P214soln4

# P214soln4 - P214~ Solutions Homework 4 by Louis Lehlond i...

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Unformatted text preview: P214~ Solutions Homework 4 by Louis Lehlond i. This problem starts by given us an explicit form for the functions f and g and with these the general solution t) takes the following form: y(;r, t) :f(r w 222‘) + + of) : Acos[k(:r ~ 7%)] + A cos[k(.r + ’01)] :A%{eik(1._nt) + €ik(r+z;f} I A%{€iklf(€—iwi +eiwt)} where I have used that v : w/k. We recognize the sum in the last term as being just, twice a cosine (from the Euler identity —see homework i) and then taking the real part is easy (since A is real); we just need to use the Euler identity again en” : cos(k:a;) + isinch). 3/(17. t) 244%{eikf2('os(wt)} : 2Acos(k17)cos(wt) (1) we recognize this as a standing wave because the x and t dependence are separated. There exists some fixed positions where y 2 0 for all time and similarly there exists fixed times where y : 0 everywhere (for all x’s). 2. For this whole problem, it is useful to define the usual variable 11+ 2 :r + vi and u- : .I,‘ a m‘ (a) We start with the following general solution : f(x — of) + 9(13 + wt) and we want to specialize to a solution that satisfy the followmg initial condition ‘11)): l ‘ t:(] strong condition as this is true for all x. Let‘s just do it, we first need to use the chain rules to transform the time derivative into 71,: derivative : 0. This is a very 8y(;r, f) A i)f(11,_) [Pun + 8901+) 811+ 8t {:0 O'IL_ at 311+ (31‘, ,4) then we use that r: in to get that 03/(17, 2‘) 8f(u_) 89(u+) _~_* * ll, ‘ + (7t 1:0 (31L (971+ b0 Now, at t : O we have that U+ : 'u_ : 17 and it is easy to convince oneself that the last expression just becomes: (7’ .17.]? 8 1‘ (9 I _ yr) snipe) (2) 0t [:0 8m (3m Setting the last equation equal to zero gives 55:) : @621). Integrating both side we get: f(lf) : 9(1") + constant The constant of integration can be safely set to zero as it corresponds to the string equilibrium position which we can take to he y : 0. Hence we have that f : 9(1), the two functions are identical! (b) Since the string is motionless initially (the transverse velocity is initially zero everywhere) we can use the result from the last part that f : 9(1). This means that initially we can view the triangular pulse as the sum of two equal triangular pulses, At t : d/v, each of this triangular pulse (f and g) have traveled a. distance d but in opposite P2] 4’ Solutions Hoinowork 4 l)_\' Louis Lohlond dirm‘tions. This is because i is a fnnvtion of .1‘ ﬂ I‘f and travels to the right. versus g which travels to the loft. So we have the foll<,)wing picture at time t z 0‘ t : (1/1‘ and i : Zd/zi. : o + A F ’t‘ A z ‘ A __,_ _ g > A + Wm . a ( a I A ~_ ! «r: 5/1)“ M/ 1L ; g3 it? A 25‘92.) V --........ / 2///// \ ///\ l: I if W / 7“, W 7 777 M m WV/ » i H‘ fr,” ,1 73 i d 34 ‘36} 56*? d jgmw ‘ i (0) This 1)r<’)l)lmn is ('(,>1iiplot,(-‘,ly analog to part, (a) but, now we have a, Spatial derivative instead of a time derivative. so 0%.); t) M (ﬂfﬁL ) €)TL_ + 8901+) 071+) . till 0.2‘ MU Hui 8m (311+ (72‘ and now we use that (Elf : 1 to get {MAJQU VA '7)f(u__) + 09(u+) 0.1", [H “ 0H,, 071+ 1") 511100 this lh‘ vvzrlnatzctl at f : 0 this 1‘C(l11(‘(‘5 snnply to (will + (’23:) : 0, lntvgrating and sotting the arbitrary constant to 0 we get that : ~g(.17)_, the two functions are identical but opposite in sign! (d) The initial velocity distrilnition (1y(.r.