WAVES AND ACOUSTICS [JEE ADVANCED PREVIOUS YEAR SOLVED PAPERS] ] JEE Advanced Single Correct Answer Type 1. cylindrical tube, open at oth ends, has a fundamental frequency ‘fin ar. The tube i dipped vertically in water so that half of itis in water. The fundamental frequency of the air column am b. 394 ef a ar 2. A transverse wave is described by the equation ‘The maximum particle velocity is equal to four times the ‘wave velocity if ad ce AR 3. A tube, closed at one end and containing when excited, the fundamental mode of frequency $12 Hz i the tube is open at both ends the fundamental frequency that can be excited is (in H. 1024 b 512 ©. 286 4.128 CIFJEE 1986) 4. A wave represented by the equation y= a 60s (kx — ca) is superposed with another wave to form a stationary wave such that point x= 0 is node. The equation forthe other a. asin(kx +) by asin(kx-e) & macosike rar) ~asin(k en) CIT-JEE 1988) S. An organ pipe P, closed at one end vibrating in its first harmonic and another pipe P: open at both the ends vibrating in ts third harmonic are in resonance with a given tuning fork. The ratio of the length of P; to that of a. 83 ba © V6 4.13 (IIT-JEE 1988) 6 Am object of specific gravity pis hung from a thin steel wire. The fundamental frequency for transverse standing waves in the wire is 300 Hz. The object is immersed in ‘water so that one half ofits volume is submerged. The new fundamental frequency in Hz is 2022)". (22) 20 2p-1 e025) «2 22!) (IT-JEE 1995) 7. The extension ina string, obeying Hook's law, is x. The speed of sound in the stretched string is. Ifthe extension in the string is increased to I. Sr, the speed of sound will be a 122¥ b. 061 y 1309 4.0.75» (ITJEE 1996) 8. An open pipe is suddenly closed at one end and with the result frequency of third harmonic of the closed pipe is found 10 be higher by 100 Hz than the fundamental frequency of the open pipe. The fundamental frequency ‘of the open pipe is a. 200Hz b. 300 Hz e240 Hz 4. 480 Hz (IIT-JEE 1996) 9. A whistle giving out 450 Hz approaches a stationary observer at a speed of 33 mis. The frequency heard by the observer in Hz is (speed sound = 330 m/s) a 409 b 29 © 517 4. 500 (IT.JEE 1997) 10, A travelling wave in a stretched string is described by the equation y = A sin (kx ~ ax) The maximum particle velocity is a Ao bak daddk dx (INT-JEE 1997) I. A string of length 0.4 m and mass 107 kg is tightly clamped at its ends. The tension in the string is 1.6 N. Identical wave pulse is produced at one end at equal intervals of time, Al. The minimum value of Ar which allows constructive interface between successive pulses is a 0055 b0.10s 0208 4.0405 (IT.JEE 1998) 12. The ratio ofthe speed of sound in nitrogen gas to that in helium gas a 300 K is a 27 » a7) © (3s @. (6/5 (IT-JEE 1999) 13, A train moves towards a stationary observer with speed 34 ‘ms. The train sounds a whistle and its frequency registered by the observer if. Ifthe tains speed is reduced to 17 ‘vs, the frequency registered if. Ifthe speed ofthe sound . 19/18 (IIT-JEE 2000) 14, Two vibrating strings ofthe same material but lengths L and 2L have radii 2r and r. respectively. They are stretched ‘under the same tension. Both the strings vibrate in theit fundamental modes. the one of length L with frequency m, and the other with frequency m,.The ration /n is given by ad bea 28 a (LIT-JEE 2000)15. ‘Two pulses ina stretched string whose centres ar initially ‘Sem apartare moving towards each other as shown in the figure, The speed of each pulse is 2 ems. After 2 s, the ‘otal energy of the pulse will be a 2e10 ». purely kinetic purely potential partly kinetic and partly potential (IIT-JEE 2000) 16, The ends of a stretched wire of length L are fixed at x = Oand x= J. In one experiment, the displacement of the wire is yj = A sin (RUL) sin ox and energy is E, and in another experiment its displacement is y 8: = AsinQu/Lysin2ov and energy is E>. Then a. £,=E; b E,=26, aE, 4. = 166, (ITT-JEE 2001) 17. A siren placed at a railway platform is emiting sound of requency 5 kHz. A passenger sitting in a moving train A records frequency of 5.5 kHz while the train approaches the siren, During his return journey in a differen train B hhe records a frequency of 6.0 kHz while approaching the same