CHAPER 14 Wae Motion 3 In a longitudinal wae motion the particles of the medium oscillate about their mean or equilibrium position along the direction

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1 Waes and hermodnamics 14.1 Introduction A wae is an disturbance from a normal or equilibrium condition that propagates without the transport of matter. In general, a wae transports both energ and momentum. he disturbance created b a wae is represented b a wae function (, t. For a string the wae function is a (ector displacement; whereas for sound waes it is a (scalar pressure or densit fluctuation. In the case of light or radio waes, the wae function is either an electric or a magnetic field ector. Wae motion appears in almost eer branch of phsics. We are all familiar with water waes, sound waes and light waes. Waes occur when a sstem is disturbed from its equilibrium position and this disturbance traels or propagates from one region of the sstem to other. Energ can be transmitted oer considerable distances b wae motion. he waes requiring a medium are called mechanical waes and those which do not require a medium are called non-mechanical waes. Light waes and all other electromagnetic waes are nonmechanical. he energ in the mechanical waes is the kinetic and potential energ of the matter. In the propagation of mechanical cal waes elasticit and inertia of the medium pla an important role. his is wh mechanical waes sometimes are also referred to as elastic waes. Note that the medium itself does not moe as a whole along with the wae motion. Apart from mechanical and nonmechanical waes there is also another kind of waes called matter waes. hese represent wae like properties of particles. 14. ranserse and Longitudinal Waes here are two distinct classes of wae motion : (i transerse and (ii longitudinal. In a transerse wae motion the particles of the medium oscillate about their mean or equilibrium position at right angles to the direction of propagation of wae motion itself. crest trough Fig oscillation mean position wae motion iww eww.ww his form of wae motion traels in the form of crests and troughs, as for eample, waes traelling along a stretched string. his tpe of waes are possible in media which possess elasticit of shape or rigidit, i. e., in solids. hese are also possible on the surface of liquids also, een though the do not possess the propert of rigidit. his is because the possess another equall effectie propert (surface tension of resisting an ertical displacement of their particles (or keeping their leel. Gases, howeer, possess neither rigidit nor do the resist an ertical displacement of particles (or keep their leel. A transerse wae motion is therefore not possible in a gaseous medium. An electromagnetic wae is necessaril a transerse wae because of the electric and magnetic fields being perpendicular to its direction of propagation. he distance between two successie crests or troughs is known as the waelength ( of the wae. fw dit,ww eẇw.,ww ching ing.c.com Get Discount Coupons for our Coaching institute and FREE Stud Material at

2 CHAPER 14 Wae Motion 3 In a longitudinal wae motion the particles of the medium oscillate about their mean or equilibrium position along the direction of propagation of the wae motion itself. his tpe of wae motion traels in the form of compressions and rarefactions and is possible in media possessing elasticit of olume, i. e., in solids, liquids and gases. mean position oscillations wae motion he distance between two successie compressions or rarefactions constitute one waelength. Sound waes in a gas are longitudinal in nature. In some cases, the waes are neither purel transerse nor purel longitudinal as, for eample ripples on water surface (produced b dropping a stone in water, in which the particles of the medium (here water oscillate across as well as along the direction of propagation of the wae motion describing elliptical paths. We need not howeer, bother ourseles with an such tpe of waes here. Again waes ma be one dimensional, two dimensional or three dimensional according as the propagate energ in just one, two or three dimensions. ranserse waes along a string are one dimensional, ripples on water surface are two dimensional and sound waes proceeding radiall from a point source are three dimensional he General Equation of Wae Motion As we hae alread read, in a wae motion, some phsical quantit (sa is made to oscillate at one place and these oscillations of propagate to other places. he ma be, (i displacement of particles from their mean position in case of transerse wae in a rope or longitudinal sound wae in a gas. (ii pressure difference ( dp P or densit difference ( dρ in case of sound wae or (iii electric and magnetic fields in case of electromagnetic waes. he oscillations of ma or ma not be simple harmonic in nature. Now let us consider a one dimensional wae traelling along -ais. In this case is a function of position ( and time ( t. he reason is that one ma be interested in knowing the alue of at a general point at an time t. hus, we can sa that, = (, t But onl those functions of and t, represent a wae motion which satisf the differential equation, = k Here k is a constant, which is equal to square of the wae speed, or k = Fig ww.pi w.pic pickmcoaching.co ching g.com Get Discount Coupons for our Coaching institute and FREE Stud Material at

