UNIT # 06 (PART - II)
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- Clemence McCarthy
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1 EXERCSE. Wae on urface of ater are combination of longitudinal and tranere ae. ( ). m, 5m/, T f 5 UNT # 6 (PRT - ) WVE THEORY. y in t JEE-Phyic ; y 5 in t co t in t co t / / : in t. 6 / m f 5 5 Number of ae () y.5 co(t x) f cont. f f/ t x y.5 co.5 co(t+x). f ae ec Hz ; 5m f m/ 5. y a in t y a co t a in (t + /) y lag y behind by phae. 6. y propagarte in +x-axi and y along e x-axi. (y y x- ) 7. y co kx in t [in (t + kx) + in(t kx)]. T T T T. and RL : R R 5 5 : R R 9 (Stationary Wae) () Node6\E : \Data\\Kota\JEE-danced\SMP\Phy\Solution\Unit 5 & 6\.Wae Motion.p65 y co (k +t) [ + co (kx + t)] t 8. y co in t co t (Progreie Wae) () in t in t in t + in t + in t + in999t in t + in t + in 999t x 9. y y in ft max y f ; ae f Gien (): max ae y (f) (f) x ; y (f) co ft y. 5. a a a a a co a a 9 max : min 6. Tenion kx 7. ' k(.5x) '.5. '. coeff of m/ coeff of k T N 69
2 JEE-Phyic 8. L TL Y T LY LY y L L 8. T f L ; d T Y d 7 m/ T f ' L f ' f 9. f f (m no. of loop in teel ire n no. of loop in aluminium ire) (m n ) m T n T L L m 8 n ;. f. 8.5 Hz. d t m m (minimum) () n 7.5Hz m T 8 f 6 L.6 78 RT. M H O M O V H O V O M air. f air f ater air 5.5 mm f 6 air. T xg gx gx. f f 5 L f L (ymmetrical about x) (x) f Hz. The right end ill hoot up on the ire. (). The equation repreent a progreie ae moing along x-axi of ingle frequecy. (x ). L 5 5 m f 5Hz 5. y re y + y' (at x ) a in (kx t) + y' put x and get y re 6. Ditance beteen poition haing node and antinode aelength. Å () 7. n n n n + + n n n n n n n n n n. f f f L f' L f' f 5. f f, 6. f f L L L L 6 L Gien that f f f ; f L (L) f L L Hz L 7. f Length of pipe in econd harmonic. () L 8. f (/ L) 6 L L. m Node6\E : \Data\\Kota\JEE-danced\SMP\Phy\Solution\Unit 5 & 6\.Wae Motion.p65 7
3 JEE-Phyic 9. m f. L /, /, 5/... 5cm, 75cm,5cm Minimum length of ater column 75 5cm () L +.6 r (cloed organ pipe) () L +. r (open organ pipe) () (L +.6 r ) (L +.r ) r r.5 L. f 5 5 ; f 56 5 f 6 min 8 min. f f > f ; f 6 Hz ; f f < f ; f 5 Hz. For onometer ire () n (n+) 95 n no of harmonic n 9 L L f 9 L 6 f L 56 Hz. f 5 f 5 7. L ; (L y) f f f y y (L y)l L 8. d log L y L log ( +.) d 9. d log log (taking antilog) 5. f f o ob log (oberer i approaching) () 5 5 Hz 5. f f app f o f o f o f.5 f o 8 m / 5. f f 5 5 Hz 5. f min f o R Hz Node6\E : \Data\\Kota\JEE-danced\SMP\Phy\Solution\Unit 5 & 6\.Wae Motion.p65 f f. 55 m/ f f 5 8 f Hz 6. C C f 56 Hz f 6 Hz (f 56) (f 6) f 5Hz (f 56) (6 f ) f 58 Hz 7