siren. The ratio ofthe velocity of train B to that of rain Ais a. 242/252 be? 4 4.6 © (IT-JEE 2002) 18. A sonometer wire resonates with a ning fork forming standing waves with five antinodes between the (v0 bridges when a mass of 9 kg is suspended from the wire. ‘When thismassis replaced by a mass M, the wie resonates with the same tuning fork forming three antinodes for the same positions of the bridges. The value of Mis a. 25kg b. Skg 12Sky 4d. 125kg (IIT-JEE 2002) 19, A police car. moving at 22 nv, chases a motoreyclist, ‘The police man sounds his horn at 176 Hz. while both fof them move towards a stationary siren of frequency 165 Hz. Calculate the speed of the motorcycle, iF itis given that he does not observe any’ beats. Police car Moroso Oo oO 90 em Se Statonary tea . cn (uss He) 33ms b22m/s cero dk I as (IIT-JEE 2003) 20. In the experiment forthe determination of the speed of sound in air using the resonance column method, the length ofthe air column that resonates in the fundamental ‘mode, with a tuning fork is 0.1 m. When this length is ‘changed ‘0 0.35 m. the same tuning fork resonates with the first overtone. Calculate the end correction, & 0012 m © 005m , 0.025 m 4. 0.024 m (IIT-JEE 2003) 21. A source of sound of frequency 600 Hz is placed inside ‘ater. The speed in water is 1500 sms and in air i is 300 m/s. The frequency of sound recorded by an observer Standing in aris a. 200Hr 1207 b. 3000 He 4. 600 Hr -JEE 2004) 22. Appipe of length closed at one end, is kept ina chamber . A second pipe open at both ends is Placed in a second chamber of gas of density ps. The ‘compressibility of both the gases is equal, Calewlate the length of the second pipe if frequency of first overtone in both the cases is equal 4,2 b 3'Ve. 4, ue (IIT-JEE 2004) e: 23, Ina resonate tube with tuning fork of frequency $12 He, first resonance occurs at water level equal to 30.3 em and second resonance occurs at 63.7 em. The maximum possible error in the speed of sound is a. S.1emis b, 1024 emis 2048 em/s 4. 153.6ems (IIT-JEE 2005) An open pipe is in resonance in 2nd harmonic with frequency f;. Now one end of the tube is closed and frequency is increased 10, such thatthe resonance again ‘occurs in nth harmonic. Choose the correct option, AIT-JEE 2005) 2S. A massless rod of length 4 ypu iC Lis suspended by two ‘identical strings AB and CD of equal length. A block of mass m is suspended from point 0. ® 7— >? such that BO is equal to “x, Further itis observed thatthe frequency of Ist harmonic in AB is equal to 2nd hharmonie frequency in CD, xis, ak vt ry s M. L Ce Z__ AUTIEE 2006) 26. In the experiment to determine the speed of sound using a resonance columncr 29, aM ‘a. prongs of the tuning fork are kept ina vertical plane '. prongs ofthe tuning fork are kept ina horizontal plane . imoone of the two resonances observed. the length of | is close tothe wavelength 4. mone of the 1wo resonances observed. the length of the resonating air column is close to half of the wavelength of sound in air [A transverse sinusoidal wav moves along a string in th positive direction at a speed of 10 emis. The wavelength of the ‘wave is 0.5 mand its amplitude iy Hem. Ata particular time tthe snapshot ofthe wave is shown in the figure. The velocity of point P when its displacement is Semis (IIT-JEE 2008) A vibrating string of certain length ! under 4 tension T resonates with a mode corresponding tothe frst overtone (Ghitd harmonic) ofan air coluran of length 75 cm inside 4 tube closed at one end, The siting also generates 4 beats per second when excited along with a tuning fork of frequency m, Now when the tension ofthe string is slightly increased, the number of beats reduces to 2 per second. Assuming the velocity of sound in ai 0 be 340 mus. the frequency w of the tuning fork in His a Md b. 336 ems . 1093 (IIT-JEE 2008) A hollow pipe of length 0.8 m is closed at one end. At its ‘open end a 0.5 m long uniform string is vibrating in its second harmonic and it resonates with the fundamental frequency of the pipe. If the tension in the site is 80 N and the speed of sound is 320 ms ' the mass ofthe string ase ble & We 4.402 (IT-JEE 2010) Two monatomic ideal gases | and 2 of molecular masses ‘mand is respectively. are enclosed in separate containers kept atthe same temperature. The ratio ofthe speed of ‘ound in gas Ito that in gas 2 is given by a b. Vim, em a. 2010) A police car withasiren of frequency 8H is moving with ‘uniform velocity 36 kav towards a tll building which reflects the sound waves, The speed of sound in airs 320 ‘avs. The frequency of the siren heard by the car driver is a. 850 kHy e775 kH b 828MM @. 