3 4 hus, the aboe equation can be written as, = (i he general solution of this equation is of the form (, t = f ( a ± bt (ii hus, an function of and t which satisfies Eq. (i or which can be written as Eq. (ii represents a wae. he onl condition is that it should be finite eerwhere and at all times. Further, if these conditions are satisfied, then speed of wae ( is gien b, coefficient of t b = = coefficient of a he plus (+ sign between a and bt implies that the wae is traelling ling along negatie -direction and minus ( sign shows that it is traelling along positie -direction.. Sample Eample 14.1 Show that the equa tion, = a sin ( ω t k sat is fies the wae equa tion =. Find speed of wae and the di rec tion in which it is trael ling. Solution = ω a sin ( ωch ω t ch k chk and = k a sin (ω ωt k We can write these two equations as, km = ω t k Comparing this with, = We get, wae speed = ω Ans. k he negatie sign between ωt and k implies that wae is traelling along positie -direction. Sample Eample 14. Which of the fol low ing func tions rep re sent a wae (a ( ( t ln ( + t (c e 1 t ww(b (d + t Solution Although all the four functions are written in the form f ( a ± bt, onl (c among the four functions is finite eerwhere at all times. Hence onl (c represents a wae. 0.8 Sample Eample 14.3 (, t = rep re sents a mo ing pulse where and are in metre [( 4 + 5t + 5] and t in sec ond. hen choose the cor rect al ter na tie(s: (JEE 1999 (a pulse is moing in positie -direction www. ww.pickmcoaching.com w.p ching.co.com Get Discount Coupons for our Coaching institute and FREE Stud Material at

4 (b in s it will trael a distance of.5 m (c its maimum displacement is 0.16 m (d it is a smmetric pulse Solution (b, (c and (d are correct options. he shape of pulse at = 0 and t = 0 would be as shown in Fig ( 0, 0 = = 0.16 m 5 From the figure it is clear that ma = 0.16 m Pulse will be smmetric (smmetr is checked about ma if and At t = 0; ( = ( From the gien equation 0.8 ( = ( = or ( = ( herefore, pulse is smmetric. at t = 0 Speed of pulse : At t =1s and = 1.5 m 0.16m = 1.5m = 0 ct t = 1s t = 0 ict Fig alue of is again 0.16 m, i.e.,, pulse has traelled a distance of 1.5 m in 1 second in negatie -direction or we can sa that the speed of pulse is 1.5 m/s and it is traelling in negatie -direction. herefore, it will trael a distance of.5 m in seconds. he aboe statement can be better understood from Fig Alternate method : If equation of a wae pulse is = f ( a ± bt 0.16m ww.pw.p 0.16m 0 t = 0 Fig the speed of wae is b a in negatie -direction for = f ( a + bt and positie -direction for = f ( a bt. Comparing this from gien equation we can find that speed of wae is 5 =1.5 m/s and it 4 is traelling in negatie -direction. Get Discount Coupons for our Coaching institute and FREE Stud Material at

