Phương pháp kỹ thuật giải nhanh bài tập trắc nghiệm Vật lí 12 (Tập 1): Phần 2

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Nối tiếp nội dung phần 1 tài liệu Kỹ thuật giải nhanh bài tập trắc nghiệm Vật lí 12 (Tập 1), phần 2 giới thiệu các chuyên đề: Sóng cơ, dao động và sóng điện từ, dòng điện xoay chiều. Mời các bạn cùng tham khảo nội dung chi tiết.

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Nội dung Text: Phương pháp kỹ thuật giải nhanh bài tập trắc nghiệm Vật lí 12 (Tập 1): Phần 2

I tnuat gtat ttnann m iisi vat L t i^, tap I - Nguyen Quang Lac Vai may FX570-ES ta bam n h u sau : A? + A^ + A^ => A2 = - A ^ = V42 -2^ = 2V3 (cm) ,y [ M O D E J § t r e n m a n h i n h x u a t h i # n c h u : CMPLX Mat khac: A^ = A^ + A^ + 2A^Aj cos(92 - 9]) |SHIFTJ|MOD£| § chuyen doi don vi do goc la R (Rad): n 71 fi! |4>/3|^HIFTJ 0. § g § ISHIFTJ 0 H • • j h li jisv'v' :jr. o 2^ =4^+(2V3)%2.4.2N/3COS [SHIFT] 0 § 0 '^^'^ ^^^^ '^^t 4 Z 7t/2 I 3; 571 Nghia la bien do bang A = 4; pha ban dau cf) = n/2 cos •cp = 0 6J 2 '.....VP. => Chon A Vay A2 = 2\/3(cm); (p = 0 => Chpn D Cau 146. Mot vat thuc hi^n dong thoi hai dao dpng dieu hoa cung phuong c6 0ii_148. M o t vat thuc hien dong thoi 2 dao dong c6 phuung trinh la phuong trinh dao dong la X j = 4 s i n ( 7 t t + a) (cm), = 4\/3cos7ct (cm). Bien do dao dong tong hop dat gia tri nho nha't khi X j = 4^2 sin27it(cm); X2 = 4>/2 cos27it (cm). Ket luan nao sau day la sai? , > 371 A . Bien dp cua dao dpng tong hpp la A = 8cm iML.'''i^..\ A. a = C.a = 0 D. a = T: 2 B. Tan so ciia goc dao dpnh tong hpp co = 27t rad/s ,. ; Huong dan gidi: C. Pha ban dau cua dao dong tong hpp 9 = — Ta c6: x, = 4 7 2 cos Ttt + a — (cm) va X j =4\/2cos(27tt) (cm). 4 2 D. Phuong trinh cua dao dpng tong hpp x = 8cos cm Bien dp dao dong tong hop xac djnh: 4 n Huong dan gidi: A^ = A j + A j + 2 A j A j cos((p2 -(pj) = A j + A j +2AjA2COS a — 2 Taco: x, = 4 V 2 s i n 2 7 i t ( c m ) va X j = 4\/2cos27tt (cm) A De bien do dao dong dat gia tri nho nha't thi Vi h a i d a o d p n g t h a n h phan deu co tan so goc co = 27irad/s nen d a o d p n g tong hpp cOng co tan so goc (a = 2n rad/s dap an B d i i n g . cos a — = - l = > a - — = 7t=:i>a = — ^ Chpn A 2 2 2 Bien d p c i i a d a o d p n g tong hpp x a c d j n h : Cau 147. Mot chat diem tham gia dong thoi hai dao dpng dieu hoa tren cung 2 2 2 A 2 - A ^ + A ^ + 2 A J A 2 C O S ( 9 2 - 9 I ) = (4V2) + ( 4 V 2 ) + 2 ( 4 7 2 ) . c o s ^ mot true Ox c6 phuong trinh: xi = 4cos cot-^ cm; X2 = A2cos(o)t + (p2)cni 3j => A = 8cm => Dap an A diing. Pha b a n d a u c u a d a o d p n g tong hpp d u p e x a c d i n h : Phuong trinh dao dong tong hop x = 2cos((ot + (p)cm. Trong do cpj -(p = -- A , sin(p, + A , sin(p, tgcp = — • — = -1: .(p = — dap an C diing. jf'. Cap gia tri nao cua A2 va cp sau day la dung? A^ C O S ( p j + A2 COS(p2 Phuong trinh dao dpng dieu hoa tong hpp co bieu thuc: V5 / A. sVScm; - B. 2V3cm; - C. sVScm; 0 D. 2V3cm;0 2 4 Huong dan gidi: ^T^^-> X = 8cos 27rt-i^ c m d a p a n D s a i -> C h p n D 4 Bieu dien cac dao dong tren gian do vec to, V o i may FX570-ES ta bam n h u sau : 7-. ta c6: OA = A i = 4cm; OB = A2;. OC = A = 2cm P O D E ] ^ tren man hinh xuat hien chu: CMPLX T u gian do vec to suy ra: | H I F T ] [ M O D E J § chuyen doi don vi do goc la R (Rad): BC2 = O C + OB2 f72 SHIFT (-) -7t/2 SHIFT (-) 0 161
- SHIFT] y g y M a n h i n h hien t h i ket qua: 8 Z - 7 t / 4 A=(V3B)^-4(52-25)>0 B = - H / 4 A d i i n g ; B d i i n g ; C d i i n g ; D sai. ^ C h o n D . Bmax = lO(cm) thay vao (1) ta dupe A = S^/Scm => C h p n A C a u 149. M g t vat t h a m gia d o n g t h a i hai dao d o n g c i i n g p h u o n g , cung tan so M p t con lac 16 xo t h a m gia d o n g t h o i 2 dao d p n g d i e u hoa c i i n g C O bien d o Ian lug-t la A i = 3cm va A 2 = 4cm. Bien do dao d g n g t o n g h o p khorig phuong, c i i n g tan so' co = 5\f2 rad/s, c6 d p I f ch pha bang ^ . Bien d p ciia 2 dao the nhan gia t r i nao sau day: A . 5,7cm B. 1,0cm C. 7,5cm D . 5,0cm dpng thanh p h a n la A i = 4 c m va A 2 . Biet d p I o n van toe ciia v a t tai t h o i d i e m Huong dan gidi: (Jpng nang bang the nang la 40( cm). Bien dp thanh p h a n A 2 bang Theo p h u o n g phap gian d o vec to Fre-nen t h i bien d o dao d p n g tong h o p A. A2 = 4 cm B. A2 = 4V3 cm C. A2 = \/3 c m D. A2 = 6 cm l u o n phai thoa m a n : | A j - A j | < A < A j + A j . Theo bai ra, ta c 6 : 1 c m < A < 7cm. Huang dan gidi: Vay A k h o n g the bang 7,5cm => Chpn C Ta C O co nang ciia vat: E = W j + W, = ^ k A ^ C a u 150. M o t vat t h y c hien d o n g t h o i hai dao d o n g d i e u hoa c i i n g phuong, K h i d p n g nang bang the nang, ta co: c i m g tan so' i = 4 H z , c i i n g bien d o 5 cm va c6 d o lech pha A(p = —. Cho 7t^ = 10. E = 2W^ = i k A 2 = > 2 - m v 2 = i k A 2 d 2 2 2 Van to'c ciia vat k h i no c6 gia toe 32>/2 (cm/s^) la A . ±307c cm/s B. ± 2 0 ; : cm/s C. ± 3 0 ^ 2 cm/s D . ±407i cm/s 2v^ A^=2-- • A = 8cm Huong dan gidi: Ta c6: co = 27:f = 871 (rad/s) M a t khac ta co: A^ = A^ + A^ + 2A1A2 COS(A(P) = 75 => A = 5 V 3 c m A ^ = A J + A ^ + 2A1A2 C O S ( A 9 = A ^ + A ^ ) A p d y n g cong thuc: ^ Al=A^-Al=8^-A^=>A2=4Scm ,2 2 v2 a' => C h p n B V = ±4071 ( c m / s) Chpn D (0 CO CO C a u 153. Cho hai dao d p n g d i e u hoa c i m g p h u o n g : X j = 2 c o s ( 4 t + (pj)cm; ^ n C a u 151. Cho h a i dao d p n g dieu hoa c i i n g p h u o n g : X j = A c o s a->t + — cm vn = 2 c o s ( 4 t + (p2)cm v o i 0 < (P2 -(Pj ^ n. Biet p h u o n g t r i n h dao d p n g tong 3 hop x = 2cos c m . Gia t r j ciia cpt la: X 2 = Bcos c o t - ^ cm (t d o bang giay). Bie't p h u o n g t r i n h dao d p n g tong hop 6 2 la X = 5cos(cot + (p)(cm). Bien dp dao dpng B c6 gia tri cue dai k h i A bSng: B. 6 66 ' 2 A . 573cm B. 5 cm C. Syflcm D . 2,5V2cm 'ff': Huang dan gidi: , Hwang dan gidi: Gian d o vecto n h u h i n h ve. Ox la tri^c goc: Tir cong thiic tong h o p dao dpng ta c6: OA = O A i + O A 2 571 Theo bai: O A i = O A 2 = 2 (cm) 5^ = A ^ + B^ - 2 A B C O S — hay A ^ - VSBA + B^ - 2 5 = 0 (1) 6 => T a m giac O A A 2 can tai O => O A A 2 = O A 2 A De p h u o n g t r i n h (1) c6 n g h i ^ m A v o i B la t h a m so t h i : O A la p h a n giac A j O A 2 ( V i O A 2 A A 1 la h i n h thoi).
