(4 pts.) (4 pts.) (4 pts.) b) y(x,t) = 1/(ax 2 +b) This function has no time dependence, so cannot be a wave.

Transcription

1 12. For each of the possible wave forms below, idicate which satisf the wave equatio, ad which represet reasoable waveforms for actual waves o a strig. For those which do represet waves, fid the speed ad directio of propagatio, ad sketch the waveform at t=0. The quatities a, b, c, A, ad v are all positive costats. x, t ax bt c (4 pts.) a. 2 b. x t 1, (4 pts.) 2 ax b c. x t x, t Asi (4 pts.) a b 2 2 d. x, t Asiax bt (4 pts.) Solutio a) (x,t) = (ax + bt + c) 2 This fuctio ca solve the wave equatio, so i that sese it is a solutio. However, it is a parabola which goes to ifiit at large x, so it does ot describe a reasoable wave. b) (x,t) = 1/(ax 2 +b) This fuctio has o time depedece, so caot be a wave. c) (x,t) = Asi(x/a + t/b) This ca be rewritte (x,t) = Asi( 1/a (x + at/b)). It is of the form f(x-vt) with v = -a/b. This is a reasoable wave form. It travels i the x directio with speed a/b. d) (x,t) = Asi(ax 2 bt 2 ) This caot be writte i the form f(x-vt). Also, if ou plug it i the wave equatio it does ot give a costat velocit at all poits i space ad time. Grader: 5 pts each. 3 for successfull pluggig i, ad 2 for explaiig if it s a good wave or ot. (The first 2 poits are for cases where partial credit is eeded, some of these ca be solve/explaied without pluggig ito the wave equatio).

2 13. (20 pts) It is observed that a pulse requires 0.1 secod to travel from oe ed to the other of a log strig. The tesio i the strig is provided b passig the strig over a pulle to a weight which has 100 times the mass of the sprig. (a) What is the legth of the strig? (b) What is the equatio of the third ormal mode? Solutio 100 g (a) v t Plug i a expressio for the velocit: v T where the tesio is equal to the weight of the mass: 100 g : 100g t The mass desit cacels: 100 gt Solve for : 2 100g t 9.8m

3 Grader: 10 pts. The ma plug i 0.1 secods right awa ad have strage lookig expressios with icorrect uits, but that is OK if the get the right aswer. (b) This is a clamped strig for which we have writte the ormal modes before. Simpl plug i 3 for : 3x A si cos t Sice we have some iformatio o the velocit, the ca be give explicitl: The velocit was simplified above: 3x A si cosvkt 3x A si cos 10 gk t Use the stadard defiitio of the waveumber: A 3 x si cos 10 g 2 t Ad the expressio for the ormal mode wavelegths for a clamped strig: A 3x si cos10 g 2 t 2 3x g A si cos 10 3t 3x A si cos 30 g t For this problem, we foud above that g ad are umericall equal, so ou ca express it like this:

4 3x A si cos30t Grader: 10 pts. If the left it i terms of g ad to avoid a apparet problems with uits, give full credit. 14. Cosider a strig extedig form x = 0 to x = as show below. At x = the strig is fixed to a rigid wall, ad at x = 0 the strig is attached to a rig of egligible mass which is free to slide o a frictio-free rod. (frictio free, so o trasverse force). 0 x a) Usig the appropriate boudar coditios, give the solutio (x,t) for the th ormal mode, ad give explicitl the frequec ad the wavelegth. (20 pts.) b) Sketch the patters for the three lowest allowed modes. (10 pts.) Hits: Thik carefull about the x=0 boudar coditio. The size of the rod ad massless rig are egligible. Perhaps a free bod diagram would be i order! The right boudar coditio is simpl that for a clamped strig: the -value is zero at all times., t 0 For the left boudar coditio, cosider a free bod diagram of the massless rig: F rod T strig

5 The massless rig feels a horizotal force from the rod which opposes the horizotal compoet of the strig tesio. The lack of frictio meas that there will be o vertical force from the rod. Sice the rig is massless, there must also be o vertical force from the strig. If there were, the massless rig would istatl respod (with ifiite acceleratio) ad move to a poit where the vertical compoet of the tesio is zero. Sice the vertical tesio is proportioal to the slope of the strig (d/dx), the slope must be zero. So the boudar coditio is: x x0 0 Just as iitial coditios ca appl to the fuctio (iitial positio) or its first derivative (iitial velocit), boudar coditios ca appl to the fuctio (costat positio) or its first derivative (costat slope). How ou reach the aswer depeds o our iitial guess for the strig shape. Sice the left side is zero ad the origi is displaced, cosie would be a good choice: Appl the left boudar coditio: x, t Acoskx cos t x x0 Ak si cos t 0 which tells us that = 0. Now appl the right boudar coditio:, t Acoskcos t 0 which is ol true if: k 1,3,5,7, m 1 k m m 1,2,3,4,5,... 2 Sice k = 2/(alwas!), the ormal mode wavelegths are: 4 1,3,5,7,... Sice = v k(alwas!), the ormal mode frequecies are:

6 v 1,3,5,7,... 2 The ormal mode solutios are therefore: v Acos xcos t 2 2 1,3,5,7,... Here are the patters for the first three modes: Grader: There are ma was to do this problem! Correct idetificatio ad use of the two boudar coditios is worth 10 poits. The boudar coditio at x = ca ol be as I stated it. For the boudar coditio at x = 0, the ma assume that the ed will oscillate with amplitude A ad get the same aswer. Note that the ma also eglect the phase if the choose wisel betwee sie ad cosie to describe the shape (some will uderstad this, ad some will get luck). The expressio for wavelegths is worth 5 pts, frequecies are worth 5 pts, ad ormal modes are worth 4pts. Note that these expressios ma var depedig o choice of sie or cosie, ad o choice of odd itegers or all itegers. The sketches are worth 2 pts each.

7 15. (20 pts) A stretched strig of mass m, legth, ad tesio T is drive b two sources, oe at each ed. The sources both have the same frequec ad amplitude A, but are exactl 180 degrees out of phase with respect to oe aother. (Each ed is a atiode). What is the smallest frequec cosistet with statioar vibratios of the strig (a ormal mode)? This is Frech problem 6-5. (20 pts.) Solutio The statioar state with smallest frequec will be the first ormal mode. Sice both eds are ati-odes, the =1 ormal mode has a wavelegth equal to twice the strig legth : Acos t Acos t You ca also get this result mathematicall b guessig a solutio: x, t A cosk xcos t Its best to guess cosie sice the strig has amplitude at x = 0. Appl the left boudar coditio: 0, t A cos0 cos t Acos t This tells us that the amplitude of the ormal mode (A ) must just be the amplitude of the drive oscillatio (A). Also, the frequec of the ormal mode must be the frequec of the drive oscillatio. Now for the right boudar coditio:, t Acosk cos t Acos t Here I have plugged i at the right boudar, ad set it equal to the value of the left boudar, but 180 degrees out of phase. This is true if:

8 cos 1 k which is true for k odd (1,3,5,7,9,...) Or ou ca hadle the odd this wa: k 2 1 all (1,2,3,4,5,6, ). Now to get the frequec: odd (1,3,5,7,9,...) v v The lowest frequec will be =1: v 1 I terms of the parameters give: 1 T m 1 T m Grader: 10 pts for gettig that the first ormal mode wavelegth is 2 (which ca be doe with equatios or graphicall), ad 10 pts for fidig the lowest frequec. If the stop too earl ad leave the expressio for lowest frequec i terms of v (velocit) rather tha T ad m, take 2 pts.