750KH/ AITJEE 2011) 32. A student is performing the experiment of resonance column. The diameter of the column tube is 4 em. The frequency of the tuning fork is 12 Hz. Theatr temperature is 38°C in which the speed ot sound 1s 336 mis. The ‘ero of the meter scale coincides with the top end of the ‘evonance column tube, When the hist resonance occurs, the reading of the water level in the column is 140m b IS2am © I6tem . 176m AIT.JEE 2012) 1 using a resonance 33. A students performing an expe column and a tuning fork of frequency 2445 He is told thatthe air im the tube has been replaced by wther gas (assume thatthe column remains filled with the gas) The eight at which resonance occurs 1 (0.350 = 0.008 m. the gas inthe tbe is (Useful information: VI67RT = 640 J! mole VIGORT = 590 5° mole "The molar masses AF in ‘0 grams are given in the options. Take the values of «2 " si pt \w foreach Multiple Correct Answer Type 1A wave equation which gives the displacement along the Ysdireetion is given by v= 10-*sin 60" + 20) where cand {re in metres and 1s time in seconds. This represents & travelling with a velocity of 30 mvs in the negative direction b. of wavelength wm fe. of frequency Mn He 4. of amplitude 10 * my travelling along the negative direction. AIT-JEE 1981), 2. Anaiecolumn ina pipe. whichis closed atone end wll be tn resonance witha vibrating tuning fork of frequency 264 HL ifthe lengthof the column incmis\ Speedo sound = 330 ms) a 3125 b, 62.50 6. 9375 4.125 (IRIE 1988)3. The displacement of particle in a string stretched in the sedirection is represented by y. Among the following expressior® for y those describing wave motion are ber -a'r 4. cosk'x' 0") (IIT-JEE 1987) 4. Velocity of sound in air is 320 m/s. A pipe closed at one end has a length of | m. Neglecting end corrections. the aircolumn in the pipe can esonate for sound of frequency 80 He b, 240 He ec. 320He 4. 400 HZ (IIT-JEE 1989) 5. A wave is represented by the equation & coskxsinoot & cost(kr tw) in| 10x + 150 +2 los + 15a +2 where xis in metres and si in seconds. The expression represents: wave travelling in the positive x-direction with @ velocity 1.5 mvs. b. a wave travelling in the negative s-direction with @ velocity 15 nv, a wave travelling in the negative x-direction having a ‘wavelength 0.2 m, 4. 4 wave travelling inthe positive s-direction having a wavelength 0.2 m, (IIT-JEE 1990) 66. Two identical straight wires are sretched so as to produce 6 beats per second when vibrating simultaneously. O1 changing the tension slightly in one of them, the beat ‘frequency remains unchanged. Let 7, and T; represent the higher and the lower initial tensions inthe strings, While ‘making the above change in tension a. Twas decreased by T, was increased 6 Ty wasincreased dT, was decreased (UT-JEE 1991) A sound wave of frequency f travels horizontally to the right It is reflected from a large vertical plane surface moving to left with a speed v. The speed of sound in medium is 44, The number of waves striking the surface per second er) sy 1b. The wavelength of reflected ware is") feew) «rn The frequency ofthe retested wave is f=" a, ‘The number of beats heard by a stationary listener to of the left of the reflecting surface ATT.JEE 1995) 8. A wave disturbance in a medium is described by are in metre and ris in second. 10, 4 2. 14 18, A node occurs at x= 0.15 m ‘An antinode occurs at x= 0.3 m 6 The speed of wave is mis 4. The wave length is 0.2 m CIITJEE 1998), ‘The (x, ») coordinates of the comers of a square pate are (0, 0). (L. L) and (0, L). The edges of the plate are ‘clamped and transverse standing waves are set up init It tL. 3) denotes the displacement of the plate at point (x ¥) al some instant of time, the possible expression(s) for vis (are) (a= positive constant aL aco s/2L eon y/2L) 1b. asinirx/Lsin(x y/L) 6. asin(zx/L)sin(2x yl) 4d. acox(2ax/L)siniz VL) (UTE 1998) A transverse sinusoidal wave of amplitude a, wavelength Aand frequency fis travelling on a stretched sting. The ‘maximum speed of any point on the sting is 1/10, where Vis the speed of propagation of the wave. Ifa = 10° m id v= 10 mvs then A and fare given by mb 2=10'm e d.f=10' He (IIT-JEE 1998) Je (4. 