hwsol10

hwsol10 - Homework 10 Good Vibes: Introduction to...

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: Homework 10 Good Vibes: Introduction to Oscillation Consider a block of mass l/1. attached to a spring with force constant k, as shown in the figure. The spring can be either stretched or compressed. The block slides on a frictionless horizontal sUlface, as shown. When the spring is relaxed, the block is located at X'=O . If the block is pulled to the right a distance A and then released, Awill be the amplitude of the resulting oscillations. Assume that the mechanical energy of the blockspring system remains unchanged in the subsequent motion of the block. Part A: After the block is released from x -A o A = .A, it will: 1rn 0 f-;o~ : '"f it W; w; J) J} IY\ be... 0 j (l cO'()S ta.. II ve. TO (OfYI~ +he. Ie f+ u YJ 1-; aj'n~ x::::--A.J a.AJt the!' i+ lIJiIJ hac#' at) Part B: If the period is doubled, the frequency is: b1 cJ <L {; n-, + ion -r- I -;:=. 1/:1.. _J thus ;f P ,e r i DcA ;S do u b }f-cA. ~ T:=:. fre-coue-l"CY is hCAJV~d.. ""7 ':2. Part C: An oscillating object takes 0.10 8to complete one cycle; that is, its period is 0.10 s. What is its frequency f? T::::. ~ l- Or :=:. -J::; -i. T /0 Hz. f ::;..1 ~ 0,/5 Part D: Ifthe frequency is 40 Hz, what is the period T? x I 1 T - 90Hz.. ::: 6_ () ;)5 s T=- Al AS R o t lQ ............... Part E: Which points on the x axis are located a distance Afrom the equilibrium position? l3of), A. and fro m 1; },o. ve i..s _ 0. dis;J h ce men+ / A I +- j) e )( -CAX Part F: Suppose that the period is T. Which of the following points on the t axis are separated by the time interval T? One.. ClAn d..e f; t- h e j 1)(2.. } 0- S L( Cr e 5 +- +0 crest ,\) o-rf, or lAy) e~ UJ va./en+ by /nferva.l, :r n II v S +,<A+ioY)) .L K anJ P sep erafecJ rerioot Part G: What is the period T ? o ~O J'" J's }'if 0:5 C j 11 cA. t-/ OY\ : .fA e.re +0 (' e."f- 4 (0,00 5 5) ::;: 1(),7;:l..5; Part H: How much time tdoes the block take to travel from the point of maximum displacement to the opposite point of maximum displacement? ihis 1 J1 f e. f' ~ ~ V 0-.- I , IS no. I f C{ fJ ~rioO<./ +-),erel<Jr~ -: T - ~og 6 Part I: What distance ddoes the object cover during one period of oscillation? }it; /5 0 f. f- h e. sci lJ ~ +Ton dis I- 0... V) C- fC fh e. o...rY'p/;fucJe (1\.= O~/d.Wl) eJ..::: LjAjQ,q~~ Part J: What distance ddoes the object cover between the moments labeled K and N on the graph? From I~ r6 N )5 . ~'1 o f 0n e / tJ}f 0 S Gj )1 0..1-/0 11_ d~ 3~~3bd Harmonic Oscillator Equations A block of mass lllis attached to a spring whose spring constant is k. The other end of the spring is fixed so that when the spring is llIlstretched, the mass is located at x = (). Assume that the +x direction is to the right. Part A: At what time tldoes the block come back to its original equilibrium position (x = 0) for the first time? /\ X I We.- I~h 0 lA/ eX (t)) =: A co s em t-) e-~ ~ ) E-:::.