frequency ofthe oscilla:?T is unaffected by the amplitude with which it has been set in motion-if th e oscillator has -' frequency of, say,2 Hz when oscillating with a small amplitude, then it also has a fre-=Juency of 2 Hz when oscillating with a large amplitude. This property of the oscilla:or is called isochronism. Note that the period is long if the mass is large and the spring constant is small. This is as expected, since in each period the spring must accelerate and decelerate the :nass, and a weak spring will give a large mass only little acceleration.

I' I

-A



jF"

I

I

+A

..

x

I

x=o FIGURE 15.10 (a) Positive d isplacement of the particle; th e force is negati ve. (b) Negativ e displ acement of th e particle; th e for ce is positive.

CHAPTER 15

478

Oscillations

When you place a heavy encyclopedia, of mass 8 kg, on a kitchen scale (a spring scale; see F ig. 15.11), you notice th at before comi ng to equilibrium , the pointer of the scale oscillates back and forth aroun d the equilibrium positi on a few time s with a period of 0.4 s. What is the effective spring constant of th e internal spring of th e kit ch en scale? (N eglect other m asse in the scale.)

Spring scale oscillates about its shifted equilibrium.

EXAMPLE 3

SOLUTION : The mass of8 kg in conjunction with th e internal spring of th e scale forms a ma ss- and-spring system, to whi ch we can apply E q. (15.2 1). Ifwe square both side s of th is equat ion, we obtain

FIGURE 15.11 A h eavy book on a spring which gives us

scale osc illates up and dow n.

2

k = 47T -

m

(15 .23

T2

With m

= 8 kg and

T k

= 0.4

s, th is becom es

= 47T 2 X

-

8 kg -

(0.4

-

S) 2

=2

X 10 3 N /m

COMMENT: In thi s example, th ere is not only th e for ce of th e spring acting or. the mass, but also the force of gravity on the mass (the weight) and friction forces. The force of gr avity determines whe re the spring will reach equilibrium, but this force has no dir ect effect on the frequency of oscillation arou nd equilibrium. T he

fricti on forces cause the oscillations to stop aft er a few cycles , but only slig h tl:" redu ce the freq uenc y (see Secti on 15. 5). For negli gible friction, th e frequ en cv depe nds exclusively on the m ass and th e spri ng constant.

Con *84. C onsider th e motion of the damp ed harm on ic oscillator plotted in Fig . 15.2 1. (a) A ccordin g to thi s plot , what fractio n of its am pli tu de does th e oscillator lose in its first osc illa tio n> (b) W h at fracti o n of its ene rgy d oes th e oscillator lose in irs first oscillation? (c) A ccording to E q . (15 .5 1), wh at is th e value ofQ for thi s oscillato r>

FIGURE 15.36 A microele ctromechanical system (MEMS) oscillator, the silicon memb rane structure suspended above the faceted silicon tren ch .

*85 . If you sta nd on o ne leg an d let t he ot he r dangle freely back and fort h starting at an init ial am plitude of, say, 20° or 30°, the amplitu de will decay to one -half of th e initial amplitu de after about four swi ngs . Regard ing the d angling leg as a d amped oscillator, what value of Q can you de d uce from thi s?

REVIEW PROBLEMS 86. A particle per for ms simpl e ha rmonic motio n alo ng th e x axis with an am plitude of 0.20 m and a period of 0.80 s. At t = 0, the particle is at max im um di stan ce from th e origin; that is, x = 0 .20 m. (a) What is th e eq ua tio n tha t describes the position of the part icle as a func t ion of time ? (b) C alculate th e positi on of the particle at t = 0.10 s, 0.2 0 s, 0. 30 s, and 0040 s. 87 . I n an electric sab er saw, the ro tational moti o n of the elec tric mo tor is co nverted into a back-and-forth mo tion of the saw blade by a mech anism sim ilar to th at show n in Fig. 15. 5. Suppose th e peg of the rot ati ng wheel mo ves aro und a circle of d iameter 3.0 cm a t 4000 revlmin and thereby mo ves the slo tted arm to whi ch the saw blade is bolted. What are the amp litu de and the frequency of th e bac k- and- forth sim ple harm on ic mo tio n of the blade ? 88. In re spo nse to a sound wave, the middle of you r eardrum oscillates back and for th wit h a freq uen cy of 4000 H z and an amplitud e of 1.0 X 10 - 5 m . What is th e m aximum spe ed of the eardr um >

89 . Suppose that tw o particles are performing sim ple h arm onic mot ion alo ng th e x axis wi th a period of 8.0 s. The first particle moves acco rd ing to th e eq uatio n x

= 0.3 0 cos (

:t)

and th e second parti cle acco rd ing to th e eq uation x'

= 0.30 sin (

:t)

where th e distance is m easured in m ete rs and the ti me in seco nds . (a) When doe s th e first par ticle reach th e mi dpoint? T he turning point? Draw a d iagram showing rhc pan icle and its sa telli te particle at th ese tim es. (b) Wh en do es the seco nd particle reach th e mid po int? The turning poin t? Draw a diagram showin g the p art icle and its satellite particle at these times. (c) By so me argum ent, esta blish that wh enever th e first particle passes through a point on th e x axis, th e seco nd pani cle passes th rough this sam e point 2.0 s later.