5 6 Sample Eample 14.4 In a wae mo tion = a sin ( k ω t, can rep re sent: (JEE 1999 (a electric field (b magnetic field (c displacement (d pressure Solution ( a, b, c, d In case of sound wae, can represent pressure and displacement, while in case of an electromagnetic wae it represents electric and magnetic fields. In general, is an general phsical quantit which is made to oscillate at one place and these oscillations are propagated to other places. Introductor or Eercise Proe that the equation = a sin ω t does not satisf the wae equation and hence it does not represent a wae. ( b ct. A wae pulse is described b (, t = ae, where a, b and c are positie constants. What is the speed of this wae? 3. he displacement of a wae disturbance propagating in the positie -direction is gien b 1 = 1 + at t = 0 and 1 = at t = s 1 + ( 1 where and are in metre. he shape of the wae disturbance does not change during the propagation. What is the elocit of the wae? A traelling wae pulse is gien b, = t ( Here and are in metre and t in second. In which direction and with what elocit is the pulse propagating? What is the amplitude of pulse? 5. If at t = 0, a traelling wae pulse on a string is described b the function, 10 = ( + Here and are in metre and t in second. What will be the wae function representing the pulse at time t, if the pulse is propagating along positie -ais with speed m/s? 14.4 Plane Progressie Harmonic Wae Consider a function = f (, represented graphicall b the solid cure shown in Fig = f(+ 0 ww.prep 0 0 = f( = f( 0 w.pic pickm ckmcoaching.co g.com Fig If we replace b 0, we get the function, = f ( 0 It is clear that shape of the cure has not changed, the same alue of occurs for alues of increased b the amount 0. In other words, assuming that 0 is positie, we see that the cure has been displaced to Get Discount Coupons for our Coaching institute and FREE Stud Material at

6 the right an amount 0 without deformation. Similarl, = f ( + 0 corresponds to a displacement of the cure to the left b an amount 0. For eample if we hae two functions: 1 = and = ( 5 and 1 = 16at = 4 then has the same alue, i.e., 16 at = or = 9. Now, if 0 = t, where t is the time, we get a traelling cure. hat is = f ( t represents a cure moing to the right with a elocit, called the wae elocit or phase elocit. Similarl = f ( + t represents a cure moing to the left with elocit. herefore, we conclude that a mathematical epression of the form, (, t = f ( ± t is adequate for describing a phsical situation that traels or propagates without deformation along negatie or positie -ais. he quantit (, t ma represent the deformation in a solid, the pressure in a gas, an electric or magnetic field, etc. When (, t is a sine or cosine function such as, (, t = Cok A sin k ( t or (, t CA Cc = A cos k ( t it is called plane progressie harmonic wae.. In plane progressie harmonic wae oscillations of are simple harmonic in nature. he quantit k has a special meaning. Replacing the alue of b + π, we get the same alue of, i. e., k π + t A k t A k t k = + π, sin = sin [ ( + π] k = A sin k ( t = (, t he quantit, π k = designated as waelength, is the space period of the cure, that is the cure repeats itself eer length. he quantit k = π represents the number of waelengths in the distance π and is called the wae number. herefore, π (, t = A sin k ( t = A sin ( t = f( t w.p Fig = f( + t O m w.pic ching ing.c.com Fig A 7 Get Discount Coupons for our Coaching institute and FREE Stud Material at

7 8 represents a plane progressie harmonic wae of waelength propagating towards the positie -ais with speed. he aboe equation can also be written as, (, t = A sin ( k ωt where ω = k = gies the angular frequenc of the wae. Further, ω = π f, where f is the frequenc with which oscillates at eer point. We hae the important relation, = f Also if is the period of oscillation then, π π 1 = = ω f t We ma also write, = A sin π hus, the equation of a plane progressie harmonic wae moing along positie -direction can be written as, π t = Asin k ( t = A sin ( k ωt = A sin ( t = A sin π Similarl, the epressions t = A sin k ( + t = A sin ( k + ωt = A sin π + represents a plane progressie harmonic wae traelling in negatie -direction. = constt t = π ω / Displacement of a particle km3 3 / t t ww ww at different instants Fig Note that t as propagates in the medium (or space, it repeats itself in space after one period, because = which shows that, the waelength is the distance adanced b the wae motion in one period. herefore, in plane progressie harmonic wae we hae two periodicities, one in time gien b the period, and one in space gien b the waelength, with the two related b = t = constt = π k Position of different particles at same instant in a string. ww.pickm chin Get Discount Coupons for our Coaching institute and FREE Stud Material at