--f2-~---'7 e +j I}') x=O cos TA ((}):::o wAen vs)f; t '" % V elo c i +1 Part B: Find the velocity vof the block as a function of time. f'A e. /:5 fA =: e.. J. ri (~ I/o.t-)'v e 0/ +A ~ ?ds,'f,'Oh ; X{t) =- V(t) -Afj, Sit' t) Part c: what I~S .., V fYI I-S (A,x ! tt W .~ Xn.ow wh ic.h occu 11 cvlf!..n m CA.X J rn lJ m wh.en Sj,,(!ff)=) JS r..s wn +A e OJ't) U yYlen+ o.f. s,. 11 t; ~ IV"lCAX/::= A ~ Part D: Find the acceleration aof the block as a function of time. o.-is the +irYI~ oleri\/tt+;ve of -rAe- 1/el0ciry ~ a- (f: ) = rJ v LO::;;; - A -!f;; 41'd C 05 (Iff. -t-) Part E: Specify when the magnitude of the acceleration of the block reaches its maximum value. Consider the following options: a. only once during one period of motion, ~ when the block's speed is zero, c. when the block is in the equilibrium position, @) when the block's displacement equals either /lor e. when the block's speed is at a maximum. -A 0- c c.e. , ey~+i J. O/J' is 9r ecd- e. s+ 111 J-, e. n otr / i ~iJd e 's '3 i<' ",-!-eSf-, e. When 0...YY) ve,Joc;+-y ,'s -zero_ 0-+ IA. e f I; +vd~ ('5 Cj l' eo.fes-f A ex. not - A Part F: Find the kinetic energyK of the block as a function of time. ~ =- y~ vn& (tU r;y( ~ ::: 1;1. /-<. A .,.~ A-;l.~ S J n '- (I::. t) fr' f' 0 +-; c e. t- A 0.. + +h,s =: .. Sj n ' ;1...( ~ 1m t ) ~ uo..t\ +i+-y ;>6S ifi ve_ J~5 a. Iw 0- yS Part G: Find J(nu"" the maximum kinetic energy of the block. /"'. rn tA)( o ceo rs .wlle/) s /n ~{fi+) ~ I A 2./).., J~fYlo.x -;L Part H: The kinetic energy of the block reaches its maximum when which of the following occurs? fA e, h; n e Vv' he f) /VlOt+iC. e- Y\ er5 Y 0+ i -5 J) (]A + 0.. WI tAX i )IYI v jYl We. V e.-Io cit,>' 15 occurs 0.. WI (AX; /Yl m. F fOyn be 10r J(b.J 1I11'OW .f-A,s wherJ fl'o..C fAe eX I~S (! IVle,1f 0 t -rAe h roc ~ ls ~el()- Problem 13.34 Your boss at the Cut-Rate Cuckoo Clock Company asks you what would happen to the frequency of the angular SHM of the balance wheel if it had the same density and the same coil spring (thus the same torsion constant), but all the balance wheel dimensions were made one-third as great to save materia!. Part A: By what factor would the frequency change? rnitiOJly Whcd fo== ~ ~Tf WJ1~J) j_K. I where and I~ is +11 e. for i.sol' C(J)1SftJ-J1T r rs rAe..- mornef'l+ o? IJ1ert"ct_ hOfP j eftS .+l1?2- ha lOJ) Ce- Wheel 1'3 } A e r 0/)'5" 1. (101 s i z- e. ? dim _ /~ ell s iOns are.. / ~~t M~~V~ M.::-~ ?3 Wer0 f' C,()(YJ?o.re +0 +0 l- ..L ~ J rr j rx ---::r:- 110:::;OV==-j 1ir~ T= 7T(~)1/.3 ~7 J;iJ ! .~-----------~J~ 0'13 := 0 r;;:- Io -There- Poro) -rAe- t-rettJuf?nCY I T -0 - ... AJ no r:t I :=- ~ (J:-)~- cho..nt; es by a .po..cfo I.c -:2..43 a.?':; of ~ It 3 (), /5. {; Part B: Would the frequency increase or decrease by the factor found in part (a)? -rM f reg Vb C! i n cr e Oesf s b1 Cl f-o f' 0.. cf or I:::. /5-6 Part C: By what factor would the torsion constant need to be changed to make the smaller balance wheel oscillate at the original frequency? rJovJ S fDr j 56 () lflfroc1uce C. o!) rhe new vo.r Ct jo..bJe J< ' .J -;- A e... s t-an +- f-hAf wi II I/D w for: 10::; f Il'L 1... Til'S fo be- rru e for -F=-~/~ o ~ r-;:;:::If) n- - - ~ [). h': rr .K 70/.21.1 J :=f Yl 0 w.J (/1/ e.. ha- Ve... -1-0 ..so Il/e for -~l=){flJ ~ f<,(J _L_ f"\, / Ie) J6/a 43 :24 3 ~, / -Xo ~ ;p; I X :::=- Part D: Would the torsion constant need to be increased or decreased by the factor found in part (c)? I hUS tAe tor SIC> J1 of JLj3, ~ Co nS+iAnt fJo-s +-0 he J e c(eo.s~ol bl CA. rD.