AOA2 = AOAj = OAA2 Q 2- Thai gian chuyen dpng thling cua v|t m t u liic ban dau den v i tri 16 xo l^'hong bien dang la: => A O A , = O A A 2 = O A 2 A A. B.^s. C.^s. D. ^- s. A A O A 2 deu ^ A O A 2 = - =i> AOA, = - 25S 20 • 15 30 2 3 ' 3 Huong dan gidi: 7t ', Theo bai: AOx = — => A^Ox = — =>(p, = - — Chpn B Vi tri can bang ciia con lac 16 xo each v i tri 16 xo khong bien dang x: 6 ^ 6 ^ 6 kx = fjmg => X = |Limg/k = 2 (cm). Cau 154. Mgt vat tham gia dong thai hai dao dong dieu hoa cung phuang v6, Chu ki dao dpng: T = 2 7 t J ^ - 0,2n (s) "^'•> •- Xj =4cos^5>/2t-njcm va X 2 = A 2 C O S 5\ll\ — c m , trong do t tinh bang Thai gian chuyen' dpng thSng ciia vat m t u liic ban dau den v i tri 16 xo giay. Biet dp Ian van toe ciia vat tai thoi diem dpng nang bang the nang ij 40cm/s. Hay xac djnh bien do dao dpng thanh phan Aj ? khong bien dang la: t = j + - ^ = -^(s) => Chpn C ' ^ ; A. VScm , ,^ B. 2V3cm C. 373cm D. 4V3cm rSu 157. * Mot con lac 16 xo gom mot vat nho kho'i lupng 0,02kg va c6 dp cung Huang dan gidi: 1 N/m. Vat nho dupe dat tren gia do c6 dinh nam ngang dpe theo trvic 16 xo. Khi dpng nang bang the nang ta c6: 2Ed = E H? so ma sat trupt giiJa gia da va vat nho la 0,1. Ban dau g i u vat 6 v i t r i 16 xo bj nen 10 cm roi buong nh^ de con lac dao dpng tat dan. Lay g = lOm/s^. Toe dp 2imv2=ikA2 Ion nhat vat nho dat dupe trong qua trinh dao dpng la A. 1572cm/s B. 25V2em/s C. 40^^ cm/s D. 50v^em/s • A = 8cm k Huang dan gidi: Do hai dao dpng thanh phan vuong pha nen Van toe ciia vat dat gia trj Ion nhat trong qua trinh dao dpng tat dan ehinl" A 2 = ^A^ - A j = 78^-4^ = 4V3cm => Chpn D v$n toe Ian nhat ma vat dat dupe trong i chu ki dau tien. Gpi x la v i tri l i do Cau 155. Mot con lac 16 xo dao dpng tat dan. Cu sau moi chu ki bien dp ciia no ma vat c6 toe dp Ion nhat. giam 2,5%. Phan nang lupng cua con lac bj mat d i sau moi dao dpng toan Ap dung dinh luat bao toan nang lupng, nang lupng ban dau ciia h§ gom the phan la nSng cue dai ban dau: ~ k A p , nang lupng ciia h§ tai thoi diem vat c6 van toe c i . ; A. 35% ^ B. 45% C. 55% D.95% ' '^ Huang dan gidi: ?i gom dpng nang ^ m v ^ , the nang ^^x^ va dp Ion eong ciia lye ma sat: 1 2 2 = ^mg Al = |img(Al + x) Nang lupng trong chu ki dau la: W = - mco A 1 kx^ +)img(Al + x) -kAl^ =-mv^ + — Nang lupng trong chu ki tiep theo la: W ^mco^A'^ A/ . 2 ^ J i A l 2 - A x 2 - 2 ^ g ( A l + x) Suy ra: = (0,975)^ * 0,95 = 95% =^ Chpn D m w A m V / Cau 156. * Mot con IMc 16 xo c6 dp ciing k = 10 (N/m), khoi lupng vat nan? Dat y = v 2 = — A l ^ - — x 2 - 2 ^ g ( A l + x ) . X e t d a o h a m y' = - — x - 2 ^ i g m m m m = 100 (g), dao dpng tren mat phang ngang, dupe tha nhe tu vi tri 16 xo d^^ 6cm so voi v i t r i can bang. H? so ma sat trupt giua con lac va mat ban bang H * y' = 0 » X = - 1 ^ ^ = -0,02m = -2em ' . ,,
Ki thuatgilii iiluiiih BI'rN Vat Li 12, t&p 1 - Nguyen Quang Lac Cfy TNHH Ml Kangvi^ K h a o sat svt bien thien ciia y ta tha'y v o l x = - 2 c m t h i y dat gia t r j cue {Chi vat d u n g h i n t h i co nang cua vat chuyen het thanh cong ciia l y c m a sat. Suy ra v dat gia t r j eye dai tai v i t r i x = - 2 c m Do do: A ^ ^ = Hmg.s = W => 0,02.0,1.10.5 = 0,5 => S = 2 5 ( m ) =>ChpnB ' ' v ^ 3 ^ = J — A l ^ - — x 2 - 2 n g ( A l + x) = 0 , 4 V 2 m / s = 40N/2cm / s £^vi 161. M p t con lac lo xo n a m ngang d a n g dao d p n g tat dan. N g u o i ta d o => C h g n C jju C h p n D tu thoi d i e m tha den t h 6 i d i e m vat qua v j t r i 16 xo k h o n g b i bien dang Ian dau Cau 159. M p t tarn v a n bac qua m p t con m u o n g c6 tan so' dao d p n g rieng bang tien la: 0,5 H z . M p t n g u 6 i d i qua t a m v a n v o i bao nhieu buoc t r o n g 12s t h i tarn van bi A. 22,93em/s B. 25,48em/s C. 38,22cm/s D . 28,66cm/s r u n g len m a n h nhat? Huang dan gidi: A. 8 buoc B. 6 buoc C. 4 buoc D . 2 buoc Chpn Ox = true 16 xo, O = v j t r i ciia vat k h i 16 xo k h o n g bien dang, chieu Huang dan gidi: d u o n g la chieu dan ciia 16 xo. K h i b u o c chan vao t a m v a n t h i chan n g u o i da tac d u n g vao t a m van mot K h i vat chuyen d p n g theo chieu am: - k x + famg = ma = m x " luc. Muo'n t a m v a n r u n g len m a n h nhat t h i tan so'ciia l y e tac d y n g vao t a m van p h a i bang tan so'dao d p n g rieng ciia t a m v a n . r _^mg^ N -k - m X \ G p i N la so buoc chan: => — = 0 , 5 = > N = 6 b u o c chan => C h p n B Cau 160. M p t con lac 16 xo c6 d p c i i n g 100 N / m , vat nang c6 k h o i l u p n g lOOg- = 0,02m = 2cm; co = j — = lOrad / s dao d p n g tren m a t phMng n a m ngang. so m a sat g i u a vat va m|it phang k Vm ngang la fa = 0,02. Cho gia toe t r p n g t r u o n g g = 10 m/s^. Keo vat k h o i v i t r i caf X - 2 = A.cos(tot + (p) =:> V = -(oAsin(a)t + (p) bang m p t d o a n 10 c m r o i tha nh?. Q u a n g d u o n g vat d i d u p e cho d e n k h i d i m r Liie to = 0 -> xo = 6 c m => 4 = Aeos cf) vo = 0 han la: => 0 = -lOasincp => cp = 0; a = 4 em A . 25 c m B.25m C. 250 cm D.250m => X - 2 = 4cos(10t) (cm) Huang dan gidi: K h i 16 xo k h o n g bien dang: x = 0 =:> coslOt = -1/2 = cos27i/3 t = n/15 s C o nang ciia vat la: W = ^ k A ^ = ^.100.0,1^ = 0 , 5 ( j )
90 [jjen '^^ dpng toan phan va h$ so'ma sat giiia qua cau voi day kim Vtb = a 28,66 cm/s 7t/15 3,14 ila => Ch(?n D A. 0,05 B.0,025 C. 0,005 D. 0,0025 Cau 163. Mot con lac 16 xo dao dong tat dan cham. Cu sau moi chu ki, bien do Huang dan gidi: dao dong cua no giam 0,5%. Nang lugng dao dong cua con lac bi mat di sau f a tim bieu thue ciia dp giam bien dp sau moi dao dpng toan phan. moi dao dong toan phan la Gpi A i , A 2 la bien dp dao dpng tai hai thoi diem each nhau nira dao dpng. A. 0,5% B. 1% C. 1,5% D. 2% £)p giam CO nang cua con lac: AW = ^ k ^ A j - A j vau .ca/.n: .r ;r V. Huangddngidi: ,;,„^_ , , , ,,,, Congcua lire ma sat: AA^^, = F ^ , { A J + A2) = nmg(Ai + A2) Ta c6: ^ = 0,5% => 1 - — = 0,005 — = 0,995 A A A Do AW = A A ^ , = > A j - A 2 = ^ - 1 A2 Nang lugng trong chu k i dau: W = - mco2 A ,!,v. 4|img )' ^ ' Dp giam bien dp trong moi dao dpng: AA = 2 ( A J - A 2 ) = 1 7 2 Nang trong chu k i tie'p theo: W = — mco A Theo bai ra: 200AA = AQ = 2cm 7 kAA Suy ra: = (0,995^ « 0 , 9 9 = 99% Tudo: AA = r T r = 0,01cm va n = - ^ = 0,005 Chpn C W A 200 4mg 1< Vay nang lug-ng bi mat di la 1% => Chon B CSu 166. Mot con lac 16 xo dat tren mat phSng nghieng goe ao = 60" so voi mat Cau 164. Mot con lac don c6 dp dai lo = 16cm dupe treo trong toa tau a ngay v i ngang, h^ so' ma sat gii>a vat va mat phang nghieng la | i = 0,1, vat c6 khoi tri phia tren cua true banh xe. Con lac dao dpng manh nhat khi van toe cua lupng m = 400 (g), lay g = 10 m/s^. Cong suat can cung cap cho con lac de no doan tau bang 15 m/s. Lay g = 10 m M 7t^ = 10, coi con tau chuyen dpng thang dao dpng dieu h6a voi bien dp A = 5cm va tan so f = 6 Hz la: deu, chieu dai moi thanh ray bang A . P = 180(mW) B.P = 200(mW) C.P = 220(mW) D . P = 240(mW) A. 12 m B. 16 m C. 18 m D.24m Huang dan gidi: Huang dan gidi: Sau khi cung cap bu nang lupng cho vat thi vat se dao dpng dieu hoa: Cong cua lire ma sat tieu hao sau moi chu ky: Chu ki cua lye cuong buc tac dung len con lac: T = — A^„^| = H.N.4.A = n.m.g.cos(a).4.A = 0,1.0,4.cos60.4.0,05 = 0,04(j) Chu ki dao dpng rieng ciia con lac: TQ = 27t J — Nang lupng can cung cap sau moi chu ky la: ,1 VS §iif4 AW = IA^, I =0,04 0) Con lac dao dong manh nhat khi T = 27iJ— = - ==> 1 = 2n,:^.v = 12m Cong suat can cung cap: P = ^ = AW.f = 0,04.6 = 0,24(w) = 240(mW) 8 => Chpn A p=> Chpn D Cau 165. Mot 16 xo c6 dp cung k = 600N/m, mot dau co'dinh, dau kia gan qun J67. Mot dao dpng tat dan c6 bien dp giam 2% sau moi chu ki. Sau 5 chu cau nho khoi lupng m = 300g, qua cau c6 the trupt tren mot day kim loai cang Vi CO nang dao dpng c6n lai chie'm so'phan tram so voi co nang ban dau la: ngang trung voi true 16 xo va xuyen qua tarn qua cau. Keo qua cau ra khoi vj l-A. 17,3% B. 28.9% C. 73,2% D. 81,5% tri can bang 2cm roi tha cho qua cau dao dpng. Do c6 ma sat nho, dao dpng Huang dan gidi: ch^m dan, sau 200 dao dpng thi qua cau dung l ^ i . Lay g = 10m / . Dp giam Theo gia thie't sau moi chu ky bien dp giam 2% tue la: ,1