1) = ON 40 + 5 + 5] represents a moving pulse, Where rand yare in mette and fin second, Then pulse is moving in + directo in 2 sit will travel adistance of 25 m (ity maximum displacement is 0.16 m 4. itisa symmetric pulse airy 1999) Ina wave motion v= asin(ks ~ dor), y-ean represent ‘a electtc field , magnetic field displacement 4. pressure (UIT-JEE 1999) Standing waves can be produced 4. on a string clamped at both the ends. ‘on a string clamped at one end and free at the other. ‘c. whem incident wave gets reflected from a wall. 4. when ro identical waves with a phase difference of ‘Fate moving inthe same direction, : UIE-JEE 1999) The function x= 4 sin“ar + B cos'an + Cin a cos oF represents SHM for which of the option(s) a. Forall value of A, Band C (except C= 0) 2B, amplitude = BV (AIT-JEE 2006) ‘A student performed an experiment to measure the speed ‘of sound in air using resonance air-column method. TWo Fesomances inthe ai-column were obtained by lowering the water level. The resonance with the shorter air-colummn isthe first resonance and that with the longer air-column is the second resonance, Then, 4 the intensity of the sound heard at the first resonance ‘was more than tht atthe second resonance,1. the prongs of the tuning fork were kept ina horizontal plane above the resonance tube. 6 the amplitude of vibration ofthe ends ofthe prongs is typically around 1 em. the length ofthe ar-column atthe first resonance was somewhat shorter than 1/Sth of the wavelength of the sound in ar. (IT-JEE 2009) 16, A person blows into open-end ofa long pipe. Asa result, high-pressure pulse of air travels down the pipe. When this pulse reaches the other end ofthe pipe. ‘4. aigh-pressure pulse stats travelling up the pipe. ifthe ‘other end ofthe pipe is open. 1 low-pressure pulse tats travelling up the pipe. if the other end ofthe pipe is open «& alow-pressure pulse starts travelling up the pipe i the other end of the pipe is closed 4. ahigh-pressue pulse tart travelling upthe pipe if the ctherend of the pipe isclosed— (IIT-JEE 2012) 17, Two vehicles, each moving with speed u on the same horizontal straight road are approaching each other, Wind ‘lows along the road with velocity w: One ofthese vehicles blows a whistle of frequency f. An observer in the other vehicle hears the frequency of the whistle to be fs. The speed of sound in still aris V. The correct statements) is are) ‘8 Ifthe wind blows from the source tothe observer. f:>f ‘Ifthe wind blows from the observer tothe source, f.> f, ‘If the wind blows from observer to the source, fs +=01m.002m,03m ‘Speed ofthe wave is piven by ye 2 0 option) Sms option c) option) 9. bc. Due tothe clamping of the square plate atthe edges Aisplacements along the x and waxes will individu ‘2210 atthe edges, Only the choices (b) and (c) predict these Aisplacements correctly. This is because sin 0~0. 1 € For a transverse sinoidal wave travelling ona string, the ‘maximum velocity is@us Also the maximum velocity is v 10 rdeims a=) 910 x2af 1 I FeO 2 The velocity »=/4 fa @ ‘We know that equation of moving pulse i yafice sn w ‘On comparing Eqs. i) and (i. we get vad m= 23 ms ameny Wave will uavel adistance of 25 min 2s, Beasbed. “The wavemoion y=a sin (x ~ at represents {a clectic fed in elecuomagntic ware (©) magnetic eld in electromagnetic wave (6) displacement in sound wave (a) pressure in sound wave, ‘Hence all the four options are comet. B.A, bye Standing waves are produced by the superposition of tw ‘of two identical waves travelling in opposite dtecton, Option 4) is accordingly not correct Option (a). b) and (c) are correct. 