()) : 933:1) was (inlllpUlGd before (see Eq. (2)) and . [:0 i , since lit‘l‘t‘ the string is originally llat \er Can use the result of part, ((1) that. : vgtr). This gives: ( 0) (ll/(131\$ 8f(a:) 09(1‘) 9 ﬁfty) ( g) If 1‘. : — :’I' —- ‘ ,‘ : -1“, 2, y ' at to OJ: ’ 01' 0:1: The initial \'(~locit,y distribution is just. #27; times the slope of f(.1‘). By inspection from the graph of given in the problem we can sketch 1);, 0). P214— Solutions Homework 4 v by Louis Leblond (e) Finally we are asked to sketch snapshots of the piano strings at If : (1/7: and t : Qd/v_ First let’s do 1 : 0. At that time7 the string is flat but we can see it the sum of two identile function of opposite Sign f and g(Jr). “’0 know what these function looks like so we have at 15:0: At 2‘ : (1/1), the pulse from f moves to the right by a distance d and the pulse from 9(3') moves to the left. I - l \ ,, / —___ ah / / / / I At t. : Eat/v we have a similar picture. 3’53 3. (problem 15.79) This problem is really just about redoing the derivation done in lecture. (3) Take a. small segment of string of length A1: and mass TIL The kinetic energy is just K : \$7711}?! and the kinetic energy per unit of length (which we called mg) is: 1 , 1 “3/ . ,t 2 7,1,k(;[::t) : : ~21: (b) with t) : A cos(lit:r — wt) we get: 1 _ uk(1:,t) : §w2A2 sinzﬂn: w wt) k A. , P2l4— SOlllllmlh Honiewn‘k 4 7 My Lonix L(‘l‘)l()l)(l // Q//// my : ill o [23% //i (ll): AA The stretched length l is just: 22 ~ 1 011/ 2 ‘) _ . . , where l have, assumed that. is much smaller then one. and I have used the lnnonnal oxlxinswn (l + .7‘)" \$3] + 72.1: (d) The work done is just the force times the change in length. The Change in length is just the stretched length minus the length nnstretched (which is Hence. 1 0y 2 W : PU ~ Ar) : 5F Am - .17 The potential envi‘gy per unit of length (called up) is the work divided by A3: and so we have 7, 2 up(;1:./f/) : gF (3.): xi! v for the sine wave of part (b) this gives: .. fl .242; 2 M, up(;1,.f) v §Fk A an) (A1, ~ at) (f) Using F : [1.w2//k2 we see easily that '11,;L.(::17,t) : 'up(;r, t). P214— Solutions Homework 4 by Louis Leblond (g) up and uk are maximum at y : 0 because this is where the velocity and the slope are maximal. They are zero at a maximum of position because the velocity and the slope are zero at these points. (h) The total energy 5 is the sum of uk and up, multiply by V to get: P(;r, t) : 5v 2 (uk + up)v 2 211mC : 2ii/LA2w2 sinQUcaL‘ — wt) : 2x/F/1A2w2 sinQ(k\$ — wt) where l have used that uk 2 up and that v : t/F/u. We recognize the last equation as the same as Eq. (15.22) in the text. That the power is 51) makes sense because ﬁrst the units match. Energy per unit of length times length per time gives energy per time which is power. Also intuitively, we expect the rate of energy ﬂow to be the energy density time the velocity of propagation and this is what we get. 4. (problem 15 .82) (921% t) 531% t) (it at (6) P(2:7 t) I ~ F (a) We start with the standard standing wave equation (equ. 15.28): y(.7:, t) = A sin(k..r) sin(wt) We can evaluate the two derivative and plug—in the formula for power : [CA cos(k;17) sin(wt) gygi 2 will sin(k1¢)cos(wt) P(:r. t) : ~kaA2 cos(k:r) sin(k.r) cos(wt) sin(wt) where we can use the following Trigonometric identity cos 6’ sin 6 : ésin 26, 1 . P(.7:. t) : *ZkaAZ 511mm) sin(2wt) (7) P214— Solutions Homework 4 by Louis Leblond (b) The ( verage power is the total energy for a time interval At divided by that time interval. '51 t2 — t1 [‘2 P(;z:, t)dt PAV 2 —~—— N ow over any period or multiple of a period the integral of sin(2cyt) vanishes as the sin oscillates between positive and negative values that cancels each