Huang dan gidi: ' 100 A Ta T CO t = 4s = 2T => S=2.4A=2.4.4=32cm => C h p n D C o nang dao d o n g sau m o i chu k y g i a m d i con lai Wi: Q^aJTl: M p t con lac d o n c6 chieu dai 121cm, dao d p n g d i e u hoa tai n o i c6 gia toe t r o n g t r u a n g g. L a y Tt'^ = 10 . C h u k i dao d p n g cua eon lac la: , „ , w. 221 = (l-0,02) -1-0,04 A. Is B. 0,5s C. 2,2s D . 2s w Huong dan gidi: 2 = ( l - 0 , 0 4 ) W = 0,96W Chu k i dao d p n g cua eon lac T = 2nJ- = 2nP^ = 2.1,1 = 2,2s Vg V 7t2 Sau hai chu ky, ca nang con lai: = 0,96Wi = {0,9ef W => C h p n C , Sau n a m chu k y , ca nang con l a i : Cau 172: M p t vat n h o k h o i l u p n g lOOg dao d p n g dieu hoa v o i chu k i 0,2 s va ca W5 = ( 0 , 9 6 f W = 0,815W = 81,5%W =^ Ch(?n D ' ' nang la 0,18 J (moc the nang tai vj t r i can bang); lay =10. T a i l i d p 372 cm, ti so d p n g nang va the nang la v Cau 168. M o t con lac d o n g o m vat c6 k h o i i u a n g m , day treo c6 chieu doi 1 = I m . Keo con lie lech k h o i p h u o n g thang d u n g m o t goc ao = 0,1 (rad) mi A. 3 B. 4 C. 2 D.l r Huang dan gidi: b u o n g k h o n g v a n toe ban d a u . Con lac chiu l y c can m o i t r u o n g d g Ian coi nhu k h o n g d o i va c6 gia t r i bang 0 , 1 % t r o n g l u g n g ciia vat. Q u a n g d u o n g ma con T a c o (0 = Y = 10n(rad/s) lie dao d p n g ke t u luc b u o n g tay cho den luc d u n g han la. 2 2 A. 4m B. 5 m C. 6 m D. 7 m Co nang W = => A = 0,06m = 6cm Huong dan gidi: Ap d u n g d j n h luat bao toan va chuyen hoa nang l u p n g ta c6: K h i d 6 ^ = ^ = ^ = i Wo = W + A«n W, W, x2 => Wo = Anwx = Fcan.Sm.nx => C h p n D mgl"° Cau 173: M p t vat n h o dao d p n g dieu hoa theo p h u o n g t r i n h x = A cos47tt (t t i n h ^ Smax = ^ = 500.1.ao = 5 ( m ) => C h p n B bang s). T i n h t u t=0, k h o a n g t h a i gian ngan nhat de gia toe eua vat c6 d p I a n mg u V / bang m p t nua d p Ian gia toe cue dai la 1000 A . 0,083s. B. 0,125s. C. 0,104s. D . 0,167s. Cau 169: M o t con lac 16 xo c6 k h o i l u a n g vat nho la m j = 300g dao d p n g dieu Huong dan gidi: hoa v o i chu k i I s . Ne'u thay vat nho c6 k h o i l u p n g m bang vat n h o c6 khoi Ta CO tai v j t r i l a i = 0 , 5 amax t h i 1x1= 0,5.A . K h o a n g t h o i gian ngan nhat tir l u p n g m ' t h i con lac dao d p n g v a i chu k i 0,5s. Gia t r i m2 bang X = A den X = 0,5.A la t = T/6 = 0,5/6 = 1/12 = 0,083. ^ C h p n A A. lOOg B. 150g C.25g D . 75 g £ | u 174: H a i dao d p n g deu hoa eiing p h u o n g , cung tan so' c6 bien dp Ian l u g t Huang dan gidi: 'a A i = 8cm, A2 =15em va l^eh pha n h a u ^ . Dao d p n g t o n g h p p cua h a i dao Ta c6: = 0,5Ti => 2 7 c ^ | ^ = 0,5.27i =^ m ' = ^ = 75gam Chpn D •^Png nay eo bien d p bang Cau 1 7 0 : M o t v a t nho dao d p n g d i e u hoa v o l bien dp 4cm va chu k i 2s. Quan;- A.7em. B. 11 em. C.17em. D . 23 em. d u a n g vat d i d u p e t r o n g 4s la: Huang dan gidi: A. 8 cm B. 16 c m C. 64 c m D.32 c m H a i dao d p n g v u o n g pha nen bien dp dao d p n g tong h p p :
A = 7 A I + A | =17cm end Dt 3. =>ChonC - ' ' S O N G C O L Cau 175: H a i con lac d o n c6 chieu dai Ian l u g t la 81 cm va 64 cm d u g c tree Q tran m o t can p h o n g . K h i cac vat nho ciia hai con lac dang 6 v i t r i can bang J H E T H O N G H O A K I E N THL/C d o n g t h o i t r u y e n cho c h i i n g cac van toe cung h u a n g sao cho hai con l i e dao I Song CO hpc va cac dac t r u n g ciia song r, , , d o n g dieu hoa v o i c i i n g bien do goc, t r o n g hai m a t p h a n g song song v d i nhau ^) Dinh nghta song ca hoc Gpi At la khoang t h o i gian ngan nha't ke t u liic t r u y e n v a n toe den liic hai day . Song C O hpc la cac dao d p n g dan h o i d u p e Ian t r u y e n d i t r o n g m o i treo song song nhau. Gia t r i At gan gia t n nao nha't sau day? >.„ trtfong vat chat theo t h o i gian. ;> tmt'f' A . 8,12s. B. 2,36s. C. 7,20s. D. 0,45s. - K h i song t r u y e n t r o n g m p t m o i t r u o n g t h i cac phan t u ciia m o i t r u o n g Huang dan gidi: chi dao d p n g q u a n h v i t r i can bang ciia c h i i n g ma k h o n g chuyen d o i theo song, PT dao d p n g xi, X 2 : xi = A cos a),t — ; X2 = A cos ( 0 , t chi CO pha dao d p n g eua c h i i n g dupe t r u y e n d i . ^ ^ 2 - .Song CO hpc k h o n g t r u y e n dupe t r o n g ehan k h o n g . .\nrn' Chon D vuong goc v o i p h u o n g t r u y e n song. Song ngang t r u y e n dupe tren be m a t chat Cau 176 : M o t vat n h o dao d o n g dieu hoa theo m o t q u y dao thang dai 12 cm. long va t r o n g chat ran. V i d u : song tren m a t nuoc,... Dao d o n g nay c6 bien do la - Song dpc: l a song eo p h u o n g dao dpng ciia cac p h a n t u m o i t r u o n g t r i i n g A. 3 cm. B. 24 em. C. 6 em. D . 12 cm. voi p h u o n g t r u y e n song. Song dpc t r u y e n dupe t r o n g chat ran, chat l o n g va Huang dan gidi: chat k h i . V i d u : song nen dan dpc theo m p t 16 xo... Bien d p = chieu dai q u y dao/2 = 12/2 = 6cm => C h p n C c) Cac dai luang dac trung cua song ca hoc Cau 177 : M o t vat nho dao d o n g dieu hoa doc theo true O x v o i bien d p 5 em, - C h u k i T cua song la chu k i dao d p n g ehung eiia cac p h a n t u vat chat k h i chu k i 2 s. Tai t h o i d i e m t = 0, vat d i qua can bSng O theo chieu d u a n g . Phuong CO song t r u y e n qua va bang c h u k i dao d p n g ciia n g u o n song. t r i n h dao d o n g eua vat la - Tan so f ciia song la tan so dao d p n g chung cua cac p h a n tCr vat chat k h i A . X = 5cos (em) B. X = 5cos 2 7 r t - ^ (em) CO song t r u y e n qua ^ = ~ 2 2j - Bien d p song a tai m p t diem la bien d p dao d p n g ciia p h a n t u vat chat tai C. X = 5 cos 2 x t + ^ (cm) D. X = 5cos (cm) diem d o k h i song t r u y e n qua. Thong t h u o n g cang ra xa t a m tao song t h i bien 2 2 eoscp = 0 va sintp < 0 (p = -nil. - Buoe song A l a khoang each g i i i a hai d i e m gan nhau nha't tren cung m p t P h u o n g t r i n h dao d o n g la x = 5eos p h u o n g t r u y e n song dao d p n g cung pha. Buoe song eQng la q u a n g d u o n g ma (cm) => C h p n A . 2) Song t r u y e n d u p e t r o n g m p t chu k i . ' ' 1 1 A. - Cong t h i i c lien h ^ T, f, v. A: >. = v.T = j ; T =j; { = v = A..f = -