4 JEE-Phyic EXERCSE c. No. of ae triking the urface f f c () freqeuncy of the reflected ae () c c f' f c f c Waelength of the reflected ae 6. Let ae equation be () x t t z e t t, x + t x + t t t, x ( t ) x + m/ 7. y in (t kx + ) co (t kx + ) at t, x, y.5, > 6 y in t kx 6. c f ' c c f c. m Node occur at x.5 m,...; () 5,,.5 m, gt df g 8. f fo fo dt m/ 9. y mm in x t ntinode occur at x,,....m,.m,.m,...; () ae peed () 5. y at x. 5 m/ Thi can be tatified by the term in (in n x ) L n x L in x t n 8 t (n,,,,...) 5 n t min /. L x (at a ditance 'x' from free end) (x). f f x gx T dx(g g) L f 5Hz and f Hz For cloed tube () V L ' V V and L f ' f f ' (L ) ae xg T gx xg x L L d g a g/ dx (contant eeryhere) ( ) No S ut + at L + g 8 t t 8L / g Node6\E : \Data\\Kota\JEE-danced\SMP\Phy\Solution\Unit 5 & 6\.Wae Motion.p65 7
5 JEE-Phyic. Total energy () f n n f (f fundamental frequency ( ) f n Frequency of n th harmonic ()) <KE> <PE> TE. n the tationary ae, the particle in the alternate loop are out of phae... ( RT k k f, f L M L M L f f D k L, f k C L 8 k L, f C /f D 8 m x Pr. mp. P co P at x (x P x co P ) and P at x / P at x / Pipe may be cloed at x and open at x m. (xx m ). Let a initial amplitude due to S and S each. (a ) k(a ), here k i a contant ( k) fter reduction of poer of S, amplitude due to S.6a. (S S S.6a) Due to uperpoition () a max a +.6a.6a, and a min a.6a.a max / min (a max /a min ) (.6a/.a) 6 Comprehenion#. Point a. Point c,d,e. point b,f. max point,d,h 5. EXERCSE T xg xg d g xg g a dx Comprehenion#. n (n,,,,...) ( R) n R n R, R R R,,... Node6\E : \Data\\Kota\JEE-danced\SMP\Phy\Solution\Unit 5 & 6\.Wae Motion.p65 7. m / m. 5Hz Separation beteen and 6 m / () 8. Comparing ith the equation () y in n x L co(t) mm or mm n x n 6.8x x or L L m. For n, L.5 m... max (n ) ( R) (n ) (n,,,...) R R R R,,... n 5. max to produce maxima at D R (D max R) 5. max to produce minima at D R (D max R) 7
6 JEE-Phyic EXERCSE V(). h gt h t and h t g, t h t t + t 8.77 ec mm.8 mm 7. For ire, L L FL F Y Y 6 5 F. f L. Hz 8. T.5 m /.5 L t. ec. T T T L.5 m /, t.5 ec / S S 6m 6.m (n ) P for detructie interference () f(6. 6) (n+) (n ) f (n+) ( m/).8 Hz, Hz, Hz, 8 Hz, 6 Hz, Hz T m/ 8 f 56 Hz; 5cm.5 m Equation of the ae () y in (t kx) 6. V bottom V top.5 in (ft f x top f.5 in(69 t.8x) top T / T 8 / bottom bottom ) m/ m/ top.6.m. x y 5 in co(t) x x.5 in t.5 in t (i) (ii) (iii) Equation of incident ae () y.5 in x t Equation of reflected ae () x y.5in t 6cm Ditance beteen adjacent node cm () 5() in x in( t).5 9 in in air. m 6, ater 6.9 m. (i) (y max ) x5cm (ii) cm 5 x 5 in in 5 5 Poition of node ( ) 5cm, cm... Node6\E : \Data\\Kota\JEE-danced\SMP\Phy\Solution\Unit 5 & 6\.Wae Motion.p65 7
7 . 5. (iii) (i) (96) x in 5 in (96t) at x 7.5 cm, t.5 ec in 5 in 96 x Y in 5 co (96 t) x in 96t 5 + x in 96t 5 6 m / m f m/ () f f ' f eat frequency () f f 5 5 f 7.87 Hz. f oberer f 68 Hz JEE-Phyic f f 8 Hz S 6. f f 8Hz S.5 Node6\E : \Data\\Kota\JEE-danced\SMP\Phy\Solution\Unit 5 & 6\.Wae Motion.p65 6. Let the pipe reonate in n th & (n+) th harmonic (n(n+)) (n+) 9 n(59) n L 8. F tring > F pipe F tring T T L 96.5 m F pipe Hz. T T 8 T 7. N 9. f 5 5Hz (i) Total beat produced in (ii) (5) Time interal in hich max intenity become minimum ( ). ec f 5. The frequency of < frequency of (<) f f 5 7 f 5 f Hz. Frequency reaching the all f ' f () Frequency receied by the oberer f f.hz.5 S 5. f f f S S 6. Frequency receied by ubnarine 7. () f f f Frequency reflected by ubrnarine, () f f 5.9 khz gm kg cm.m S f S.5m/ S 5 5 kg/m, T 6 N T f L. m/ tring fork.7 m/ 75