-ero, Problem 13.43 . . .Ffi An apple weighs 1.20 N. When you hang It firom tIIe en d 0if a l ong spring oJ. orce constant 1.44 N/m an d negligible mass, it bounces up and down in SHM. if you stop the bouncing and let the apple swing from side to side through a small angle, the frequency of this simple pendulum is half the bounce frequency. (Because the angle is small, the back and forth swings do not cause any appreciable change in the length of the spring.) What is the unstretched length of the spring (i.e., without the apple attached)? When fA-.- S"jJl'ly13 i:s osci lIO+in:] ( houJII CI'n J) W o :::- w Whe,Y\ JFi-j 'S .f--A e- - . SPl'i)15 lA. c t-; n 3 /,' I~e LA- p e nc!U lum w/=/i we.. KnoW Jw fi= -;: J~ I Ij~ - w '-- WO/:z SoJviYl:J {or L }!f::; ~? L '-tIN. L I L = Ltw I~ - 5 I- A f<.- S 1-r e. + c ;, ~ot 'LW I'" leY13 f-;' L-=:. L o + w. h L(j 3W I~ Problem 3.45 After landing on an unfamiliar planet, a space explorer constructs a simple pendulum of length 52.0 She finds that the pendulum makes 97. 0 complete swings in a time of 132 s. What is the value of g on this planet? CIIl. II ferioel 7 f- Corn le+es q7 jy'/l?).S 5]# ifl~5 ) / W he>... +-- is l- he ~ /J;;Zs 17svvlY\3 S -fOf I ~ 3 I S wi s - 1lj =-T We. ~ f\6W f-Ae e.~uQtion T . S-' I := ,;:;-(,/ l::j vV e., 1--\ hOW hnow)1 ) c:r L, J j-),e,e iOle, vU e. Co...() So I v e 101 +A e un 3 I- 3 (; s::=. an + h is jJ laneJ, ;;ar '?.: 5:J r:. ~ j ~. 36-.5);t.::::;)7T Q.5Jrn 5 (I . 365) j ~C3--rr)'J.(O. 5:1:") .:::- Problem 3.50 A 1.80-kg connecting rod from a car engine is pivoted about a horizontal knife edge as shown in the figure. The center of gravity of the rod was located by balancing and is 0.200 m from the pivot. When it is set into small amplitude oscillation, the rod makes 100 complete swings in 120 s. Calculate the moment of inertia of the rod about the rotation axis through the pivot. d = 0.200 III L j Me ; n +A e p Y' e 1//0 u.5 pI'"()b Ie m.J the- period hi we- Lo.Y) CalcvlaJe faflinj T..:::: 1;).05 IOOSWiYl35 J,.J, S eco nc;is F'o{ (). p),yS)'CO../ j/rLncJ..vluyYI/ fA- peJ"'oo< 0+ s)'Yla..l/ C5scilJ6.J-ions ohe'/5 -the. rela+;ol1: :HT TW ~ n e e-cJ. /-m :d.. r + 0 .s 0 } ue- -fo == ;;(71 /; :r. .r ab 0 U + 1-), e j7;' ~ 110 I-.- /,25 (I-8M3) (9. g)'YI/s ~ (0. ;), ) m l ~\~ "J rT / _ I -(~1Jft,)(9.8'71"4){O, ;).YY1) I == 0:-;;l S) "( / - 0 1-1 )( q 8 ")s ~ (0 - ;l"J <) (~ IT):l. U.::: 0 - );),9 h'J yY1 <7 ...
View Full Document

This note was uploaded on 05/22/2008 for the course PHYS 2A taught by Professor Hicks during the Winter '07 term at UCSD.

Ask a homework question - tutors are online