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90. A part icle of 6.0 kg is executing simple har monic mo tion along the x axis under the influence of a spring. T he part icle moves according to th e equ ation x

= 0.20 cos (3.0t)

where x is measured in meters and t in seconds . (a) Wh at is th e frequency of the motion? W hat is the spring con sta nt of th e spring? What is the maximum speed of the mo tion? (b) Suppose we replace th e parti cle by a new part icle of2 .0 kg (but we keep the same spr ing), and suppose we start th e mo tion with th e same amp litu de of 0.20 m . What will be th e new frequency of the motion ? What will be th e new maxim um speed? 91. The mo tion of th e pisto n in an automobile eng ine is approximately simple har monic. Suppose that the piston travels back and forth over a distance of 8.50 em and has a mass of 1.2 kg. W hat are its maximum accele ration and maximu m speed if the eng ine is turni ng at its high est safe rate of 6000 rev/min? What is th e maxim um force on th e piston? 92 . A Small Mass M easurement In strument (SMMI) was used in Skylab to measure th e masses of biological samples, small ani mal s, chemicals, and other such item s used in life-sciences experiment s whil e in orbit (see F ig. 15.3 7). The sample to be measured is strapped to a tray supported by leaf springs, and the mass is de term ined from the observed period of oscillation of the tray-and-mass. To calibr ate th is instrument, a test mass of 1.00 kg is first placed on the tray; the peri od of oscillation is th en 1.08 s. Suppose th at when th e test mass is rem oved and an unknown sample is placed on the tray, th e per iod becomes 1.78 s. What is the mass of the sample? Assume th at the mass of th e tray (and th e straps) is 0.400 kg.

(a) What value of the spring constant can she de th ese da ta? (b) If she then takes a child of 20 kg in her arms ar-; _: stands on th e scale, what will be the new frequer: oscillation of th e point er? 95 . Ropes used by mo untain climbers are quite elastic, ::.:: :.. behave like springs. A rop e of 10 m has a spring COil " -k = 4.9 X 103 N/ m. Supp ose th at a mount ain climber hangs on thi s rope, whic h is stretched vert ically down. . th e freq uency of up- and -down oscillation s of the rno clim ber? 96. C onsider a particle of mass In moving alon g th e x axis

_

th e influence of a spring of spring constan t k. T he equii": po int is at x = 0, and th e amplitude of th e motion is "i. (a) At what point x is the kin etic energy of the particle ec to its pot ent ial energy?

=!

A, what r: :..-:(b) W hen th e particle reaches the poi nt x of its energy is potential, and what fraction is kinetic97. A simple har monic oscilla tor consists of a mass of 3.0 kg' .r: ing bac k and forth alo ng a hor izontal frict ionless track wipushed and pulled by a spring with k = 6.0 X 10 2 N/ m. Suppose that in itially the mass is released from rest at a 6:;ranee of 0.2 5 m from the equ ilibrium point. W hat is the energy of this harmon ic oscillator? W ha t is the maximum speed it att ain s when passing th rough th e equilibrium poir.:: 98. A simple har mon ic oscillat or of mass 0.80 kg oscillates wi _ frequency of 2.0 H z and an amp litude of 0.12 m . Supp ose th at , while the mass is instantaneously at rest at its turn ing po int, we qu ickly shi ft th e fixed end of the spring to a new fixed position, 0.12 m farth er away from the mass. How doe; thi s change the amplitu de of the motion? The frequen cy? energy ?The maximum speed? The maximum accelera tio n? 99. A pendulum has a length of 1.5 m. What is th e perio d of th is pendulum ? If you wanted to construct a pe ndulu m with exactly half this period , how long would it have to be? 100. A n "inte rrupted" pe ndulum consists of a simple pend ulum of length ! that encounters a nail placed at a d istanc e below th e point of supp ort. If th is pendulu m is released from one side, it will begin to wrap around the nail as soon as it passes th rough th e vertical position (F ig. 15.38). What is the period of th is pendulum ?







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FIGURE 15.37 Small Mass M easurement Instru ment. 93. A simple harm onic oscillator has a frequency of 1.5 H z. W hat will happen to the frequency if we cut the spring in half and atta ch both halves to the mass so that both sprin gs push jointly?

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I

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I

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94. A physicist of 55 kg stands on a bathro om scale (a spring scale, with an intern al spring). She observes that when she mount s the scale sudde nly, the poin ter of th e scale first oscillate s back and fort h a few time s with a frequen cy of 2.4 Hz.

I I I I I

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FIGURE 15.38 An "int errupted" pendulum.

11111.11

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Answers to Checkups 101. A phy sical pendulum consists of a uniform spherical bob of ma ss M and rad ius R suspended from a massless string of length L (see F ig. 15.39). Taking into account the size of th e bob, show th at the period of small oscillations of th is pendulum is f R2 + (R + L)2 T = 27T ) g( R + L ) ::...5_ - - , - - - -_



505

102. A un iform rod ofl eng th L is swing ing abou t a pivot at a distan ce x fro m its cen ter (see F ig . 15.40). Fi nd th e period of oscillat ion of thi s physical pen dulum as a functi on of x. For wh at choice of x is th e period sho rtest?

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