8 JEE-Phyic 8. (i) f Hill f (ii) Hz km/h ound t t t hr here t time the ound to reach the hill (t ) Let t time for the echo to reach the train (t ) echo peed of echo () 6 km/hr train + echo d train t ( + 6) t t hr Ditance from the hill here echo reache the train () d (t + t ) Frequency reaching the hill () f ' f t Frequency of echo ().95 km. f max f EXERCSE V () Hz f min f Hz. co here ( )t t (i) (ii) For ucceie maxima () t 6.8 ec For detection of ound () co co ± t t,.57. Frequency reaching the all f ' f b () Frequency reaching the motorit after reflection from all ( ) t f f ' t f t 58 6 Hz 6 9. f guard f o f o m f f ' f Frequency directly reaching the motorit () f f m b eat frequency ( ) b f f f f ( + m ) b m b Node6\E : \Data\\Kota\JEE-danced\SMP\Phy\Solution\Unit 5 & 6\.Wae Motion.p65 76
9 JEE-Phyic. (i) For the particle P (P) 6. For air () (ii) y y t x ( ) cm/ (along negatie x-axi) (x-) Equation of ae () y in (t + kx + ) at t, x, y, L L f 5 (L ) L cm For CO (CO ) L, 6 L f 6 5 (L ) L. cm. in, cm f cm / 5Hz cm x y in t (iii) Energy carried in one aelength () E 5 ( ) ( ) 6 5 J 7. (i).65 m, d f.65 t infinity, path difference (). the man approache, the path difference change a (),,,, Hence only minima ill appear to the man. () (ii)for d x x and d x x x 9.8 m, x.99 m Node6\E : \Data\\Kota\JEE-danced\SMP\Phy\Solution\Unit 5 & 6\.Wae Motion.p65 5. mplitude of reflected ae () k k r i k k mplitude of tranmitted ae () t k k k i Equation of reflected ae () y r 6.67 co (x + 5 t) Equation of tranmited ae () y t.67 co (x 5t) 9. x cm (i) Total no of aelength L 5 () Number of loop formed 5 () (ii) Maximum diplacement at x 5 (x) (iii) y 6 in 5 6 in cm x co ( t) 6 in ( x) co ( t) 6 in( x) in ( t) m/ KE max dx 6 in( x) dx 77
10 JEE-Phyic here T. KE max 6J x (i) y 6 in co (t) y + y x y in t, y in x t. (i) Combination of ae producing tanding ae : Z + Z (). (ii) (iii) Combination of ae prodcuing a ae traelling along x y line :Z + Z (xy ) Poition of node in cae (i)x (n + ) k ((i)) cae (ii) x y (n+) k Y 5 m / 8 5 L L. m 5 5 k 5. EXERCSE V-. The fundamental frequency for an open pipe i () f ; f open ; f cloed fopen f f f cloed open pipe : l cloe pipe. the tring i tied beteen to rigid upport hence there ill be node at both end. The longet aelength for node at both end ill be the one for hich) ( ) cm l f 5,5. (i) Equation of the ae () 6 co(5x) in(5, t) (ii) y 6 in (5t 5x) y 6 in (5 t + 5x). mplitude after reflection () k k r i k k mplitude after tranmiion () k t i k k cm 5. longet 8 cm y in 6t x On comparing the gien equation ith the general equation of ae, e get ( ) y y in t kx 6 ; k Wae peed () 6 k m/ 6. The frequency of the ibrating tring ith repect to tuning fork i either (56+5) Hz or (56 5) Hz ( (56+5) Hz (56 5) Hz) f ire l T Node6\E : \Data\\Kota\JEE-danced\SMP\Phy\Solution\Unit 5 & 6\.Wae Motion.p65 78