- N a n g lu-ong song: Qua t r i n h t r u y e n song la qua t r i n h t r u y e n nang luong ^ Giao thoa song N a n g l u o n g cua song tai m p t v i t r i chinh la nang l u g n g cua cac p h a n t u \ sokhdi nietn chat dao d p n g tai d o va nang l u o n g cua song ty le t h u a n v o i b i n h p h u o n g hi^^ ^ Giao thoa song la su t o n g h o p hai hay nhieu song ket h o p t r o n g k h o n g dp dao d p n g . an t r o n g do c6 n h u n g cho bien dp song tong h p p dupe tang c u o n g hoac eo D o i v o i song phSng nang l u o n g song g i a m ty le v o i khoAng each. D o i voj ^hiJnS '^'^^ '^^ ^^"^ ^'^"^ song cau nang l u o n g giam ty le v o i b i n h p h u o n g ban k i n h . Song p h a n g ne'u b6 H a i n g u o n ket h p p la hai n g u o n eo eung tan so va c6 d p Ipch pha k h o n g qua m a t m a t t h i bien d p d u p e xem la k h o n g d o i nen nang l u p n g t r u y e n qug thay doi theo t h a i gian. m p i d i e m la n h u n h a u . H a i song ket h p p la hai song do hai n g u o n ket h p p phat ra. l ' 2. P h u o n g t r i n h s o n g s J,) phuong trinh song tong hop tai M f t, , - P h u o n g t r i n h song la bieu thuc cho phep xac d i n h l i d p u ciia t i m g phan * Tncong hop hai nguon CO do lech pha bat ky v t u X theo t h o i gian t. p h u o n g t r i n h song tai hai n g u o n eung p h u o n g Si, S2 c6 dang: - Gia s u p h u o n g t r i n h song tai O c6 dang: U Q = a.cos(a)t) = acos(cot + c})i) va U2 = aeos(a)t + c|)2) ' ' ' - P h u o n g t r i n h song tai M each O m o t doan d (song t r u y e n t u O den M ) : Phuong t r i n h song tai M do hai n g u o n t r u y e n t o i : (Ot- - 2 7 1 - 27rdi' 27rd2^ aeoso) = acos COt + CP] - va u 2M = aeos " i M = aeos C0t + (p2 ~ P h u o n g t r i n h song tai M ' each O m p t doan d ' (song t r u y e n t u M ' den O): Phuong t r i n h song tong h p p tai M : Uf^. = a cos CO = acos a)t + 2 7 t — d2-di Acp = " i M + " 2 M =2acos cos X 2 - D p l^ch pha dao d p n g giira hai d i e m M va N bat k i t r o n g m o i truong d-i-di Acp MN Bien d p dao d p n g tai M : A,^ = 2a cos v o i Acp = cp2 - cpi t r u y e n song each n g u o n O Ian l u p t la dM va dw: Acp = 27i = 27t Tren doan thSng noi \\x Si den S2 (S1S2 = 1) so cue dai, cue tieu giao thoa d i + H a i d a o d p n g c i m g p h a : A(]) = k 2 n (voi k e Z) qua chinh la so gia t r i k n g u y e n thoa m a n : + H a i d a o d p n g n g u p c p h a : A(|) = (2k + 1).TT (voi k € Z) 1 Acp , 1 Acp (keZ) + H a i d a o d p n g v u o n g p h a : Acjj = (2k + 1). ^ (voi k e Z) — +—< k < —+ 1. In X 2n 1 1 Acp , 1 1 Acp - P h u o n g t r i n h tren c h o thay s o n g c6 t i n h t u a n h o a n theo t h o i g i a n va theo + -X
Tren doan thang noi tir S i den S2 (S1S2 = 1) so'eye dai, cue tieu giao tho, • rnpt eye dai giao thoa; ne'u hai n g u o n dao d p n g n g u o c pha t h i d u o n g t r u n g qua ehinh la so'gia trj k nguyen thoa man: trifC la r " 9 t eye tieu giao thoa. T r o n g doan n o l g i u a hai n g u o n S1S2 khoang each g i i i a eac v a n eye d a i + So'cuc dai: - — < k < — (k e Z ) ; hoac eye tieu l i e n tie'p bang n h a u va bang ^ ; K h o a n g each g i i i a v a n eye d ^ v a • • X X + So cue tieu: - — - —
- D a y d a i /co d i n h rriQt dau, m p t d a u t y do: I = ( 2 k + 1 ) — (Chieu dai ,1 N e u d u n g d o n v j la Ben (B) t h i : L = l o g : ^ day bang m o t so' le Ian m p t p h a n t u b u o c song). + Sobung:k + l f'j^L- N e u d u n g d o n v i la dexiben (dB) t h i : L = l O l o g — + S6'nut:k + 1 IQ ^. p. d) Do van toe truyeh song ^) Cac dac trtmg sinh U cua dm thanh - D u n g hien t u p n g song d u n g , tit t h i n g h i e m , ta d o dup-c b u o c song > . D p cao: La dac t r u n g sinh l i eua am, p h u thupc vao t a n so' a m . A m cao bang each d o d u p e X/2 khoang each giua hai n u t song l i e n tie'p hoac hai byng (tharih) la a m co tan so' am I o n ; a m thap (tram) la am eo tan so' a m n h o . D p thap song l i e n tie'p va d o d u p e tan so' f. |,ay cao ciia a m con d u p e h i e u qua s u t r a m hay b o n g cua am. - D u n g eong t h u c v = Xf ta t i m d u p e v a n toe v . - D p to: La m p t dac t r u n g sinh l i ciia am, p h y thupc vao m i i c c u o n g d p a m 5. Song am {pxbng d p a m va tan so am). a) Song dm + Gia t r i c u o n g d p am I be nha't ma tai n g u o i con cam n h a n d u p e g p i la nguong nghe. Gia t r j cua n g u o n g nghe p h u thupc vao tan so'. - Song a m la song co hpc ma tai con n g u o i c6 the cam n h a n d u p e . Song am + Gia t r j I nao d o d i i I o n l a m tai nghe co cam giac nhue n h o i , d a u d o n CO tan so n a m t r o n g khoang t u 16Hz den 20.000Hz. N g u o n a m la ba't k i vat nao thi gpi n g u o n g d a u . N g u o n g d a u k h o n g p h u thupc vao tan so'. phat ra song a m . + M i e n I n a m t r o n g khoang n g u o n g nghe va n g u o n g d a u g p i la m i e n - Cac a m co tan so tren 20 OOOHz g p i la sieu a m (tai d o i cam t h u dupe). nghe dupe. M i e n nay r p n g hep p h u thupc tan so'. - Cac a m co tan s o d u o i 16 H z gpi la ha am (tai cho cam t h u dugfc) - A m sac: La m p t dac t r u n g sinh l i ciia am, p h u thupc vao tan so' am, bien b) M6i trubng truyen dm vd van toe truyen dm dp song a m va cac t h a n h p h a n ca'u tao eiia am, t i i c la p h u thupc vao d o t h i dao - Song a m t r u y e n d u p e t r o n g cac m o i t r u o n g vat cha't d a n h o i n h u ran, dpng cua a m . long, k h i . Song a m k h o n g t r u y e n d u p e t r o n g ehan k h o n g . N h i r n g vat l i ^ u nhu + A m CO ban va hoa am: M p t nhac cu phat ra m p t a m co tan so fo (am co bong, n h u n g , n h i r n g tarn xo'p t r u y e n am kem, ehiing d u p e d u n g de l a m vot ban hay hoa a m t h u nha't) t h i bao g i o cung phat ra d o n g t h o i cac hoa am t h i i 2, l i f u each am. 3,... CO tan s6'2fo, 3fo,...Do hien t u p n g do, am phat ra t u m p t nhac cu la s u tong ' - V a n to'e t r u y e n a m p h u thupc vao t i n h d a n h o i , m a t d p v a nhiet d p cua hpp cua am co ban va cac hoa am, no dupe gpi la nhac am, t u y no co tan so'cua m o i t r u o n g . V a n toe t r u y e n am t r o n g chat ran Ian h o n v a n to'e t r u y e n a m trong i m CO ban fo n h u n g d u o n g bieu d i e n cua no k h o n g eon d u o n g h i n h sin dieu chat l o n g , v a n toe t r u y e n am t r o n g chat l o n g Ion h o n v a n toe t r u y e n am trong hoa ma la m p t d u o n g cong tuan hoan phue tap co chu k i , ta g p i n o la d o thj chat k h i . dao d p n g cua am. H p a am nao co bien d p I o n nha't se quyet d j n h d p cao cua am - Song a m t r u y e n t r o n g chat k h i la song dpc; song a m t r u y e n t r o n g chat phat ra. ran bao g o m ca song dpc va i o n g ngang. + Dao d p n g am tong h p p co t i n h tuan hoan, d a n g cang p h u e tap t h i so c) Cac dac tntng vdt ly cua dm thanh hpa am cang n h i e u , am sac cang p h o n g p h i i . C i m g m p t am sol d o n h i e u d u n g - T a n so am: f t u l 6 H z d e ' n 20.000Hz. ^ khae n h a u phat ra sac thai am t h a n h ma ta nghe khac nhau la d o so' hpa a m - C u o n g d p am va m u c c u o n g dp am: toan khac nhau. + A m d u p e phat ra t u m p t tieng no, tieng go vao t a m k i m loai,.. g p i la + C u o n g d p a m I tai m p t d i e m la nang l u p n g t r u y e n t r o n g m p t d o n v! am, chving k h o n g co tan so' xae d j n h , d o t h i cua ehung la d u o n g cong t h o i gian qua m p t d o n v j d i ^ n tich dat v u o n g goe v o i p h u o n g t r u y e n am to' ' ^ o n g t u a n hoan. E E P diem do. I = — = — . D o n v j : W / m ^ (oat tren met v u o n g ) ^guon nhac dm vd hgp cong huong S-t 47tR^t 47iR^ - N g u o n nhac a m la n h i r n g n g u o n phat ra nhac a m , tuc la phat ra n h u n g + M u c c u o n g d p am L la dai l u p n g do bang loga thap p h a n cua ty s^' ^ CO tan so xae d j n h , m o i nhac cu la m p t n g u o n nhac am. g i i i a c u o n g d p a m I tai d i e m d a n g xet va c u o n g d p am chuan lo (lo = 10"^^ W / m - H o p cong h u o n g la m p t vat rong co kha nang epng h u o n g d o i v o i nhieu u n g v o i tan so' f = 1 OOOHz). so khac n h a u v . tang c u o n g n h i r n g am co cac tan so d o .