8 Important points in Wae motion Read so far 1. As we hae read in Art. 14.3, an function of and t which satisfies equation number (i of the same article or which can be written in the form of equation number (ii represents a wae proided it is finite eerwhere at all times. What we hae read in Art is about plane progressie harmonic wae. If f ( a ± bt is a sine or cosine function, it is called plane progressie harmonic wae. he onl special characteristic of this wae is that oscillations of are simple harmonic in nature.. he general epression of a plane progressie harmonic wae is, = A sin ( k ± ω t ± φ or = A cos ( k ± ωt ± φ Here φ represents the initial phase. 3. I hae seen students often confused whether the equation of a plane progressie wae should be, = A sin ( k ωt or = A sin ( ωt k Because some books write the first while the others write the second. It hardl matters whether ou write the first or the second. Both the equations represent a traelling wae traelling in positie ω -direction with speed = he difference between them is that the are out of phase, i. e., phase k difference between them is π. It means, if a particle in position ng = 0 at time t = 0 is in its mean position and moing upwards (represented b first wae then the same particle will be in its mean position but moing downwards (represented b the second wae. Similarl the waes, = A sin ( k ω t and = A sin ( k ω t are also out of phase. 4. Particle elocit ( p and acceleration ( a p in a sinusoidal wae : In plane progressie harmonic wae particles of the medium oscillate simple harmonicall about their mean position. herefore, all the formulae what we hae read in SHM appl to the particles here also. For eample, maimum particle elocit is ± Aω at mean position and it is zero at etreme positions etc. Similarl maimum particle acceleration is ± ω A at etreme positions and zero at mean position. Howeer the wae elocit is different from the particle elocit. his depends on certain characteristics of the medium. Unlike the particle elocit which oscillates simple harmonicall (between + Aω and Aω the wae elocit is constant for gien characteristics of the medium. Suppose the wae function n is, (, t = A sin ( k ω t (i Let us differentiate this function partiall with respect to t and. (, t = Aω cos ( k ω t (ii (, t = Ak cos ( k ω t (iii Now, these can be written as, (, t ω = (, t k Here, w.pick ching ing.c.com (, t = particle elocit P ω k =wae elocit 9 Get Discount Coupons for our Coaching institute and FREE Stud Material at

9 10 and (, t = slope of the wae hus, P = ( slope (i ie.., particle elocit at a gien position and time is equal to negatie of the product of wae elocit with slope of the wae at that point at that instant. he acceleration of the particle is the second partial deriatie of (, t with respect to t, t ap = (, = ω A sin ( k ωt = ω (, t i. e., the acceleration of the particle equals ω times its displacement, which is the result we obtained for SHM. hus, a P = ω ( displacement ( or We can also show that, = (, t ω (, t t k (, t (, t = which is also the wae equation. Figure shows the elocit ( P and acceleration ( a P gien b Eqs. (i and ( for two points 1 and on a 1 p string as a sinusoidal wae is traelling in it along positie -direction. p,a p a p At 1 : Slope of the cure is positie. Hence from Eq. (i particle elocit ( P is negatie or downwards. Similarl displacement of the particle is positie, so from Eq. ( acceleration will be negatie or downwards. Fig At : Slope is negatie while displacement is positie. Hence P will be positie (upwards and a P is negatie (downwards. Note he direction of P will change if the wae traels along negatie -direction. Sample Eample 14.5 he equa tion of a wae is, (, t = 0.05 sin π ( t π m Find : (a the ew waelength, the frequenc and the wae elocit (b the particle elocit and acceleration at = 0.5 m and t = 0.05 s. Solution (a he equation ma be rewritten as, π (, t = 0.05 sin 5π 0πt m 4 ww.pickmcoa Coaching.com Comparing this with equation of plane progressie harmonic wae, (, t = A sin ( k ω t + φ we hae, (i Get Discount Coupons for our Coaching institute and FREE Stud Material at