11 JEE-Phyic Node6\E : \Data\\Kota\JEE-danced\SMP\Phy\Solution\Unit 5 & 6\.Wae Motion.p65 On increaing tenion; the beat frequency decreae to Hz, o probable frequency of the ire ith repect to fork none i ( Hz ) ut on increaing the tenion of ire, the frequency of the ire mut hae to increae. So, if the original frequency of the ire i aumed to be 6 then it reduce to 58 herea if it i aumed to be 5 it ha increaed to 5. e ere expecting increae, o the correct frequency of the piano ire i (56 5) Hz. ( Hz5 (56 5) Hz) 7. f the frequency of fork i Hz then probable frequencie of fork i either 96 Hz or Hz. on attaching ome tape on fork, be at frequency increae, thi i poible only if the frequency of fork i 96 Hz. ( Hz 96 Hz Hz 96 Hz) 8. Gien that () oberer ound 5 pplying Doppler' effect, e get () f f ; 6 / 5 6 f f 5 f f 6 % f n' n (95) ; f 6 f m. ntenity change in decibel ( ). log log RT RT n, xn, x T x M M. y.5 co (x - t) comparing the equation ith the tandard form, () x t y co T / and /T /.8 5. and. C. n' + eteen & b/ b/ & C b/ C & 9 n n b/ from (iii) eq n. of motion u + a 5. y. in T + a 9 a n n T m.5. t x..5 k. T...5 a 98 m. 6.5 N 79
12 JEE-Phyic 6. y(x t) [ ax bt] e EXERCSE V- /K b a in e x direction.. Ma per unit length of the tring, () 7. y (x, t) a in (t kx) y (x, t) a in (t kx) ut ntenity () n a a n a n ntenity depend on frequency and amplitude So tatement- i true tatement- i fale ( ) 8. y in (t kx) & y in (t + kx) y uperpoition principle () y y + y in (t kx) + in (t + kx) in t co kx mplitude () co kx t node diplacement i minimum () co kx co kx kx (n ) x (n ) x (n ) here n,,... m.5 kg/m. Velocity of ae in the tring. () T.6 m.5 8m/ For contructie interference beteen ucceie pule () t min ()(.). 8 (fter to reflection, the ae pule i in ame phae a it a produced, ince in one reflection it phae change by, and if at thi moment next identical pule i produced, then contructie interference ill be obatined.) ( ). f f f f f 6 9. n n c ( / ). Fundamental frequency () V T y train S f.5 eq S (S cro ection rea) (). 78. Hz 7.7 and f f f 7. Fundamental frequency i gien by () f 9 f 6 8 T (ith both the end fixed) () Fundamental frequency ( ) [for ame tenion in both tring] [] here ma per unit length of ire. (.) ( denity ()) (r ) r r r L r r L r Node6\E : \Data\\Kota\JEE-danced\SMP\Phy\Solution\Unit 5 & 6\.Wae Motion.p65 8
13 JEE-Phyic Node6\E : \Data\\Kota\JEE-danced\SMP\Phy\Solution\Unit 5 & 6\.Wae Motion.p65. Energy E (amplitude) (frequency) ( E () () mplitude () i ame in both the cae, but frequency, in the econd cae i to time the frequency () in the firt cae. ( () () Therefore () E E 5. fter to econd both the pule ill moe cm toard each other. So, by their uperpoition, the reultant diplacement at eery point ill be zero. Therefore, total energy ill be purely in the form of kinetic. Half of the particle ill be moing upard and half of donard. C C ( m ) 6. Uing the formula () e get, and Here () f ' f (i) 6. 