II. C A C D A N G B A I T A P V A V I D y MINH HQA X 50 ^n toe truyen song la: ^ ^ j ^ ^ 50cm / s =5> Chpn D Dong^: XAC D I N H C A C D A C T R l / N G CUA S O N G C O H Q C Limy: 1-:' : * Phuong phap giai: - Su dung cac cong thuc: to = 27tf = ^= v T = . O day hoe sinh c6 the nham mpt trong cac truong hpp nhu sau: + Khoang each 2 ngpn song ke nhau la 0,5m nen A = I m = 100cm Vi 1: Mot nguoi quan sat mpt chiec phao tren mat bien thay no nho cao Van toe truyen song la: v = ^ = = lOOem / s => chpn B len 6 Ian trong 10 giay va thay khoang each giiia hai ngpn song ke nhau la 0,2m. Van toe truyen song bien bang: + Co 10 ngpn song di qua truoc mat trong 9s nen chu k i song la: lOT = 9s A. 12cm/s B. lOcm/s C. 24cm/s D. 20cm/s ^ T = 0,9s Huong dan giai: =>Van toe truyen song la: v = ^ = = 55,56cm / s => chpn A Khoang thoi gian giiia 6 Ian nho lien tiep la 5 chu ki => 5T = 10 => T = 2s Khoang each 2 ngpn song ke nhau la 0,2m => A = 0,2m = 20cm + Khoang each 2 ngpn song ke nhau la 0,5m nen A = I m = 100cm n A, 20 Co 10 ngpn song di qua truoc mat trong 9s nen chu k i song la: lOT = 9s • Van toe truyen song la: ^ ^ ^ ^ ^ 10cm / s => T = 0,9s. Khi do van toe truyen song la: v = — = - — = 111,11cm / s ^ z^ChpnB T 0,9 Lira y: => chpn C - O day hoc sinh c6 the nham nhu sau: + Khoang thoi gian giOa 6 Ian nho lien tiep la 6 chu ki => 6T = 10 => T = 5/3s Dang 2: B A I TAP V E P H l / O N G TRINH SONG V A D O L E C H PHA "k 20 G I 0 A HAI DIEM TREN PHLfONG TRUYEN SONG Van toe truyen song la: v = — = — = 12cm / s => ehpn A * Phuong phap giai ^ 3 - Gia su phuong trinh song tai O c6 dang: Uy = a.cos(a)t) + Khoang each 2 ngpn song ke nhau la 0,2m => A = 0,4m = 40cm - Phuong trinh song tai M each O mpt doan d (song truyen tir O den M ) : " -> Van toe truyen song la: ^ = ^ = ^ IQcm / s => ehpn D u ^ = acosco t — = aeos cot - 2 7 1 - + Khoang thoi gian giCra 6 Ian nho lien tiep la 6 chu ki => 6T = 10 => T = 5/3s Khoang each 2 ngpn song ke nhau la 0,2m A = 0,4m = 40cm - Phuong trinh song tai M each O mpt doan d (song truyen tit M den O): ( d^ Van toe truyen song la: v = — = ^ = 24cm / s => ehpn C acosco t + - = aeos COt + 27t- 3 - Dp lech pha dao dpng giira hai diem M va N bat ki trong moi truong V i du 2: Mpt nguoi quan sat song tren mat ho thay khoang each giua hm ngpn song lien tiep la 0,5m va c6 10 ngpn song di qua truoc mat trong 9^ t''uyen song each nguon O Ian lupt la dM va dw: Acp = 27t- Van toe truyen song tren mat nuoe bang: + Hai dao dpng cung pha: Acp = k27T (k e Z) A. 55,56cm/s B. lOOcm/s C lll,llcm/s D. 50em/s + Hai dao dpng ngupe pha: Acp = (2k + 1 ).TT (k € Z) Huong dan giai: Khoang each 2 ngpn song ke nhau la 0,5m nen A = 0,5m = 50cm + Hai dao dpng vuong pha: Acp = (2k + 1). ^ (k e Z) Co 10 ngpn song di qua truoc mat trong 9s nen chu k i song la: 9T = 9 s ^ T = ls
* Cdc vi du tninh hpa: Huong dan gidi: 27cd V i d^ 1: Song t r u y e n tai m a t chat long v o i buac song 0,1m. P h u a n g trhih . P h u a n g t r i n h song t?ii M c6 dang: U j ^ = a cos cot-- (cm) dao d o n g t a i n g u o n O c6 d a n g UQ = Scoscot ( m m ) . P h u o n g t r i n h dao d p n g tai d i e m M each n g u o n O m p t doan 6cm theo p h u a n g t r u y e n song la: . Thay U j ^ = 2cm, t = T/2 va d = A/3 vao p h u a n g t r i n h t r e n ta dugic: A . u ^ = 5cos(cjt + l , 2 7 t ) ( m m ) B. u , ^ = 5cos(cot + 1 2 0 7 i ) ( m m ) 2 = acos => 2 = acos ( r t - — ) = — = > a = 4 c m => C h p n B. C. U [ ^ = 5cos(a)t - 1207T) ( m m ) D . u ^ = 5cos(ajt - l,2n) ( m m ) T 2 3.x 3 2 Huong dan gidi: V i d u 3: M o t song co hpc ca bien d o a, b u o c song A. Biet v a n toe d a o d g n g P h u a n g t r i n h song tai M d o O t r u y e n den c6 dang: eye d a i eiia p h a n ttr m o i t r u o n g bang 2 Ian toe d p t r u y e n song, b i e u t h i i c d i i n g la: ( 2nd' 271.6^ u ^ = 5 cos cot = 5 cos cot = 5cos(cot - l , 2 7 t ) ( m m ) A. A. = Tia I ^; V 10 J B. X. = 27ia Huong dan gidi: C. A. = 2na/3 D . A, = 37ia/2 => C h o n D . In Ltfu y: .a j^^y J ^ V d d m a x J ^ ^ T ^ J _ . T = na ^ C h o n A . - 6 d a y hpc sinh c6 the n h a m n h u sau: ^ 2 2 2 + N h a m ve h u o n g t r u y e n song: V i d u 4: N g u o n song co t r u y e n tren m p t spi d a y d u a c m o ta b a i p h u o n g P h u o n g t r i n h song tai M do O t r u y e n d e n c6 dang: trinh u = acos(27rt - 0 , l 7 t x ) , t r o n g d o u va x d u p e d o b a n g cm, t d o b a n g s. 2nd 27C.6 U j ^ =5cos| cot + = 5 cos cot + = 5cos(cot + l , 2 7 i ) ( m m ) Tai m o t t h o i d i e m d a cho d p lech p h a dao d p n g ciia h a i p h a n t u t r e n day 10 each n h a u 2,5 c m l a => chpn A . + N h a m ve d o n v i chua d o i t h o n g nhat: A.^ B.^ 6 8 4 P h u a n g t r i n h song tai M d o O t r u y e n den c6 dang: Huang dan gidi: ( . 27rd^ 271.6^ Uj^ 5cos cot + = 5 cos cot + - = 5 cos (cot + 1207t) ( m m ) 27tX 10 - P h u a n g t r i n h song tai m o t d i e m M co dang: u = a cos c o t - - => c h p n B. - Theo gia t h i e t u = a cos (27it - 0, ITIX) + N h a m ca h u o n g t r u y e n va d o n v i chua d o i : P h u a n g t r i n h song tai M d o O t r u y e n den c6 dang: 27rx nen u = a c o s ( 2 7 i t - 0 , l 7 t x ) = acos cot-- 0 , l 7 i x = — ^ A . = 20em UM=5COS r cot 2nd^ = 5 cos cot- 27t.6^ = 5cos(cot-1207t) ( m m ) V 0,1 J D p l$eh pha dao d p n g ciia hai phan t u tren day each n h a u 2,5 c m la: chon C. 2iid 27t.2,5 7t, Acp = . ^ V i d\ 2: N g u o n song co t r u y e n theo m o t d u a n g thang, n g u o n dao d p n g v o i 20 4^ ^ p h u a n g t r i n h UQ = acoscjt (cm). M p t d i e m M tren p h u a n g t r u y e n song each Chpn D. n g u o n d = A/3 t a i t h o i d i e m t = T/2 c6 l i d p = 2cm. C o i b i e n d p song V i d u 5: M o t song t r u y e n tren m a t nuoe bien co b u a c song X = 2 m . K h o a n g k h o n g b j suy giam, b i e n d p song a la: each giira h a i d i e m g a n n h a u nhat tren c i m g m p t p h u a n g t r u y e n song dao