10 π wae number k = = 5π rad/m = 0.4 m Ans. he angular frequenc is, ω = π f = 0π rad/s f =10 Hz Ans. he wae elocit is, ω = f = = 4 m/s in + direction k Ans. (b he particle elocit and acceleration are, = 5 π π ( 0π ( 0.05 cos π t 4 =. m/s Ans. 5 π π = ( 0π ( sin π 4 Sample Eample 14.6 Fig ure shows a snap shot of a si nu soi dal tra el ling wae taken at t = 0.3 s. he wae length is 7.5 cm and the am pli tude is cm. If the crest P was at = 0 at t = 0, write the equa tion of tra el ling wae. ng.comπ ng =140 0 m/s Ans. Solution Gien, A = cm, = 7.5 cm π k = = 0.84 cm 1 he wae has traelled a distance of 1. cm in 0.3 s. Hence, speed of the wae, 1. = = 4 cm/s 0.3 Angular frequenc ω = ( ( k = 3.36 rad/s Since the wae is traelling along positie -direction and crest (maimum displacement is at = 0 at t = 0, we can write the wae equation as, (, t = A cos ( k ωt or (, t = A cos ( ω t k as cos ( θ = cos θ herefore, the desired equation is, 1 (, t = ( cm cos [( 0.84 cm ( 3.36 rad/s t] cm Ans. 5 w.pickmcoac Coach oaching.com com s 11 P t = 0.3s cm 1. cm Fig Get Discount Coupons for our Coaching institute and FREE Stud Material at

11 1 1. he equation of a traelling wae is, t (, t = 0.0 sin m Introductor Eercise 14. Find : (a he wae elocit and (b the particle elocit at = 0. m and t = 0.3 s. Gien cos θ= 0.85 where θ = 34 rad. Is there an relationship between wae speed and the maimum particle speed for a wae traelling on a string? If so, what is it? 3. Consider a sinusoidal traelling wae shown in figure. he wae elocit is + 40 cm/ s. Find : (a the frequenc (b the phase difference between points.5 cm apart (c how long it takes for the phase at a gien position to change b 60 (d the elocit of a particle at point P at the instant shown. (cm P ranserse waes on a string hae wae speed 1.0 m/s, amplitude 0.05 m and waelength 0.4 m. he waes trael in the + direction and at t = 0 Co Co= the = 0 end of the string has zero displacement and is moing upwards. (a Write a wae function describing the wae. (b Find the transerse displacement of a point at = 0.5 m at time t = 0.15 s. (c How much time must elapse from the instant in part (b until the point at = 0.5 m has zero displacement? 14.5 Speed of a ranserse Wae on a String (cm One of the ke properties of an wae is the wae speed. In this section we ll see what determines the speed of propagation ation of transerse waes on a string. he phsical quantities that determine the speed of transerse e waes on a string are the tension in the string and its mass per unit length (also called linear mass densit. We might guess that increasing the tension ' showed increase the restoring forces that tend B θ' to straighten the string when it is disturbed, thus A ' increasing the wae speed. We might also θ guess increasing the mass should make the motithat on more sluggish and decrease the speed. Both these guesses turn out to be right. O We will deelop the eact relationship between d wae speed, tension and mass per unit length Fig ww.pi w.pickmcoachi oaching.com g.c om Fig Forces on a section of a transersel displaced string. Get Discount Coupons for our Coaching institute and FREE Stud Material at

12 Under equilibrium conditions a string subject to a tension is straight. Suppose that we now displace the string sidewise, or perpendicular to its length, b a small amount as shown in figure. Consider a small section AB of the string of length d, that has been displaced a distance from the equilibrium position. On each end a tangential force is acting. Due to the curature of the string, the two forces are not directl opposed but make angles θ and θ with the -ais. he resultant upward force on the section AB of the string is, F = (i Under the acting of this force, the section AB of the string moes up and down. Rewriting Eq. (i we hae, F = (sin θ sin θ Since θ and θ are almost equal, we ma write F = d ( sin θ If the curature of the string is not er large, the angles θ and θ are small, and the sines can be replaced b their tangents. So the upward force is, d F = d (tan θ =. (tan θ. d d But tan θ is the slope of the cure adopted b the string, which is equal to d d. Hence F d d = d d d h d d = d his force must be equal to the mass of the section AB multiplied b its upward acceleration d dt. If µ is the linear densit of the string, the mass of the section AB is µd. We use the relation F = ma and write the equation of motion of this section of the string as, or Comparing this with wae equation, w the wae speed, µd d = d dt ( d d d ckm ckmd km d d = dt µ d = = µ Alternate Method Consider a pulse traelling along a string with a speed to the right. If the amplitude of the pulse is small compared to the length of the string, the tension will be approimatel constant along the string. In the reference frame moing with speed to the right, the pulse in stationar and the string moes with a speed to the left. Figure shows a small segment of the string of length l. his segment forms part of a (km ching ing.c.com 13 Get Discount Coupons for our Coaching institute and FREE Stud Material at