5 peed of ound () peed of train ( ) peed of train () Soling equation (i) and (ii) 7. Let f frequency of tuning fork. (f )...(ii) 5 9g f Mg ( ma per unit length of ire) ( ) Soling thi, e get M 5 kg n the firt cae frequency correpond to fifth harmonic hile in the econd cae it correpond to third harmonic. ( ) 8. The motorcyclit obere no beat. So, the apparent frequency obered by him from the to ource mut be equal. ( ) f f Soling thi equation, e get m/ 9. Let be the end correction. () Gien that, fundamental tone for a length.m firt oertone for the length.5m. (..5 m ) (. ) (.5 ) Soling thi equation () e get.5m.5 m. The frequency i a characteritic of ource. t i independent of the medium. ( ). f c f (both firt oertone) c L L L c a. The frequency i a characteritic of ource. t i independent of the medium.( ). f (nd harmonic of open pipe) f n () (n th harmonic of cloed pipe) (n) Here, n i odd and f > f (nf >f ) t i poible hen n5 becaue ith n 5 8
14 JEE-Phyic n5 n 5 f 5 5 f. The quetion i incomplete, a peed of ound i not gien. Let u aume peed of ound a m/. Then, method ill be a under. ( m/) M C Q. Since, the edge are clamped, diplacement of the edge u(x,y) for ( u(x,y) ) y (,L) C (L,L) (6..7)cm or.65 m peed of ound obered () f m/ Error i calculting elocity of ound ().8m/ 8 cm/ 5. f T f f CD T T CD...(i) Further p T (x) T CD ( x) x x (a T T CD ) x /5 6. Take yin(t kx) 7. o P y t y co(t kx) P y ( ).5 T L m P y () (5) here m, 75 cm.75 m No according to gien condition n T L m o n T L m + Hz..75 x m 5 8 O x (,) (L,) Line O i.e. y < x < L i.e. x L C i.e. y L OC i.e. x < y < L < x < L < y < L The aboe condition are atified only i nalternatie (b) and (c). ( (b)(c)) Note that u(x,y), for all four alue e.g. in alternatie (d), u(x,y) for y, yl but it i not zero for x or xl, Similarly in option (a) u(x,y) at xl, yl but it i not zero for x or y hile in option (b) and (c), u(x,y) for x, y x L and yl. ( u(x,y) (d) y, yl u(x,y) x xl (a) xl, yl u(x,y) x y (b) (c) x, y x L yl u(x,y). Maximum peed of any point on the tring a () a(f) (Gien : m/) af f f a m (Gien) a Hz Speed of ae () f ( m/) m Node6\E : \Data\\Kota\JEE-danced\SMP\Phy\Solution\Unit 5 & 6\.Wae Motion.p65
15 JEE-Phyic. For a plane ae intenity (energy croing per unit area per unit time) i contant at all point. ( Speed of pule ( ) t t and x.5 m y y.6m.6m ut for a pherical ae, intenity at a ditance r from a point ource of poer P (energy tranmitted per unit time) i gien by : (Pr ) S S x x.5 m t (b) x t alue of y i again.6 m, i.e., pule ha traelled a ditance of.5 m in in negatie x- direction or e can ay that the peed of pule i.5 m/ and it i traelling in negatie x-direciton. Therefore, it ill trael a ditance of.5m in. The aboe tatement can be better undertood from figure (b). (t x.5 my.6 m x.5m.5 m/ x-.5 m(b) ) x P or r. The