Huong dan gidi: phuong trinh song tai M do hai nguon truyen tai: Khoang each giua hai diem gan nhau nhat tren ciing mpt phuong truy^Y, 27id, ^ 27td2^ song dao dpng ngupe pha nhau la: d = X/2 = 1 m. => Chpn D MM = acos cot + (pj - va U j M =aeos cot + - V i dv 6: Mpt song truyen tren mat nuoc bien c6 buoc song X = 2m. Khoang phuong trinh song tong hpp tai M : each giiia hai diem gan nhau nhat tren cung mpt phuong truyen song dao d-, - d i Acp d, +di (D, + ( p , dong l^ch pha nhau n/2 la: UM="iM+"2M=2acos n— + — cos X 1 A . 0,75 m. B. 3 m. d, - d i Acp C . 0,5 m. D . Mpt gia tri khac. Bien dp dao dpng tai M : A,^ = 2a cos •K— - + — voi Acp = cp2 - cpi Huomg dan gidi: Dp l?ch pha dao dpng ciia hai phan tu tren phuong truyen song each nhau Tren doan thang noi tu Si den S2 (S1S2 = 1) so cue dai, cue tieu giao thoa di qua chinh la so gia trj k nguyen thoa man: d l a : A(p = — ( r a d ) V . * M J^HW*.** t mii If " i t / ! X 1 Acp , 1 Acp /, „^ — + — < k < —+ — (keZj ;i 27t X In Theo gia thiet A(p = — => = — => d = — = 0 , 5 ( m ) ^ 2 2 4 ^ ^ 1 1 Acp , 1 1 Acp /, . 0 « +— Chpn C 2 271 2 27t ^ ' V i d u 7: Song truyen tren day voi van toe 4 m/s c6 tan so' song thay doi tu 2 2 H z den 26 H z . Diem M each nguon mpt doan 28 cm luon dao dpng vuong Truang hop hai nguon cung pha: pha voi nguon. Buoc song truyen tren day la: - Dp l^ch pha Acp = 2kTt A. 160 em. B. 1,6 cm. C . 16em. D . 100 cm. - Phuong trinh song tong hpp tai M : d2-di d, +di cp. +cp, = " MM 1 M + " 2 M = 2a COS COS Huong dan gidi: D p l^ch pha dao d p n g ciia hai phan tu tren phuong truyen song each nhau [fii. • Bien dp dao dpng tai M : A|y^ = 2a COS dla: A(p = ^ ^ = ^ ^ . f ( r a d ) X V ^ ' Theo gia thiet Tren doan thang noi tu Si den S2 (S1S2 = 1) so'cue dai, cue tieu giao thoa di n , 27id, n / ^ { k + 0,5).y (k + 0,5).400 (k + 0,5).50 qua chinh la so'gia trj k nguyen thoa man: Acp = - + krc => f = - + k7i => f = — = =^ -^ — ^ 2 V 2 2d 2.28 7 + So'cue dai: — < k < ~ (k 6 Z); • • X X (k + 0,5).50 M a 2 2 H z < f 22 H z < ' - — < 26 H z 2,58 < k < 3,14; 7 i + So cue tieu: - - - - < k < - - - (k e Z) > ' •• - " § ^ 2 ^ 2 ^ ^ V i k nguyen nen k = 3 f = 25Hz. V 400 ^ - Trong doan noi S1S2 khoang each giiia cac van cue dai hoae cue tieu lien Vay buoc song truyen tren day la: X- — = = 16cm => Chpn C tiep bang nhau va bang ^ . Dang 3: B A I T A P V E G I A O T H O A S O N G C O H Q C Chuy: * P h u a n g phap giai: - K h i giai bai toan giao thoa, can kiem tra dp l§ch pha eua hai nguon. Phuong trinh song tong hop tai diem M: - K h i hai nguon dao dpng voi bien dp khac nhau thi phuong trinh song Phuong trinh song tai hai nguon cung phuong Si, S2 c6 dang: tong hop tai M bat ky dupe xae djnh nhu tong hpp hai dao dpng (hai dao dpng *hanh phan la hai song do hai nguon truyen den M) va bien dp dao dpng tai M U j = aeos(cut + cpi) va U j = acos(a)t + 92)
duQC ti'nh: = yja^ +al+ la^aj cos(A(p) t r o n g d o A(p l a d p l^ch p h a ciia haj y f d y 2: T r o n g m p t t h i n g h i e m giao thoa song tren m a t nuoc, hai n g u o n ket Pi^p S i va S2 dao d o n g v o i tan so' 16 H z , c u n g pha. Tai d i e m M each hai song t h a n h p h a n tai M . pguon Ian l u g t la d i = 30 c m va d2 = 25,5 em, song eo bien d g cue d a i . G i i i a - V o i bai toan t i m so d u a n g dao d p n g v o i bien d o cue d a i , cue tieu giijg va d u o n g t r u n g true ciia S1S2 c6 t h e m m o t g g n l o i nua. V a n toe t r u y e n hai d i e m M , N bat k y each hai n g u o n Ian l u g t la d j ^ , 62^, d j ^ j , d j ^ t h i lam song tren m a t n u o c la n h u sau: A^24 em/s B. 36 cm/s C. 72 m/s D . 7,1 cm/s + Dat A d ^ j = d 2 M - d , M ; ^ d N = d 2 N - d i N ; Gia s u Ad^^ > Ad^j Huang dan gidi: ! + N e u hai n g u o n dao d o n g eung p h a : . H a i n g u o n dao d o n g cung pha, tai M song co bien d p eye d ^ i nen ^ , , d 2 = k.A. „^:^P-- * Cue d a i : A d ^ < k > i < A d M . G i i i a M va d u o n g t r u n g true ciia S1S2 c6 t h e m m g t g g n l o i nua nen k = 2. * ' * Cue t i e u : Ad^J < ( k + 0 , 5 ) ? L < A d M * V?y 30 - 25,5 = 2.A + N e u hai n g u o n dao d p n g nguge p h a : . Buoc song A = 2,25 cm * C y c d a i : Adf^ < ( k + 0 , 5 ) > i < A d | ^ => Van toe t r u y e n song tren mat nuoc la v = A.f = 2,25.16 = 36 (em/s) => Chgn B L u u y : Co the n h a m n h u sau * Cue tieu: Ad^j < k A . < A d ^ - k = 3. Vay 30 - 25,5 = 3.A; S o g i a t r i n g u y e n ciia k thoa m a n cac bieu thue tren la so d u a n g can t i m . Buoc song A = 1,5 em ^ v = A.f = 1,5.16 = 24 (cm/s) =>Chgn A Vi 3: O be mat mgt chat long c6 hai nguon phat song ket hgp S i va S2 * Cdc vi du mirth hoa: each nhau 20cm. H a i nguon nay dao dgng theo phuong thSng dung c6 V i d v 1 : T r o n g t h i n g h i e m ve h i ^ n t u g n g giao thoa song, n g u o i ta tao tren phuong trinh Ian lugt la u^ = 5eos (407tt + n/6) (mm) va U j = 5cos(407tt + 77t/6) m a t n u o c hai n g u o n A va B dao d o n g cung p h u o n g t r i n h : u ^ = 5cosl07tt (mm). Toe do truyen song tren mat chat long la 80 cm/s. So diem dao dgng (cm) va U g = 5cos(107rt + 71) (cm), v a n toe t r u y e n song tren m a t nude la voi bien do cue dai tren doan th^ng S1S2 la 20cm/s. D i e m M tren mat nuoc c6 M A = 7,2 cm, M B = 8,2 c m c6 p h u o n g A. 11. B. 9. CIO. D . 8. t r i n h dao d o n g la : Huang dan gidi: R ' - 1 V 80 ^ D u o e song X = — - — = 4cm A . u ^ =5N/2cos(207tt-7,77t)cm B. u ^ = 5 > ^ c o s ( l 0 7 t t + 3 , 3 5 j i ) c m ^ f 20 C. U[^ = 1 0 V 2 c o s ( l 0 7 i t - 3 , 3 5 7 i ) c m D . u^^ = 5 N / 2 c o s ( l 0 7 t t - 3 , 3 5 7 t ) c m So'cue dai giao thoa tren S1S2 - i < k< ^ - i _>, - 5 , 5 < k < 4,5 • ^ X 2 X I ,^ Huang dan gidi: V i k 6 Z nen k = - 5 , - 4 , - 3 , - 2 , - 1 , 0 , 1 , 2, 3 , 4. - Buoc song A = v/f = 20/5 = 4 (cm) Co 10 gia trj ciia k, vay c6 10 d i e m tren doan S1S2 dao d g n g v o i bien d g cue - P h u o n g t r i n h song tong h g p tai M : ^=>ChonC. dj+dj (pi+92 d y 4: H a i n g u o n song co S i va S2 each n h a u 42 c m dao d g n g theo p h u o n g "M="iM+U2M=2acos cos (ot-n— - +— ~ trinh: U j = U 2 = 4cos507rt (cm) Ian t r u y e n v o i toe d g v = l,Om/s. So d i e m X 2 8,2-7,2 ' ^ o n g dao d g n g tren doan thang no'i S1S2 la ( k h o n g ke hai n g u o n ) : u^^ =2.5. cos = 5V2cos(l07tt-3,357t) cos _ A . 10 B. 12 C. 16 D.20 4 2 Huang dan gidi: r:> C h p n D ^.R. , , , v 100 , 1 Duoe sone X = — = = 4cm * ^ f 25 1 O/l