13 14 l a r = R θ ΣF r l θ R O R θ θ (a Fig (a o obtain the speed of a wae on a stretched string, it is conenient to describe the motion of a small segment of the string in a moing frame of reference. (b In the moing frame of reference, the small segment of length l l moes to the left with speed. he net force on the segment is in the radial direction because the horizontal components of the tension force cancel. circular arc of radius R. Instantaneousl the segment is moing with speed in a circular path, so it has a centripetal acceleration. he forces acting on the segment are the tension at each end. he R horizontal components of these forces are equal and opposite and thus cancel. he ertical components of these forces point radiall inward toward the centre of the circular arc. hese radial forces proide the centripetal acceleration. Let the angle subtended ed b the segment at centre be θ. he net radial force acting on the segment is ΣF F r = Coθ sin θ = θ Where we hae used the approimation ation sin C θ θ for small θ. If µ is the mass per unit length of the string, the mass of the segment of length l is m = µ l = µ Rθ (as l = Rθ m From Newton s second law ΣFr = ma = R or θθ = ( µ Rθ = R µ Speed of Wae Motion 1. Speed ed of transerse wae on a string is gien b, = µ O (b w.pi ching g.com Here, µ = mass per unit length of the string = m l Get Discount Coupons for our Coaching institute and FREE Stud Material at

14 Note = ma la = m V (A = area of cross-section of the string A (V = olume of string = ρa (ρ = densit of string Hence, the aboe epression can also be written as, = ρa. Speed of longitudinal wae through a gas (or a liquid is gien b, B = ρ Here, B = Bulk modulus of the gas (or liquid and ρ = densit of the gas (or liquid Now, Newton who first deduced this relation for assumed that during the passage of a sound wae through a gas (or air, the temperature of the gas remains constant i. e., sound wae traels under isothermal conditions and hence took B to be the isothermal elasticit of the gas and which is equal to its pressure P. So, Newton s formula for the elocit of a sound wae (or a longitudinal wae in a gaseous medium becomes, P = ρ If, howeer, we calculate the elocit of sound in air at NP with the help of this formula b substituting P = N/m Coρ and ρ = kg/m then comes out to be nearl 80 m/s. Actuall the elocit of sound in air at NP as measured b Newton himself, is found to be 33 3 m/s. Newton could not eplain this large discrepanc between his theoretical and eperimental results. La place after 140 ears correctl argued that a sound wae passes through a gas (or air er rapidl. So adiabatic conditions are deeloped. So, he took B to be the adiabatic elasticit of the gas, which is equal to γp where γ is the ratio ckc of C P (molar heat capacit at constant pressure and C V (molar heat capacit at constant olume. hus, Newton s formula as corrected b La place becomes, P = γ ρ For air, γ = So that in air, w = 1.41 P ρ which gies m/s as the elocit of sound (in air at NP which is in agreement with Newton s eperimental result. We will carr out the deriation of formula, B = ρ w.pickmcoaching.com ching in the chapter of sound (Chapter Get Discount Coupons for our Coaching institute and FREE Stud Material at