hape of pule at x and t ould be a hon, in figure (a). (xt(a) ) r 5. n cae of ound ae, y can repreent preure and diplacement, hile in cae of an electromagnetic ae it repreent electric and magnetic field. ( y ) Node6\E : \Data\\Kota\JEE-danced\SMP\Phy\Solution\Unit 5 & 6\.Wae Motion.p65 y(,).8 5.6m y x x t (a).6m From the figure it i clear that y max.6m Pule ill be ymmetric (Symmetry i checked about y max ) if at t ( y max.6m y max t y(x) y( x) From the gien equation and ().8 y(x) 6x 5 at t y(x) y( x).8 y( x) 6x 5 Therefore, pule i ymmetric.() 8 6. Standing ae can be produced only hen to imilar type of ae (ame frequency and peed, but amplitude may be different) trael in oppoite direction. ( ) Compreheni on#. n one econd number of maxima i called the beat frequency. ( ) Hence, f b f f. Speed of ae () 9 Hz R.5 9 m/.6
16 JEE-Phyic. t x, yy + y co 96t co t Frequency of co (96t) function i 8Hz and that of co (f) function i Hz. ( co (96t) 8Hz co (f) Hz) n one econd co function become zero at f time, here f i the frequency. Therefore, firt function ill become zero at 96 time and the econd at time. ut econd ill not oerlap ith firt. Hence, net y ill become zero time in. ( coff 96 y ) Compreheni on#. S + 6 m/ S m/ SUJECTVE. (i) Frequency of econd oertone of the cloed pipe () 5 L L L 5 5 m x; P xx; P± P in Kx Subtituting peed of ound in air m/ ( m) L 5 5 m 6 m/ m/ m/ m/. For the paenger in train, there i no relatie motion beteen ource and oberer, a both are moing ith elocity m/. Therefore, there i no change in obered frequencie and correpondingly there i no change in their intenitie. ( m/ ). For the paenger in train, oberer i receding ith elocity m/ and ource i approaching ith elocity m/. ( m/ m/) ' and f 8 775Hz ' f 85Hz Spread of frequency () f ' f ' Hz 5 L 6 m 5 5 (ii) Open end i diplacement antinode. Therefore, it ould be a preure node or at x; P ( x P) Preure amplitude at xx, (xx ) can be ritten a P ± P in kx here 8 k m / Therefore, preure amplitude at () L 5 / 6 x m or (5/) m ill be P ± P in 8 5 ± P in 5 P P ± (iii) Open end i a preure node i.e. P (P) Hence, P max P min Mean preure (P ) ( P max P min (P )) Node6\E : \Data\\Kota\JEE-danced\SMP\Phy\Solution\Unit 5 & 6\.Wae Motion.p65 8
17 JEE-Phyic Node6\E : \Data\\Kota\JEE-danced\SMP\Phy\Solution\Unit 5 & 6\.Wae Motion.p65 (i) Cloed end i a diplacement node or preure antinode.( ) Therefore, P max P + P and P min P P. mplitude of incident ae () i.5 cm Tenion () T 8N P Q R L.8m L.56m Ma.6 kg Ma. kg Ma per unit length of ire PQ i (PQ) m.6 kg/m.8 8 and ma per unit length of ire QR i (QR) m. kg / m.56.8 (i) Speed of ae in ire PQ i (PQ ) T 8 8m/ m / 8 and peed of ae in ire QR i (QR ) T 8 m/ m / 8 Time taken by the ae pule to reach from P to R i (PR) t V V 8 t. (ii) The expreion for reflected and tranmitted amplitude ( r and t ) in term of, and i are a follo :( ( r t ) i r i and t i Subtituting the alue, e get () r 8 8 (.5).5 cm i.e., the amplitude of