y(x,5) (cm) x (m)

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1 Spring 99 Problem Set Optional Problems Physics February 9, 999 Handout Wave Propagation and Reection. pulse has shape shown in gure as it travels at 00 Asawtooth along a stretched wire. At t = 0, leading edge of pulse 6 m m/s a ed end of wire. Make asketch to show shape of wire from s later, with signicant dimensions of pulse shape indicated 0.0 your sketch. in leading edge of pulse travels (.s)(00m/s)=0m so m from wall and traveling in opposite direction after. s. it crest of pulse m from wall after. s and again traveling in opposite direction. Since pulse inverted on reection, crest now pointing down. trailing edge of pulse hasn't hit wall still m away, so it will interfere with part of pulse that yet...it two parts of 6 m pulse: tailing m has to summed with sum center m of pulse. You can do th analytically inverted trailing slope of reected pulse at t = 0: s. Summing describes gives y =0:so at =0;y = 0 and at =,;y =,:6. A se way to do th to realize that pulse must satfy boundary quicker =0;y = 0, and that more than m from wall, re no condition Energy in Wave Motion. moving your hand starting at t = 0,you want to produce a wave pulse By a string that moving to right and has following shape shown on string has tension Nand= :0 kg/m. below. interference, so that at =,;y =,:6. has hit wall and traveling in opposite direction. We have to y =0:+0: describes trailing slope of unreected pulse at t =0:s, y=0:,0:

2 Plot motion y(0;t) which should be applied to left end of a) to produce th pulse. string velocity of wave v= p = = p N=kg=m = leading edge of pulse (at = 0m) was made starting at m=s. t = 0 and gure shows pulse at time t = 5s. leading time of pulse m long which with v =m=s means it took 0:5 s edge produce. falling edge of pulse m long so it took s to to at top m long and was produced by holding string produce. for :5 s. Putting th toger we have that to produce pulse ed must: you Rae end to a height of cm at a constant rate for 0:5 s (or a. ofcm=s) velocity Hold end ed at a height ofcmfor:5s.. Lower to zero at a constant rate for s(oravelocity of,cm=s).. string 0m long. At far end string attached to a b) ring. Sketch string at t = 5 seconds, or 5 s after you frictionless One way to construct solution to think of a ctitious that starts at = 0 m and t = 0 s. It mirror image to that pulse you created at t = 0 s but it travels to left with same speed. which <0 m, th ctitious pulse represents true reected pulse. It For add to unreected pulse to produce total wave. For t> s will total wave just reected wave. Hence y(,5) (cm) (m) begin to move your hand to make pulse.

3 Sketch string at t = seconds, or s after you begin to move c) hand to make pulse. your Adding toger incident pulse and ctitious pulse t= s gives at Calculate and sketch energy density of string as a function of d) along string when leading edge of pulse 0 m from position hand (as shown in rst gure). Be sure to include units in your your and label your sketch carefully. answer, ) = v ( ) ( =(=)( u ) +(=)( ) = ( ) = ( u ) =(N)(slope of wave) ( 0, ) =0, J=m 6 >< (, 0, ) = 0, J=m 9 0 >: gure below shows a traveling wave propagating on a string at t=0. wave propagating to right (positive -direction). time tension in string = N, and string has a mass per unit of = kg/m. string has a length of 0 m, and has a free end length Hence u = 0 elsewhere. Traveling Wave on a String at =0 m. For a traveling wave y(,0)(cm) v so energy density can be written (m) We can epress th as

4 Draw a graph of transverse velocity (chunk velocity) of wave a) t=0, as a function of. Label your aes clearly and be sure to give at r = =,v N =m=s kg=m >< 0:0 0 <,0:0 <<6 = >: = =,v >< >: cm=s 0<,:0 <<6 :0cm=s Draw a graph of power in wave at t=0, as a function of. b) your aes clearly and be sure to give units. Label power P (; 0) in general given by P (; 0) =, (; =, P 0), =,v (0:0) =:60, W 0 < >< =6:0, W <<6 (,0:0) >: units. Since th a traveling wave n but since wave a traveling wave n wave speed v given by s v = slope of dplacement vs position graph = v for given traveling wave with =N, v=m=sand Hence in part (a) we have determined 0:00 >6 and thus P(; 0) = 0:0W >6 0:0cm=s >6 power (mw) at t=0 Hence graph looks like Hence graph looks like transverse velocity (cm/s) at t= (m) (m) - -

5 Draw a graph of amplitude of wave att= sec, as a function c). Label your aes clearly and be sure to give units. of For a free boundary condition wave may be thought as a superposition of wave traveling to right, and a ctitious of Draw a graph of transverse velocity (chunk speed) of wave at d) sec, as a function of. Label your aes clearly and be sure to give t= For a traveling wave, transverse velocity calculated from = v Draw a graph of power in wave att= sec, as a function of. e) your aes clearly and be sure to give units. Label t) (; (; t) ) (; ) (; Hence, using graphs in parts (c) and (d) to get (;) (;) and transverse velocity (cm/s) at t= s wave traveling to left, starting at = 0matt=0. mirror-image ctitious wave represents true reected pulse for 0 m. (See (m) - Optional Problem (b)). Hence graph looks like y(,)(cm) (m) For any wave, power given by P (; t) =, At t= s, we have units. P(; ) =, power graph looks like power (mw) at t= s 6 5 sign depends on direction of travel of wave. wave where in part (c) of th problem not a traveling wave: it super- shown of two such waves, incident and reected waves, moving in position directions. Consequently, above relation cannot be applied opposite directly to get chunk speed. Rar, relation must be applied to right and left traveling waves that superimpose to make separately resultant wave. Using proper sign depending on wave's direc- th will give chunk velocities for each of two traveling waves. tion, se two chunk velocities are added to get total wave's chunk n, (m) - velocity. When th done, resulting graph looks like

