jyromessman ene ~~~ assertion and Reason ns on page 14 een + Wave speed is given DY V fault ‘ t amouency fis doubled x will become two time: F a edium wave Reason ditions of 1m + Forgiven cond speed remains constant Assertion ¢ Two strings sho same tension, Speed of transvers will be more, wo in figure have the a .e waves in string-1 unt TT «vee Le Here 1: is mass per unit length of ees ae string. Single Correct ‘8. The equation of a transverse wave propagating in a string is given by Y= 0.02 sin (x + 304) where, x and yare in metre and t is in second, tlinear density of the string is 1,3 x 10° kg/m, then the tension in the string is (2) Q12N (b) 12N (0) 12N —(d) 120N 4 Arope of length and mass M hangs freely from the ceiling, ifthe time taken by a tranverse wave to travel from the bottom to the top of the ri isT, then ti tocoverfirsthaltlengthis” ‘ope isT, then time va-1 ww 4) T @> Ae strings, shown in figure are made of same Same cross-section. The pulleys ena len te string AB sy and in 6, Auniform string of length 20 m is suspen "gid support. A short wave pulse is introgit ft 'm 8 Th lowest end, It starts moving up the string utday taken to reach the supports (Take, g = ig ®t ) (2a (a)2nv2s (b)2s (c)2v2s ay $ Multiple Correct Options 7, AblockM hangs vertically at the bottom eng uniform rope of constant mass per unit lent top end of the rope is attached to a fixed, rigid y support at 0. A transverse wave pulse (Pulsey wavelength Ao is produced at point Oon they The pulse takes time Tog to reach point a, the pulse of wavelength 2 is produced at point a4 2) without disturbing the position of Mit taeyg Tho to reach point O. cn Which of the following options is/are correct 0} Pulse-1 Pulse-2 (a) The time Tao =To (b) The velocities of the two pulses (Pulse 1 and Pulse 2) are the same at the midpoint of rp (©) The wavelength of Pulse 1 becomes longer it reaches point A (d) The velocity of any pulse along the rope is independent of its frequency and wavelength Subjective Questions 8. Anon-uniform wire of length L and massM has? jetjable linear mass density given by y =k whe ae, from one end of wire and k isa cons oa ne ime taken by a pulse starting at one et other end when the tension in the wit 8: Aloop of rope is whirled at a high angular vl So that it becomes a taut ik rch RA develops in the whirling ae orcadel Miscellaneous Examples & a . a rae M =2kg is suspended from a string AB of mass 6 kg as shown in fi in figure. A tran A Find its wavelength while reaching at point B. (2022 Main) 8 Concept While moving from B to A, tension will increase. So, joving fr . So, wave speed will also increase (as v = /T/ 1). Frequency will remain unchanged because it depends on source. Therefore, wavelength will also increase (as 4=“ or Av). f or hovT (asp: and f are constants) Bere Ee dn VT iy (6) = 12 em An Ex. 3 mA wave moves with speed 300 m/s on a wire ‘which is under a tension of 500 N. Find how much tensio must be changed to increase the speed to 312 m/s? Sol. Speed of a transverse wave on a wire is, T H Differentiating with respect to tension, we have dv AL 2quT v a. Dividing Eq. (i) by Eq. (i, we have WAT oy ar=on— vo2T v Substituting the proper values, we have er ALa (2) (500) (312-300) 300 =40N i.e. tension should be increased by 40 N. cals24 “7 le fe lo of 4.5 g. How muct -h the pulley? Gisturbance produced at the floor to reac Take g=10ms amass 5. Acopper wire 2.4mm in diameter is 3 m long and is used to suspend a 2kg mass from a beam. Ifa transverse Type. Based on transverse wave speed on a string. Concept (i) We know that transverse wave speed is given by Hn Yes {9 tenson suitor, then vis Calculate the time taken by the wave in e Pulse in travellin From one point to another point by the direct relation be ta distance speed niform, then v will be non-uni will be non-unifor ime can be obtained by integration, 1m A uniform rope of 7 28s 0.1 kg and length 2.45 m also uniform and we can i, If tension is non-u and in that case ti vreres ana —— eis sent along the wire by st irbanct od Seco how fast will the disturbance ja a Stopper is 8920kB/™ 4 ‘horizontal rope is attached tg driven tuning fork that vibrates My, — s over a pulley and supports, tat o mma The near mass dens ofthe rope {a} what ste speed of transverse wave gg to) what isthe waveleneth? low would your answers to parts (a 6, One end of electrically other end passe: / Answers 1. 0.055 3, 79ms"* and 63ms * 4, 0.02 approximately 5. 22m/s 6: (a) 46:3 m/s (b) 0.196 m (c) both increase by Ving 2. 32ms* See hints on page 39. te a SI Typed Problems : or Atx= fosxo8 a fone 5 m V=V05x98 =2.21 m/s ee ms Fiffecan Se° that velocity ofthe wat! Bes cient Points. So, if at point x the ‘ance dx in time ae dt, then bapa” dt = p!