14. bd. Ford = Band C= 28 X= Bos 2or + B sin 2ox= 30a) 207 ‘This is equation of SHM of amplinde /38 W A= Band C=B, then X= B + Bsin 20 This is abo equation of SHM about the point X = 8. Function oscillates between X= O and X= 28 with amplitude 8 ad, (@) lovensity of the fundamental is % ‘more than that of the overtones “Therefore the 1 resonance was having more intensity () The prongs should not he in the horizontal postion but vertical ‘over the resonance tube (©) The amplitude of vibeaons are very small 1 be observed, (8) The antnodes are formed always 2 litle above the open end of the tube ‘This is called end correction Thiseffet will be there for overtones aso. ‘Length of the air column i es than 4/4, 16, hyd. At open end phase of pressure wave charge by 40 “compression return as rarefaction. While a closed end phase ‘of pressure wave does not change so compression rem as compression, be If wind blows fom source to observer fehl ‘When wind los fm obser onan source ni Views Vewnu) In both eases, f> fi acd. sede smnote 7 iS 10 ae ‘The fred end is node while the fee end is an aninode ‘Therefore, atx = 0 is a node and at r= 3 m isan antinode Possible modes of variation are =One bd where n=O, 1.2.3. 4a a2 Bnei nei pe2he 28 Oaths 2 Banh” 6 +130" 3 ‘We have stationary wave equation y= A sin kr cos =k = 100204) 0020+ = For n Linked Comprehension Type For Problems 1-3 Late da ‘4. Number of maxima in| is called the beat frequency. Hence, 2c, Speed of wave ve vuln O53" Da6x BRA reOyener = 2A 60596 soos 4 Frequency of cos (96 m function i 48 Hz and that of as (42) function is 2 Hz. In| sos function becomes zero at 2 times, where fs the frequency. Therefore ist funtion will become. ‘er a 96 times andthe second at 4 times, Bu second wil ot ‘overlap withthe ist Hence, net wil become zero 100 times it ts For Problems 4-6 4650.6 A.D gg = 304 20= 360 vs yg 0-30 310 0v5 Sa. For the passengers in uain A, there is no relative motion between source and observer, as bth are moving with velocity 20 ms. Therefore there is 8 change in observed frequencies, tnd correspondingly there is 80 change in their intensities, Therefore, the correct option i (s). (6.8. Forthe pasengersin tain 8, observer is receding with velocity 30 mvs and source is approaching with velocity 20 m/s Therefore, pea offices. f=" = Oe Matching Column Type 1 bravest ads Boh dsin be 1. Sound waves are longitudinal waves 4 deere ina 4 1, Sound waves are longitudinal waves Le iam Ai Sting waves ar transverse waves Integer Answer Type Le) . [E=10m » 0 ant Distance between the successive nodes = 2/2 = Sem 2.) pw 7,278, where cs speed of sound w rom Eqs. (i) and (i) d= 2 mvs = 7 keh, £4 Two waves have phase difference 92. 3Resultant amplitude A=24,cos30"= 43 Asle®? Resa nent = My Fill in the Blanks Type 1. y= Asin Lr placement amps = 4 (Oteximam dplcomen) Parc vel “ vat awcovorte) ‘Therefore, velocity amplitude = Aw = A(2nv) (Maximum velocity) Panicle acceleration, oat Accs = Au" sin(er x1) “Therefore, acceleration (maximum acceleration) amplitude = Aa = AQRS) 0 =20.05m 0 0" ‘The antinode will beat a distance of 125m Bceva and cn Tin fr Fava Vie ‘where T= tension in the string and m= mass per unit length of ‘When $0.7 kg mas is suspended for fundamental mode A= 21 (Soaxe Rous 0 ‘when mass is submerged in water, new tension T= weight ~upthmust = 507 g 0.0075 » 1000 + ¢ B22 w iw [BZ oy ay S2 wT V507 7 "¥507 = 260 =240 He 4. Letihe wave velocity i v toward ight (Displacement a= 0, «= 3) [Displacement at = 25,x=4 + v2) Wasa ae “05 ms ‘The negative sign indicates that wae is taelling towards left 5 InFig w iy 6 The first frequency the driver of bus hears is the orginal frequency 200 Hz The second frequency the driver hears is the frequency of sound reflected from the wall. The two Frequencies of sound heard bythe driver ae ‘Original frequency (200 H2) Hi, Frequency of sound refete fromm the wall") mr" | moe cee yes aa “The frequency of sound retlected fom the wall ses so 2225 ]ossso ve+ Frequency of beats 205.93- 200 593=6 He 7. w= 300 mvs and n=25 Fora pth difereace of & the phase diference is 2 Fora path diference of 6m, the phase diference is an x6 2 ‘The amplitude isthe same at both poins, hence amplitude Aiference is zero. =a True/False Type 1am er ae sn. Vail OTB _ (AB Sie Vuln ” GPE ~ V25 2 Fake. The intensiy of sound at a given poi isthe energy per second received bya unit area perpendicular othe direction of propagation. Also imensity varies as distance from he point source as = [Nove of te parameters are changing incase ofa clea night ‘oraclear day. Therefore the intensity il remain the same 