reflected ae ill be.5cm. Negatie ign of r indicate that there ill be a phae change of in reflected ae. Similarly. (.5 cm r ) t (.5).cm 8 i.e., the amplitude of tranmitted ae ill be.cm (. cm). Speed of ound () m/ Let be the length of air column correponding to the fundamental frequency. ( ) Then.5.m.m.m.m (.5). m (.5).m.6m.8m.8m n cloed pipe only odd harmonic are obtained. No let,,, etc., be the length correponding to the rd harmonic, th harmonic, 7th harmonic etc. Then, (,,.5.m 5.5. m and 7.5.8m 9 or height of ater leel are ().5.6m 85
18 JEE-Phyic (.6.)m, (.6.)m, (.6.) m,(.6.8)m. Height of ater leel are ().m,.m,.6m and.8m. Let and a be the area of cro-ection of the pipe and hole repectiely. Then (a). Velocity of ound in ater i () m / Frequency of ound in ater ill be ( ) 5.5 f Hz f 5 hz (i) Frequency of ound detected by receier (oberer) at ret ould be ( ) ( ).6 m and a ( ). 6 m a Source f m/ Oberer (t ret) m/ r Velocity of efflux () Continuity equaiton at and gie : () dh a gh dt Rate of fall of ater leel in the pipe. ( ) dh a dt gh gh Subtituting the alue e get () 6 dh. H dt.6 f r f r f.69 5 Hz 5 5 ( ) Hz 5 (ii) Velocity of ound in air i () a RT M (.)(8.)( 7) 8.8 m/ Frequency doe not depend on the medium. Therefore, frequency in air i alo f 5 Hz ( f 5 Hz) Frequency of ound detected by receier (oberer) in air ould be ( ) dh (. ) H dt eteen firt to reonance, the ater leel fall from.m to.m. (. m. m) dh.. H (. )dt dh (. H ) dt [.. ] (. ).t t t f a 5 f 5 a Hz 5 f. 5 Hz 5. (i) Frequency of econd harmonic in pipe frequency of third harmonic in pipe ((i) ) (a ) M RT RT M Node6\E : \Data\\Kota\JEE-danced\SMP\Phy\Solution\Unit 5 & 6\.Wae Motion.p65 86
19 JEE-Phyic M M M 6 M 9 (a T T ) 5 7 and 5 5 / 6 7 / 5 9 M 5 6 M 9 89 (ii) Ratio of fundamental frequency in pipe and in pipe i : ( ) f / f / RT M RT M Subtituting a (a ) M. M (a T T ) M 89 from part (i), e get M f 5 89 f 6. Fundamental frequency () f (.6r).6r Here, de f a in kx dx f T ( ) and k Subtituting thee alue in equation (i) and integrating it from x to x, e get total energy of tring ((i) xx ) 8. From the relation e hae and f ' f. f T.8 f T E a T...(i)...(ii) Here, T elocity of ource/train Soling equation (i) and (ii), e get T m/ 9. Maximum particle elocity ( ) m/ Maximum particle acceleration () 9m/ Velocity of ae ( ) k m/...(i)...(ii)...(iii) From equation (i), (ii) and (iii), e get rad/.m and k5m Equation of aeform hould be () y in(t + kx + ) y(.m) in [( rad/)t + (.5m )x + ] Node6\E : \Data\\Kota\JEE-danced\SMP\Phy\Solution\Unit 5 & 6\.Wae Motion.p65 7. Speed of ound () f (+.6r) () (8) [(.6) + (.6) (.5)] 6m/, k The amplitude at a ditance x from x i gien by a in kx (xx ainkx) \\\\\\\\\\\\\\\\\ a x x \\\\\\\\\\\\\\\\\ Total mechanical energy at x of length dx i (dxx) de (dm) (dx) (a in kx) (f). L cm; m gm m L gm / cm kg / m ; T.5 N g / m.5 m/; f Hz m f m 5 cm 87
Physics 11 HW #9 Solutions
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