_ax_ rie or toa loa Pe 98 Shears &Wave Motion = 23 summary (2) In solids, =¥ =Young’s modulus of elasticity Wave Speed ve fe (i) Speed of transverse wave on a stretched wire i ae £ = modulus of elasticity a (b) In liquids, €=8 =Bulk modulus of elastic Vn Vps . {iy Speed of longitudinal wave i ie (0) In gases, > E = modulus of elasticity FERTHoveondcts20mlongrowiha ttl mansot Nate frm gre D9 ky is fastened to a fixed support. A cord attached to (a2 the other end passes over a pulley and supports an object a with a mass of 5.0 kg. The rod is struck a transverse blow Sol, (a) =100 em, f, =750Hz Stone end. Find the time required for the pulse to reach the other end. (g =9.8 m/s”) vst Sok ORL, = 150000 ems =1500ms Ans. Assume that tension in the rod is constant. Sol. Tension in the rod AB,T = mg A or T =(5.0) (9.8) =49N Mass per unit length of rod, 2 0,075 kg/m 12 Speed of transverse wave on rod, ve {i = | A2 =25.56m/s uw Voor AB 12 25.56 = ‘Awire of uniform cross-section is stretched etween two points 100 cm apart. The wire is fixed at one tend and a weight is hung over a pulley at the other end. A ‘weight of 9 ke produces a fundamental frequency of 750 Hz. (a) What is the velocity of the wave in wire? (b) if the weight is reduced to 4 kg, what is the velocity of wave? =, The required time is, t=—— =~, =0.478 Ans. v Emncfi # a v. Vi Vrs Ym va A 3500. V9 v2 = 1000 ms~* [Mneaeiny Exercise 2 1, Figure shows a string of linear mass density 1.0 gcm”* on which a wave pulse is travelling. Find the time taken by the pulse in travelling through a distance of 50 cm on the string. Take g =10ms"*. Ans. kg 2. Asteel wire of length 64 cm weighs 5 g.IFits stretched by aforce of 8N, what would be the speed of a transverse wave passing on it? 3, Two blocks each having a mass of 3.2 kg are connected by wire CD and the system is suspended from the ceiling by another wire AB. The linear mass density of the wire AB is 410 gm” and that of CDis8 gm” *. Find the speed of a transverse wave pulse produced in AB and in CD. —— |22 Fig. (b) tion ofthe displac axis is small. The net force f the tension in the string ) string with Assume thatthe nina ed ae respect to the horizontal ‘element of the string, is the sum of restoring forces fF, and Fi) ae the force of tension is cancelled, so the f Toe com onez tothe sum of the y-components ofthe net force is equal to t h force. The magnitude of the x-component of the force is ‘equal to the horizontal force of tension of the string T as shown in Fig. (b. Tocbtan the -components ofthe force, note that tan@, & and tan; =". The tan Bis equa othe slope of function ata pont, whichis equal tothe partial derivative of y with respect to x at that point. Therefore, Fa is Fae ‘equal to the negative slope of the string at x, and E is equal to the slope of the string at x,. i (% F, - i -(2) wll The net force on the small mass element can be written as of) -@)] Using Newton's second law, the net force i i 1 force is ‘mass times the acceleration, The linear densitn to? is the mass per ke it ape zat ete string and the mass of the ”) (x 2 mY) (x J - a 12), (2), ] ta wae( 2) Fa viding by TAx and taking the limit as Ax ay lim (2,13) hoy x0 Ait ea ite SPEEA Med tA inet, COE antd ess You will learn the above mathematical part ; part this equation will the ae Mathematics. Fret PProaches zero, rd wave Waves ana inermodyny Longitudinal Wave Speed in Three States speed of longitudinal wave mou 2 885 (oF aliquid) Here, B =Bulk modulus of the gas (or liquid) and p ; Now Newton, who first deduced this relation for v, as that during the passage of a sound wave through a gac the temperature of the gas remains constant, i.e. sound, travels under isothermal conditions and hence took toy. isothermal elasticity of the gas and which is equal to its pressure p.So, Newton's formula for the velocity ofa sou wave (or a longitudinal wave) in a gaseous medium becins jaa BLS If, however, we calculate the velocity of sound in air at wp with the help of this formula by substituting. p=1.01x10° N/m? and p =1.29kg/m? then v comes out to be nearly 280 m/s. Actually, the veloc of sound in air at NTP as measured by Newton himself, is found to be 332 m/s. Newton could not explain this large discrepancy between his theoretical and experimental results. Laplace after 140 years Correctly argued that a sound wae Passes through a gas (or air) very rapidly. So, adiabatic conditions are developed. So, he took B to be the adiabatic elasticity of the 8as, which is equal to yp where is the raft OfG, (molar heat capacity at constant pressure) and, (molar heat capacity at constant volume). Thus, Newton's formula as corrected by Laplace becomes. For air, 7 =1.44, so that in air, v oie ss 331.6 m/s as the velocity of sound (in ait) t! In agreement with Newton's experimental Speed of longitudinal wave in a thin rod or wire i gen ee Pp