3. True. Speed of sound waves in water is greater than that in ait So waters rae medi for sound waves 44 False. If he sound reaches the observer after being feflested from a tationary surface and the medium is also stationary. the image ofthe source will become the source of reflected sound. ‘Thus, in both the cases, one sound coming directly from the Source andthe other coming after reflection will ave the same Frequency (since velocity of source wrt, observer is same it ‘both the cases). Therefor, no beats wil be heard Subjective Type 1. ease of vibeuion in column, the anodes ath open ‘arose t smal ace ouside the open en. Tis smal ances Known send erection The alee of end conection 034 where disthe dane of te ke tn case of tube open at bth dst effecive length of tbe tircoben, sls 2ewheree = 03d e+ 2034) = d=130m=333m [Now one end is closed, the effective length of the robe Telse ire 320100 2° Fak 0.3x33~ Sone (6327 He ret Direction of ound Acoustic image ‘Sources the second reaches the observer alter being reflected from a Stationary surface andthe medium is also stationary, the image ofthe source inthe reflecting surface will become the source ofthe reflected sound, ev) = s0em Foc fintoverone of string Linear mass density of the sting = Mass of sting per unit 25x10" = SHO" 9, lenges = 2510" — 01 kgm fro. fr Vm 025\ 007 Fundamental frequency of sed organ is Yr c0dashp= 16m cee vf 320 5-9 -200m ws w In this case 8 beawsecond is heard. The beat frequency creases with the decreasing tension. This means that Bet frequency decreases with decreasing This means that beat frequency is given bythe eXPSSO0. =f, fp, o7 = [Larsxo2s2s2 5 = 2080255; os retornse =2700N 4, aera yf omer i 4-8 -221 oo1ign Very owe i F [= som lia Fondo gen . Aim, Booinehna2 ca ny i702 Since tuning forks is in resonance, therefore frequency of tuning fork i 400 He. The observer is hearing one beat per secon when the runing fork is moved aay wih a constant speed v. [1m —» ae ‘The frequency of tuning fork as heard by the observer standing stationary near sonometer wire an be found withthe help of Doppler effect. 400 He Dasston find Fundamental mode ‘ot bration sing 1 f wh 1 100%300 f° 50+ ¥ Since the beat frequency is | and asthe tuning fork is going away from the observer, ils apparent frequency should be 399 He, 99 = 00% 300 300+, 5. Tees ofwaein sigs pventy v= [F whee Tis G the tension and tis the mass per unit length, Since the tension inthe string will ncreae as we move up the string (asthe string has mass), therefore the velocity wil also imrene ws tenner arm (ore pace nga oe Since frequency remains the sae Yay Pam = 2% 0,06 =0.12 0 16 The frequency ofthe wire i fundamental mode 1 fF nays) ‘When temperature of wires changed. the tension developed T= YAa(a@) and smass per unit ength = Ap “) im, ¥=2% 10! Nim? oreo prey 1 2x10" ail 11 vibec 17. The given quatin for standing waves in the rng is =n (% cos 96m ‘ yetsin (£2 cos 6m) ‘“ i. The amplitude ofthe waves is given by Az 4 sin 15 “Therefore, the maximum displacement o x=Semis Vi = 21.78 or "Sarg (where r= 0.1.2.3...) 15 = £90,015 m,0.306m, ii, Ditferemiaxing Eq, i) with respect of, we get velocity of panicle = & 2-496 sn © )sin(96x0) Substinting += 7 Sem and = 025 seman sin( 2725) sn(96n 025) 2sin A cos B= sin (A + B) + sin A ~ B) Equation () may be expressed as 922 [anf Econo} asin aoe] ~rif EE oats sue ‘Therefore the component waves are given by 2sin{ 6m + )8. The apparent frequency from tuning fork T, ss heard by the on cbsene wile ® Decionctond 2 Dinan sued pap f ey yay fi -y ; sae? oats soe The apactfeueny om ning fk T ahead bythe observer aillbe f= fo) «i (ara) Given that beat frequency is 3. f, Was) Ga) joftsssesfese nes Asvomunence 22 asaoue 2 a1Smi two progressive waves having the same amplitude and time Petiod, but aveling in opposite direction sith same velocity superimpose, the produce tang waves ‘The following two equations qualily the above criteria and hence produce standing wave 2 cos kx cos or “The reslant itensity will be 2e0 when 24 cos Ax = 0 conde cos At, iy ‘The equations of transverse waves 24 =Acos an) 2) =A.os ty an) Combine to peoduce a wave waveling i the direction making ‘an angle of 45° withthe postive and positive yates “The esuliam waver 122A con