other when you integrate. This is different from (equ 15.25) which is for a transverse wave. In that case, the power is proportional to sin2(wt) which is always positive and gives a non-zero PAV. Hence we have found that a standing wave does not transfer energy on average. (0) i. at t 2 0, P(a:,0) 2 y(:r, O) 2 0, I will not bother graphing that. ii. at t 2 £7, sin(2wt) 2 1 and sin(wt) 2 t}? so P(;I:, 0) 2 ~ikaA2 sin(2k.r,) and y(;r, 0) 2 % sin(k:1:) ﬁg 4 ’l i /”‘N W‘ x ,r ’vA \, st? / \i‘ ff, \ all/‘22. “A a WK; , \\\, r}; “,3 XV”; KKK /, ’ ‘ # iii. at t 2 sin(2wt) 2 0 and sin(wt) 2 1 so, Pm, O) 2 0 and y(;E,O) 2 AsinUm) t; «\ (i ‘\3 E x 33 ﬁ m, \N‘KJ \ w! .é iv. at t 2 3—3, sin(2wt) 2 ~l and sin(wt) 2 i so P(:17,0) 2 ika/lg sin(2k:z7) and y(\$,0) 2 sin(lm:) ix ,~ I. _ Willi._, 9: a i P214— Solutions Homework 4 by Louis Leblond (d) At antienodes, PE is always zero because the slope is always zero7 KE energy varies between zero when string is maximally displaced and its largest value when string is at equilibrium (hop izontal). At nodes, KE is always zero because the string never moves; PE varies between zero when string is horizontal and its largest value when slope is greatest (maximal displacement) As string moves away from equilibrium energy flows away from anti-nodes and towards nodes. 5. We have that _1 2 0y 2 A 1 .12 03/ ,1 “A W 5“ t if“ a W 8 9' 5 : U](’ ‘i‘ UP P I AF :1 51’ where the last equation 8:) for the power is for a traveling wave ONLY. At t = 0 we have that : and so uk and P are zero. up can be obtain by looking at the slope of the wave. ll 1 l r‘ as \. \. / A..__._§__*_s \m «W»%sww»~f} -a i " [N f“ r l i; ’ a? P214— Solutions Homework 4 by Louis Lehlond (b) at t : (17/1) we have two traveling waves in opposite directions, for each one we have uk : Furthermore we know each wave should have %8. Hence we have Up. (c) at t : 2(1/2), the picture is very similar to the previous case. ’“ﬂ 3 7? P i i y {:5 ' Sci g (j g l l 4‘ WWW“; 35*": WW; WWMMMMM; ,_::i:-‘_~gz——»~«gm:~r MT 3 i (3 6. (problem 1631) This problem require no math. just an understanding of the concept of phase. 4‘“ ‘rﬂ (a) In order to have constructive interference at point Q we need the wave at A to arrive at B with a phase of zero. That means, we need (at least) to have a complete wavelength between A & B. Therefore the minimum wavelength is 2 meters for constructive interference. (b) To have destructive interference, we need the wave starting at A to arrive at B with a phase shift of 7r. If we have % a wavelength between A 85 B then that should do it. Therefore the P214— Solutions Homework 4 by Louis Leblond minimum wavelength for destructive interference at Q is 4 meters (since %A : 2 meters). 7. (Problem 16.36) (a) The beat frequency is just the difference between the two frequency of each violin strings. With a beat frequency of 1.5HZ and the ﬁrst Violin string at 440Hz, then the retuned string must be at 441.5HZ or 438.5HZ. (b) We have that ’1; = /\ f : V; or T : iLXZ f 2. The fractional Change in tension is: AT_T—To Pan? T_ To — f02 i. for 441.5Hz (tension increased), AT: 2 W : +0.6830% ‘ . ' r 2_ I, 2 ii. for 43845112: (tension decreased), ATT : (438*)?f4lmaSQJ‘UHZ) _ —0.6807% ...
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## This note was uploaded on 09/25/2008 for the course PHYS 2214 taught by Professor Giambattista,a during the Spring '07 term at Cornell.

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P214soln4 - P214~ Solutions Homework 4 by Louis Lehlond i...

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