S S 1 S S 1 27cd So cvrc tieu giao thoa tren S1S2 — L i . _ - < k < - - -> - I K k < 10 . Bien d p dao d g n g tai M : A ^ = 2a sm- V i k e Z n e n k = -10,-9, ...,9 Co 20 gia trj ciia k, vay c6 20 d i e m tren doan S1S2 k h o n g dao d p n g -> chpn D + AMmax = 2 a k h i d = (2k + l ) - ; k € Z V 14 V i dv 5: Tai hai d i e m A va B tren mat n u o c c6 2 n g u o n song dao d o n g cung + AMmm = 0 k h i d = k — ; keZ pha, bien d p Ian l u p t la 4cm va 2cm, buoc song la 10cm. D i e m M t r e n mat 2 n u o c each A 25cm va each B 30cm se dao d p n g v o i bien d p la Dieu kien deed song ditng tren day A. 2cm B. 4cm C. 6cm D. 8cm . Day c o ' d j n h h a i d a u : 1 = k ^ «'.ln Huang dan gidi: jv M fcu H a i n g u o n dao d p n g c u n g pha nen bien dp dao d p n g tai M : + So'byng: k + So'nut: k + 1 = V^i + ^2 + ^^\^2 COS(A(P) + So'bo song: k 27r(d,-d,) + Q u a n h# giira k v o i f: 1 = k — = k — . V o i Acp = — — — = n . Suy ra A M = 2 cm. ^ C h p n A ^ 2 2f k2 £2 V i dy 6: H a i d a u M va N ciia m p t mau day thep nho h i n h chi> U dupe dat + Buoc song dai nha't Amnx = 21 k h i k = 1 (chi co 1 bo song). cham vao m a t nuoc. Cho mau day thep dao d p n g d i e u hoa theo phuong - Day CO d j n h m p t dau, m p t d a u t u do: 1 = (2k + 1 ) ^ v u o n g goc v o i m a t nuoc. Bie't M N = 6,5cm dao d p n g v o i tan so' f = 80Hz ; toe ..A , d p t r u y e n song v = 32cm/s ; bien dp song k h o n g d o i A = 0,5cm. Ve m p t vong + S6'bung:k + 1 t r o n Ion bao ca h a i n g u o n song vao trong. Tren v o n g t r o n ay c6 bao nhieu + Sonut:k + l d i e m CO bien d p dao d p n g cue dai. Luu y: A.33 B. 32 C.64 D. 66 - Khoang each giiia 2 n u t song hay hai b u n g song bat ky: d - k - ^ , k = 1,2,3... Huong dan gidi: Buoc song: A = v/f = 32/ 80 = 0,4cm - K h o a n g each giOa 1 n i i t song v o i 1 b u n g song bat ky: So cue d a i tren M N la: < k < - => -16,25 < k < 16,25 => c6 33 d i e m tn n d - ( 2 k + l ) ^ , k = 0,1,2,3... ^, X X •[ .A M N dao d p n g v a i bien dp cue dai. K h i d o v o n g t r o n bao q u a n h hai n g u o n - T h o i gian hai Ian day d u o i thSng lien tie'p: At = • va N cat cac van cue dai 6 66 d i e m => C h p n D. Be r p n g cua m p t b u n g song la: 4a Dang 4: BAI TAP V E SONG Dl/NG M p t spi day thep cang th^ng, dat gan m p t d a u n a m cham dien thang, d o n g dien qua n a m cham co tan so'f t h i day se dao d p n g v o i tan so 2f. * Phuong phdp: - M p t spi day no'i v o i n g u o n d i f n xoay chieu tan so f, day dat t r o n g Phuong trinh song tai mot diem M bat ky: khoang giira hai ban cua m p t n a m cham h i n h chCr U t h i day se dao d p n g v o i - Xet d o a n day A B = 1, dau B co d i n h dau A dao d p n g v o i p h u o n g t r i n h : tan so cung la f. u = acosu't. y - Tan so d o day d a n phat ra (hai d a u day c o d i n h ) : f = k — ; k e N - P h u o n g t r i n h dao d p n g tai d i e m M bat k y each d i e m p h a n xa co' djnh 1' o • 27id ( ^ 2nt] - Tan so d o 6'ng sao phat ra (mpt dau bjt k i n , m p t dau de h o m p t d a u la mptdoand: Uj^=2a.sm .cos cot X y X ) ^ut song, m p t d a u la b u n g song): f = (2k + 1 ) - ^ ; k e N
iv> inuuigmi ruiunn B I I I \ L I LJ., lup i - i \ g u y e n jjuang i.ac irmn miv UVVH KHattg^rtfT Cdc vi du minh hga: y i du 4: Nguoi ta t^o song dirng tren mot soi day cang ngang giira hai diem CO djnh. Song dung dugc tao ra tren day Ian lugt voi hai tan so' gan nhau jiha't la 200Hz va 300Hz. Tan so' kich thi'ch nho nha't ma van tao ra song dung tren day la : A. lOOHz B. 50Hz C. 200Hz D. 150Hz Huang dan gidi: Huang dan gidi: Dieu kien de c6 song dirng tren day trong truong hop hai dau co'djnh: - Dieu ki^n de co song dung tren day trong truong hgp hai dau co djnh: X , 21 1 = n — => A = — l = k.^ = k ^ 2 n 2 2f 21 21 212 Vi 1 va V khong doi nen khi so bo song it nha't (tuc n = 1) thi tan so'f se nho Voi 3 mui t a c 6 n = 3=> X = — = — = — = 0,8m => chpn B. n 3 3 V V i d u 2 : Mot day dan c6 chieu dai 90cm, khi gay phat ra am co ban tvrong nha't fmin = ^ • ung CO tan so f. Muo'n cho day nay phat ra am co ban f = l,2f thi phai b a n i + Hai tan so gan nhau nha't la 200Hz va 300Hz se ung voi so' bo song lien phim cho day ngan lai con chieu dai 1' bang A. 60 cm B. 75cm C. 50cm D. 65cm tiep la k va (k + 1), tiic la 1 = k ^ va 1 = (k +1)—-—. T u hai phuong trinh Huang dan gidi: 2.200 2.300 V tAy se CO 1 = 200 Dieu kien de co song dung tren day trong truong hop hai dau co dinh + Thay vao bieu thuc fmin ta dugc fmin = lOOHz. -> Chgn A. I = n . ^ . A m CO ban do day dan phat ra ling voi n = 1 => 1 = ^ = ^ ( * ) i Vi d u 5: Mot day AB hai dau co dinh. Khi day rung vol tan so'f thi tren day CO 4 bo song. Khi tan so tang them 10 Hz thi tren day co 5 bo song, van toe Tuong t u de co tan so' f thi ' ~ ^ ~ ^ (**) truyen song tren day la 10 m/s. Chieu dai va tan so rung cua day la: A. 1 = 50 cm, f = 40 Hz. B. / = 40 cm, f = 50 Hz. r f It T u H v a r ) s u y r a i - = 1 =:> j ' = _ = _ If 1 90 = _ ^ _ = 75cm ' Chon B C./ = 5cm, f = 50Hz. D. / = 50 cm, f = 50 Hz. ^' ^ ' y I f f i,2f 1,2 1,2 Huang dan gidi: V i du 3: Song dung tren day AB co hai dau co dinh, chieu dai 40cm. Tan so - Day AB hai dau co'djnh nen chieu dai cua day la • i dao dgng cua day la 50Hz, van toe truyen song tren day la 4m/s. Tren day co: l = k . ^ = k . ^ = ^ f = k.-^- A. 10 bung, 11 nut B. 10 bung, 10 niit 2 2f 21 C. 5 niit, 5 bung D. 9 bung, 10 nut - Khi day rung voi tan so f thi tren day co 4 bo song: f = 4 . ^ (1) Huang dan gidi: V 4 Buoc song: X = v.T = — = — = 0,08m = 8cm - Khi tan so tang them 10 Hz thi tren day co 5 bo song: f +10 = 5.-^ (2) ^ f 50 Dieu ki^n de co song dij-ng tren day trong truong hgp hai dau co'djnh: Tu (1) va (2) suy ra ^ = 10 ^ 1 = 0,5m - 50cm . l = k.A JThay vao (1) suy ra f = 40 Hz =:> Chgn A 2 du 6; Mot sgi day M N treo lo lung, dau tren M gan vao am thoa dao 21 2 40 dgng voi chu ki T = 0,02s, toe do Ian truyen song la 2m/s. De tren day co So bung song: k = — = —— = 10 => so byngsong la 10, so niit song la 11. ^ n g dung, voi 13 niit va 13 bung (ke ca hai dau), thi sgi day phai co chieu X 8 ^ Chon A. bang bao nhieu? - . _ A . 20 cm B. 30 cm C. 50 cm D. 25 cm