dx — i + A cosy =) eatscnlltMool ll “The resultant intensity willbe zero when aa contlt=} kay 0 Mx) das Caen, 10.4, The frequency ofthe whistle as heatd by obserser om the ill =599H2 |. Let echo from the hil is heard by the driver at wish bs a 4 distance «from the hil. The sound was produced when the Source was at a distance 1k from il Ha “The time taken by the echo to reac from hill 1 ‘ 11300530) * (7300-99) Where yy = Hime taken by sound fom Ato H with velocity 1200 + 40) yg time taken by sound from H 10 B with veloc 0) From i) and i) ¢) #7 11200 40 ~ (1720090) (1200 — 30) Which gives v= 0.902 m “The frequency of eto as heatd by the diver ean be calelated by considering that the source is the acoustic mage. at Direction of wind eee rouge BS sso Theanine or (Seto nas 2K) ‘The angular frequency ofthe detector source is equal. ence thei time periods will also be the sae,a = When te detectors at C moving towards D the source sat A wing lft wards In this sittin that the Frequency heard Velocity of source v,= ant = 60 ms Velocity of detector ee (GH0=30) ‘Again when the detetor sat C moving wards B, the source isatA” moving rightward, In this situation thatthe Frequency hears masini + Af] a9] 7 vs eet ass Testud gun of plane popesive wave i cos 4207) ; The pris opens cotarin wo res ab= Since the wave rflectsby an obstacle, it will iffera phase dit ference of x. The intensity ofthe reflected wave is O.64 times ofthe incident wave, Intensity of incident wave 1 = A? Intensity of reacted wave f= 0.64 =F = a > 0641 «A 0644" A" = A= 08A the equation of resultant wave will be OR AcoMat +n) ==08 Acosars br) «iy wave equation can be found by superposition Acos ar b1)+ OBA cos ax bo}tiv) The particle velocity can be found by difleretiatng(¥) equa- a a -=-Ab sin (ax # 1+ 08 Ab sintar— = Ab sins + bt) O38 Absiniax br) 1 -Abjsin ax cos + cos at sn b+ 038 sin ax ‘608 bt = O8 os at nn br vp “ADIL sin ax cos br + 0.2 cos a in be ‘The manimum velocity wil cur when simati = 1 and cos bt "under these condition cos ax = O and sin 0 | Ab and Wal =O 4. y= [Aces (ax + bn} ~[0.8Ac0s ax bo} 1a 08 eos ax + 89 +024 cos ar +b ~(08.A cos (axon) (08. cos (ax +) #08. cos (ar +B} +02 Acos (axe ae ( 024 cos ar + bn] “ (6A sin asin B+ 0.24 cosa + bt) where (-1.6 4 sin a sn isthe equation of a standing wave and 0.2 cosa + isthe equation of traveling wave. ‘Aontinodes of the standing waves are the positions where the Sisplacement amplitude is maximo, =08. 4 +1) (arb) ie, sinaretasin{ne stor] (sf Lethe signal waves be given by sin 2ne\t,y2=A sin Joy. The resultant disturbance i given by co 2104 sin 2x EO Le = aa- ase Then, 72 0s mo} — 0) sin? Thus the esl dsrbance as ampli 24 os may ow For maxima: cos ma - ayy= 1 or moo) Cleaty ime interval between successive mania arose shet's oa "10 The reais en by Toa; DAs cos 8 When 8 0. inst s masiMU fy When = 22 sntemty f= 24° When 8 = intensity fg When = 322 imensiy = 24Wen = 28 ety ng =A The deccrtening in 2 2.0 or enh ue ow rd ne te Fels suto"s 1. tun he of ing i me fe hcg patton ante thie es ante The Tice te oe Because the distance between two consecutive nodes is (2/2) while between anode ad antinod i 2/8, hence “bebe « 04m Further, is given that x10! Nim? and p= 8 10*kgin* 0" “Bg 50 Hence, from v= ni, 5000-12500 He a 0s Now incident and reflected waves along the rod are M2 Asin(ar~ kx) andy, =A sin (ars ke + 9) “The resultant wave wil be vey, ype [sin (or ke) +sin ax + kx + 6)) jaler) Because there is an antinode a the free end of the rod, hence amplitude is maximum at = 0.0 2Acol r+ 3 con(tx042)= Maximum=1 ie. 0=0 And fun = 24: 10° m (given) y= 210% coskesino 10%e05 2 Jinan Putting values of And n, we gt 210 *eosS:rxsin25000xr Now. because (ora point 2 cm from the midpoint x = (0.50 # 0.02), hence 10 eosSn(0'5 20002) sin 250007 IS, The speed of whistle ra=1 5x20 30m" 6. Bs Prom toms ‘When the source i instamtancousl atthe positon A. then the frequency heard bythe observer wil be 300) = ssstte ' 30-30 ‘When the source is instantaneously atthe poston B then the ‘equency heard by the observer wil be a | i} ‘Hence the range of frequencies heard by the observer is 403.3 Ha to.48a Ho. Let and fsb lengths of open organ pipe and closed organ Pipe respectively i ee 0| 30-30 AO He Fist overtone of open