6 Boundary Conditions and Energy. following `snapshots' of transverse dplacement and Consider velocityoftwo dierent waves (A) and (B) on strings stretched transverse a horizontal component of tension equal of 5 N. horizontal scales with assume that string itself conserves energy, so that energy can We be lost or gained at a boundary (at edges of graph). For each only (A) and (B), say wher string gaining energy, losing energy, wave, maintaining same energy. Calculate (approimately) rate at or (; t) dened as power transmitted in + direction. P Left end : Both slope ( ) and transverse velocity ( Right end : Both slope and velocity are positive and refore leaving string power (; =, P t) ; string gaining energy. We estimate slope end to be n, rate at which string gaining right energy du dt P ( m;t)= = =, (5N)(0:05)(m=s) = 0:J=s Left end : slope negative but transverse velocity positive refore power entering string and (; =, P t) ; positive. Th means that energy owing to right, i.e., gaining energy. rate at which string gaining energy string du dt (0 m;t)=, = P =,(5N)(,0:05)(m=s)=0:J=s Right end : Both slope and transverse velocity are zero ) no leaving or entering right end of string. energy Reected and Transmitted Waves 5. goal of th problem to derive an epression for reected and are given a string with tension. left half of string has You per unit length and right half of string has mass per unit mass. Consider a pulse of amplitude A I and width I propagating length string, incident from left. When pulse hits dcontinuity on mass per unit length changes from to, part of pulse where transmitted with amplitude A T and width T, and part of pulse negative. Th means that energy owing to left, i.e., of wave at are in m, and vertical scales are in cm and m/s. Case (B) transmitted wave amplitudes. which string gaining or losing energy for both cases, (A) and (B). Case (A) reected with amplitude A R and width R. are =_y) zero ) no energy leaving or entering left end of string. a) What are T and R in terms of I? 6

7 reected pulse will have same width as incident so R = I. transmitted pulse width will change because pulse, velocity of pulse changes: v = p =, v = p =. Since f of incident and transmitted waves must be same, we frequency v = f I = f I v = f T = f T v T T = (v =v ) I =( p = ) I Use fact that string continuous at interface to derive a b) between A I, A R, and A T. relation At interface, string must be continuous. refore, dcontinuity, superposition of incoming and reected wave at equal amplitude of transmitted wave in order not to break must string. refore, A I + A R = A T. Energy conservation at dcontinuity requires that total energy c) incident pulse, U I, equals sum of energies in reected in R ) and transmitted (U T ) pulses: that, U I = U R + U T. Use (U epression for energy (given in lecture) to derive approimate U I = ka I= I energy in initial pulse, where a constant that depends on eact shape of pulse. Th must k U R + U T = ka R= R + ka T = T From part (a), R = I, T A I = A R + v v A T v I A v + v Friction in Wave Equation 6. stretched string immersed in a vcous uid. string has mass A unit length and stretched to tension. We will derive awave per that takes account of vcosity between string and equation Assume that vcous force on a chunk of string proportional uid. both to length of chunk and to its chunk velocity. (Think why and when th might be a reasonable assumption.) Let be about proportionality constant. Draw a free body diagram (FBD) for a segment of string. It should a) similar to one used in class, with an etra force due to look =(v =v ) I, and putting se two toger gives: have But we know that A T = A I +A R so we can solve for A R and A T eplicitly: A T =, v v I A v + v So A R = v = I vcosity. FBD looks like epressions for A R and A T following two halves of string: in A T = in terms of A I and wave speeds y τ θ τ sin θ v I A v + v R v, v A = I A v + v τ sin θ θ τ f equal : + 7

8 eaggerated view. We are concerned with small dplacements in an direction compared with length of string. Hence actual vertical of string segment well approimated by horizontal length length. b) Derive equation of motion for string in a vcous uid. For small angles, we may use approimation sin. n, sum of vertical components of tension (with a y y y =,, +,,,, taking limit as! 0we obtain equation of motion for n y, = If vcous force opposing velocity, slowing it down, what c) sign of in your equation, positive or negative? What are It reasonable to have friction on chunk proportional to velocity since th a drag force. It proportional to M increased surface area. Here has dimensions of be to due LT and, positive so that, sign in wave equation results in friction that it against direction of motion of string (slowing it down). acts where f drag force due to vcosity. Note that diagram length of chunk since longer chunk, greater friction will convention of upwards positive): tan, tan, f =, f + From Newton's Law, force equals mass times acceleration:, f + drag force due to vcosity given by = (velocity ofchunk length of chunk) f = Hence + Hence dividing by on each side yields dimensions of?

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Page1of17 PHYS214 Prelim I October 6 1998 Name: Signature: Section # and TA: The exam is out of 100 points. INSTRUCTIONS By signing this exam you certify to adhere to the Cornell academic integrity code.