15 16 3. Speed of longitudinal wae in a thin rod or wire is gien b, Y = ρ Here, Y is the Young s modulus of elasticit. Sample Eample 14.7 One end of 1.0 m long rub ber tube with a to tal mass of 0.9 kg is fas tened to a fied sup port. A cord at tached to the other and passes oer a pul le and sup ports an ob ject with a mass of 5.0 kg. he tube is struck a trans erse blow at one end. Find the time re quired for the pulse to reach the other end. (g = 9.8 m/s Solution ension in the rubber tube AB, = mg or = (5.0 (9.8 = 49 N Mass per unit length of rubber tube, 0.9 µ = = kg/m 1 Speed of wae on the tube, = 49 = µ = m/s he required time is, AB t = = = 0.47 s Ans. 56 Sample Eample 14.8 A uni form rope of mass 0.1 kg and length.45 m hangs from a ceil ing. (a Find the speed of transerse wae in the rope at a point 0.5 m distant from the lower end. (b Calculate the time taken b a transerse wae to trael the full length of the rope. Solution (a As the string has mass and it is suspended erticall, tension in it will be different at different points. For a point at a distance from the free end, tension will be due to the weight of the string below it. So, if m is the mass of string of length m l, the mass of length of the string will be,. l m = g g = µ l or g µ = = = µ oas At = 0.5 m, = =.1 m/s g m l = µ pickm ckmco Coaching.com (b From Eq. (i we see that elocit of the wae is different at different points. So, if at point the wae traels a distance d in time dt, then (i B A m Fig Fig Get Discount Coupons for our Coaching institute and FREE Stud Material at

16 dt = or t d dt = = t l 0 0 d g d g l = =.45 g 9.8 =10. s Ans. Introductor Eercise Calculate the elocit of a transerse wae along a string of length m and mass 0.06 kg under a tension of 500 N.. Calculate the speed of a transerse wae in a wire of 1.0 mm cross-section under a tension of 0.98 N. 3 3 Densit of the material of wire is kg/m Energ in Wae Motion Eer wae motion has energ associated with it. In wae motion, energ and momentum are transferred or propagated. o produce an of the wae motions, we hae to appl a force to a portion of the wae medium. he point where the force is applied moes, so we do work on the sstem. As the wae propagates, each portion of the medium eerts a force and does work on the adjoining portion. In this wa a wae can transport energ from one region of space to other. Regarding the energ in wae motion, we come across three terms namel, energ densit ( u, power ( P and intensit (I. Now let us take them one b one. Energ densit (u B the energ densit of a plane progressie wae we mean the total mechanical energ (kinetic + potential per unit olume of the medium through which the wae is passing. Let us proceed to obtain an epression for it. Consider a string attached to a tuning fork. As the fork ibrates, it imparts energ to the segment of the string attached to it. For eample, as the fork moes through its equilibrium position, it stretches the segment, increasing ng its potential energ, and the fork imparts transerse speed to the segment, increasing its kinetic energ. As a wae moes along the string, energ is imparted to the other segments of the string. atw www. ww.pickmcoaching.co w.pic ing.com Kinetic energ per unit olume We can calculate at the kinetic energ of unit olume of the string from the wae function. Mass of unit olume is the densit ρ. Its displacement from equilibrium is the wae function = A sin ( k ω t. Its speed is d where is considered to be fied. he kinetic energ of unit olume K is then dt 1 1 d K = ( m = ρ dt 17 Get Discount Coupons for our Coaching institute and FREE Stud Material at

17 18 Using = A sin ( k ω t, we obtain d = ωa cos ( k ωt dt So the kinetic energ per unit olume is 1 K = ρ ω A cos ( k ωt (i Potential energ per unit olume he potential energ of a segment is the work done in stretching the string and depends on the slope d. he potential energ per unit olume of the string is related to the slope and tension b (for small d slopes 1 d U = ρ (ii d where = wae speed = ω k Using d = ka cos ( k ω t, we obtain for the potential energ d 1 U = ρω A cos ac ( ωt (iii ack Co Which is the same as the kinetic energ. he total energ per unit olume is E = K + UU = ρω A cos ( k ωt (i We see that E aries with time. Since the aerage alue of cos ( k ω t at an point is 1, the aerage energ per unit olume (called the energ densit u is u = 1 ρω A Note (i Equation ( is the same result as for a mass ρ attached to a spring and oscillating simple harmonic wae. Howeer for a mass attached to a spring the potential energ is maimum when the displacement is maimum. For a string segment, the potential energ depends on the slope of the string and is maimum when the slope is maimum, which is at the equilibrium position of the segment, the same position for which the kinetic energ is maimum. mum. At A : Kinetic energ and potential energ both are zero. At B : Kinetic energ and potential energ both are maimum. (ii Equation (ii has been deried in miscellaneous eample number ww.pic om A B Fig ( Get Discount Coupons for our Coaching institute and FREE Stud Material at