ongan pipe First overtone of closed expan pipe = Aecrtng gun 2-3 w Asi; isthe fondamental frequency of closed organ pipe u ta 0075 m an S110 From Ea. (0). May, aes fame Taking positive sign Lazer 300 2 mm =0.993 m=99.3 cm 323" iw. ‘Taking negative sign 22= 3x11 faa, ws 1.006 m = 100.6 em17, Motors receives two sound waves one dretly from the sound source and other reflected from the fixed wall Let the apparent frequencies of these two waves a received by motorist ate and frespectvely For DirectSound: The apparent frequency eceived by motor Direction of oun fea ee wall For reflected sound: The frequency of reflected sound as ‘heard by the motorist canbe calculate by considering that the source i the acoustic image, Daeton of sound ve Acoustic image of source i . Hence, beat frequency as heard by the motorist (vem ven) FL Py ven} 18.4, The fundamental frequency ofthe closed ongan pipe =v In closes organ pipe only odd harmonics ae present. Second overtone of pipe = SS. Given Svat = 440 On solving, we get sx330 Detio * 440 “16 1. The equation of variation of pressure amplitude at any distance «from the node is AP = aP; cos kx Pressure variation is maximum at node and minimum (zero) at atinode m=0.9575 m=93.75 em Distance of centre C frm N15 8 aa Atantinode, the pressure variation s minimum (ero), therefore at antinode pressure remains equal to Py (always) “Therefore, a antin0de Pea = Prax vw. [Let m’ be the mass pe unit length then 1m} = 0.0125 kg/m, m'; = 0.078125 kg/m Wice POR is under a tension of 80 N = Tp. A sinusoidal wave pls is sent from P. 1. = velocity oF wave on PQ +E Vers “Tova ime taken for wave pulse to reach from Pto R 48256 2h aE. wR 1b Amplitude of reflected wave: ) +006 058-0145 “GR 32-80 3240, ‘Ais ve so reflected wave is invented ‘Amplitude of transmited wave [2 Joe(B2E sone sa+40, +H 20. Let be the length of air column corresponding tothe funda mental frequency 04m 7 a af f i 4225)" 4225)"a. closed pipe only odd harmonics are obtained. Now ltl! te. be the lengihs corresponding tothe 3rd harmonic, ‘Sth harmonic. Th harmonic ete Then 2m Sih harmonic s{f]=225-24-20m i, h mnemonic 9 S31,-36m a ‘or heights of water level are (360.4) m, (36 - 12), 6— 20) mand (3.6~ 2.8), ‘Therefore height of water level are 3.2 m, 2.4 m. 24m, 1.6m) and 08 m from bottom, [Let A and a be the area of cross-sections ofthe pipe and hole respectively. Then A= m2 x10) 21.26% 10% and a= (10? =3.1410°%m Velocity ofefMux, y= gm Continuity equation atop ofthe tbe and oritice / fat = a{ =H aH = a= ‘Therefore rate of fl of water level inthe pipe, ( Substituting the values, we get pul J=4 Joe Sei i AXIO" TOK 12610 S110 WH Between fist two resonances the water level falls trom 3.2 m 24 Jt = wie Situation 1 ‘Speed of sound inside the water ‘The frequency of sound inside the water vias ote MS ote ea 1445x107 In this case the medium through which the sound is travel ing is also in motion, So we will use the relation of Doppler fornia 2 [tel i. rol] ‘Apparent frequency, Again applying equation i) f oo eae) 0310" He Ma-3=10, 22. If Lis the length of each pipe A and B, then fundamental frequency of pipe A (open at both ends) Pipe a PpeB Piped Pipe vww 1 When pipe pipe B becomes closed at both ends, fundamental frequency of ” Using Eg. (i (ii) and (v), we get neve 8 oa 23, Fundamental frequency of air cola closed at one ends "se" 40030) Given = 480 Hz, D= Sem = 5% 107 m L=t6em= 16% 10 m vada (1+ 03D) = 448016 « 10° + 0.3 «5 « 10°] wis 43803175 10 mis = 336 mvs ‘The amplitude of vibration ata distance «from x Amasinks “Mechanical energy atx f length dis given by ae ‘Total energy ofthe sting Eafe 2g Tan ere! Letibe speed of the tain be v, (Case I The tain i approach Lette the natural frequency ofthe whistle, Using Doppler fect elation ‘Apparent frequency fh =) where», = Speed of sound = 30 mis (given) jee Ste a) ‘ 20-45 : cost titan 5) tee ante th gen vane ® Dee > 3300- 1yp = 2700-94, 600-205; =4 47 = 30m ‘The wave form of a transverse harmonic disturbance y = a sin(an kx = @) Given gn = 200 3s ‘a0? = 90 is? few Velocity af wave v= 20 mis ity 0 Dividing = ws30rmds ww ‘Sbstitating the value of win ) we get aI a= 2 n01m “ » Now k= wi Awe wy 3072 From (i) (9) ad he wave orm is ye otsie{su-2 3020] = (0.1 m)sinl 30 ras £1.50" + 6]