18 Power (P Power is the instantaneous rate at which energ is transferred along the string (if we consider a transerse wae on a string. Its alue depends on the position on the string at time t. Note that energ is being transferred onl at points where the string has a nonzero slope ( / 0, so that there is a transerse component of the tension force, and where the string has a nonzero elocit ( / 0 so that the transerse force can do work. Let us obtain an epression for power transmitted through a string. In unit time, the wae will trael a distance. If s be the area of cross-section of the string, then olume of this length would be s and energ transmitted per unit time (called power would be, P = ( energ densit ( olume P = 1 ρω A s Note his power is reall the aerage power. he instantaneous power is gien b, P( t, = F ( t, ( t, which comes out to be, ρω A s sin ( k ωt or ρω A s cos ( k ωt depending on the displacement equation (, t. he aerage alue of sin or cos function, aeraged oer an whole number of ccles is 1. Hence the aerage power will 1 behin ρω A s. Intensit (I Flow of energ per unit area of cross-section of the string in unit time is known as the intensit of the wae. hus, power P I = = area Cf of cross-section s Cof or 1 ckmi km= I = ρω A M Again this is the aerage intensit transmitted through the string. he instantaneous intensit ρω A sin ( k ωt or ρω A pico ics cos ( k ωt depends on and t. Note (i Although the aboe relations for power and intensit hae been discussed for a transerse wae on a string, the hold good for other waes also. (ii Intensit due to a point source : If a point source emits wae uniforml in all directions, the energ at a distance r from the source is distributed uniforml on a spherical surface of radius r and area S π r. If P is the power emitted b the source, the power per unit area at a distance r P from source is. he aerage power per unit area that is incident perpendicular to the ws ww= wπ = 4 the ww4π 4π r direction of propagation is called the intensit. herefore: P I = 4 πr MCa o ww.pickmcoaching.com chin ing.c 19 or I 1 r Get Discount Coupons for our Coaching institute and FREE Stud Material at

19 0 Now, as amplitude A therefore, be written as I, a spherical harmonic wae emanating from a point source can A ( r, t = sin ( kr ωt r Sample Eample 14.9 A stretched string is forced to trans mit trans erse waes b means of an os cil la tor cou pled to one end. he string has a di am e ter of 4 mm. he am pli tude of the os cil la tion is 10 4 m and the fre quenc is 10 Hz. en sion in the string is 100 N and mass den sit of wire is kg/ m 3. Find : (a the equation of the waes along the string (b the energ per unit olume of the wae (c the aerage energ flow per unit time across an section of the string and (d power required to drie the oscillator. Solution (a Speed of transerse wae on the string is, = (as µ = ρs ρ S Substituting the alues, we hae 100 = a1 ac0 π MC (4 ( = C3 Cm m/s rad ω = πf f = 0 π = 6.83 rad s s = 1.44 m 1 k = ω Equation of the waes along the string, (, t = A sin ( k ω t w.p.p 3 3 = ( 10 4 m sin ( 1.44 m rad t s Ans. (b Energ per unit olume of the string, 1 u = energ densit = ρω A Substituting the alues, we hae u = ( ( 6.83 ( 10 3 = J/ m Ans. w.pickmcoaching.co ching.com Get Discount Coupons for our Coaching institute and FREE Stud Material at

20 1 (c Aerage energ flow per unit time, P = power = 1 ρω A ( s = ( u ( s Substituting the alues, we hae π 3 P = ( ( ( = J/s Ans. (d Power required to drie the oscillator is obiousl W. Ans. 5 Introductor Eercise Spherical waes are emitted from a 1.0 W source in an isotropic ic non-absorbing medium. What is the wae intensit 1.0 m from the source?. A line source emits a clindrical epanding wae. Assuming the medium absorbs no energ, find how the amplitude and intensit of the wae depend on the distance from the source? achin ching.c ng.com Get Discount Coupons for our Coaching institute and FREE Stud Material at