Huang dan gidi: V i d u 3: M p t cai gieng sau 20m t i n h t u thanh den mat nuac. M p t em be - Buac song: A, = v T = 2.0,02 = 0,04m = 4cm d u n g gain thanh gieng, d o so y da d a n h r o i ehie'c bat. Biet v | n toe t r u y e n am - D i e u k i ^ n de c6 song d u n g tren day: 1 = k | + ^ (*) v o i k e N trong k h o n g k h i la 320m/s, lay g = lOm/s^. T i n h t h o i gian t u liic d a n h r o i den luc nghe thay tieng bat cham vao nuoc. Thay>. = 4 c m v a o n t a c 6 : l = 2k + l (**) A . 2,4s. B. 2,1s. C. 2s. " D . 2,0625s. -> De CO song d i r n g tren day v o i 13 b u n g va 13 n i i t t h i k = 12. Huang dan gidi: K h i d o tir (**) ta c6: / = 2k + 1 = 2.12 + 1 = 25(cm). - » C h p n D . . - T h o i gian t u k h i ehie'c bat r o i den k h i cham vao m§t n u o c la Dang 5: B A I T A P V E S O N G A M * Phttang phdp: - T h o i g i a n t u liic bat cham vao mat nuoc den k h i nghe thay tieng bat - Cac bai toan t h u o n g gap o dang nay la xac d i n h cac dai lug-ng dac trung h 20 cho song am n h u van toe am, m u c c u o n g dp am, tan so a m , tan so cac hoa cham vao mat n u o c la t , = — = = 0,0625s am ... V 320 Cac cong thuc t h u o n g su d u n g la: - V a y tong t h o i gian la t = t j + t j = 2,0625s -> C h p n D . V i du 4: M p t n g u o i d u n g biia go nh? vao d u o n g sat va each d o 1376m, eo - V a n toe: v = — = k.f; v = - mpt n g u o i ap tai vao d u o n g sat t h i nghe thay tieng go som h o n 3,3s so v o i T t E E P tieng go nghe t r o n g k h o n g k h i . V a n toe am t r o n g k h o n g k h i la 320m/s. Van - C u o n g d g am: I = —- = r- = r toe am t r o n g sat la S.t 4:iR2t 47IR2 I I A . 1238m/s. B. 1356m/s. C. 1336m/s. D . 1376m/s. - M u c c u o n g d o a m : L ( B ) = l o g — hay L ( d B ) = 101og — Huang dan gidi: ^0 Vi 1: M o t song am d u p e phat ra t u m o t n g u o n a m c6 d a n g h i n h cau. Biet - G p i t la t h o i gian n g u o i kia nghe dupe tieng go t r o n g d u o n g sat ke t u tai d i e m each n g u o n 2 m , c u o n g do a m la 0,06W/m2. c i a s u nang l u g n g phat flic go. ra dupe bao toan. Xac d j n h cong suat phat am ciia n g u o n . * - V o i v J , V2 la v a n toe am t r u y e n t r o n g sat va t r o n g k h o n g k h i A.5W B.4,5W C.4W D.3W Ta c6: V j t - V2 ( t + 3,3) = 1376 ^ t = I s ; v i = 1376m/s. Chpn D. Huong dan gidi: V i d^ 5: M p t n g u o i d u n g gan 6 chan n i i i h i i len m p t tieng. Sau 8s nghe tieng N a n g l u p n g dupe phan bo deu tren dien tieh mat song la mat eau: S = 47:R m i n h v o n g lai, biet van toe am trong k h o n g k h i la 340m/s. K h o a n g each t u C o n g suat phat am la: P IS = 0,06.4TT.2^ « 3 W -> C h p n D . chan n i i i den n g u o i d o la V i du 2; M p t la thep m o n g , m p t dau eo d j n h , dau eon lai d u p e k i c h thich de _ A . 1333m. B. 1386m. C. 1360m. D . 1320m. dao d p n g v o i ehu k i k h o n g d o i va bang 0,08s. A m d o la thep phat ra la Huang dan gidi: A . sieu a m . B. ha am. Gpi / la k h o a n g each t u n g u o i d o den chan n i i i . C. nhae am. D . am ma tai n g u o i nghe dupe. Huang dan gidi: Theo de ra v t 21 => 1 = y = 1360m => C h p n C 12,5HZ Tan so cua am phat ra la f = ^1 = 12,5Hz V i du 6: H a i a m eo m u c c u o n g d p am h o n k e m nha 40dB. T y so c u o n g dp T 9ni cua c h i i n g la: A m do la thep phat ra la ha a m C h p n B. _ A . 400 B.100. C. 40 D . 10000
Huang dan gidi: ' Huang dan gidi: T a c6: L = l O l o g — v 6 i I i t h i L j = l O l o g ^ (*) Cac dap an B, C, D deu d i i n g . Chon A ^0 ^0 Ket l u a n nao sau day khong diing ve s u t r u y e n song co A. Song n g a n g la song m a p h u o n g dao d p n g ciia p h a n t u m o i t r u o n g v o i h t h i L2 =101og^ (**) ^ v u o n g goc v o i p h u o n g t r u y e n song. B. Song dpe la song m a p h u o n g dao d p n g ciia p h a n t u m o i t r u o n g t r u n g T u n va D - ) . L2 - L i = l O l o g ^ - lOlogiL = l O l o g ^ v o i p h u a n g t r u y e n song. ^0 ^1 ,i C. Song t r u y e n t r o n g m o i t r u o n g ran l u o n la song dpe. V i L 2 - L i = 40 -> 40 = l O l o g - ^ l o g ^ = 4 =^ ^ = l O " = 10000 ^ C h p n D. D. Song t r u y e n t r o n g m o i t r u o n g k h i l u o n la song dpe. ^1 ^1 h Huang dan gidi: V i dy 7: H a i hoa am lien tiep do mq>i day d a n phat ra c6 tan so h o n kem T r o n g m o i t r u o n g ran g o m ca song dpe va song ngang. -> C h p n C . nhau 56Hz. H o a a m t h i i 4 c6 tan so: C g u S : V a n toe t r u y e n song p h u thupe vao yeu to nao? A.28HZ B.224HZ C. 86Hz D. 168Hz Huang dan gidi: A. N a n g l u p n g song. B. T a n so dao d p n g . Theo bai ra: n f - (n - l ) f = 56 - » tan so a m co ban f = 56Hz C. M o i t r u o n g t r u y e n song. D. Buoc song. Vay tan so hoa a m t h u t u f4 = 4f = 4.56 =224 H z . -> C h o n B. Huang dan gidi: Van toe t r u y e n song p h y thupc vao m o i t r u o n g t r u y e n song. - > C h p n C III. B A I T A P L U Y E N T A l ' C O H l / ^ N G D A N C A C H G I A I Cau 6: M o t song co hpc co bien d p A, buoc song X. Toe d p dao d p n g eye dai C a u 1: Song n g a n g t r u y e n dupe t r o n g cac m o i t r u o n g nao sau day : ciia p h a n t u m o i t r u o n g bang 3 Ian toe dp t r u y e n song. T i m h ^ t h u e lien h? A. Ran, l o n g . giua A va X. B. L o n g , k h i . A.A=2TIA/3. B . A = 27XA. C . A = 37TA/4. D . A = 3nA/2 C. Ran va m a t tren m o i t r u o n g chat long. Huang dan gidi: D. RMn, l o n g , k h i . T ' A o^ -1 AcoT 27iA ^. . Huang dan gidi: Ta CO A c j = 3— =>X = => C h p n A Song ngang t r u y e n d u p e t r o n g m o i t r u o n g ran va tren m a t chat long. T 3 3 => C h p n C 5. i X a u 7: N g u o n phat song S tren mat nuoe tao dao d p n g v o i tan so f = lOOHz gay C a u 2: Song dpe d u p e t r u y e n t r o n g m o i t r u o n g nao d u o i day: cac song t r o n Ian r p n g tren mat nuoe. Bie't k h o a n g each giira 7 g p n l o i lien A. RSn, long, k h i . B. R^n, long. tiep la 3 c m . Toe d p t r u y e n song tren m a t nude bang C. K h i , chan k h o n g . D. R^n, chan k h o n g . A. 25 cm/s. B. 50 cm/s. C. 100 em/s. D . 150 em/s. Huang dan gidi: Huang dan gidi: Song dpe t r u y e n d u p e t r o n g cac m o i t r u o n g rin, long, k h i . Chpn A K h o a n g each giCra 7 g p n l o i lien tiep la 3 cm: 6A = 3 em => A = 0,5 c m Cau 3: Ket l u a n nao sau day khong dung ve s u t r u y e n song co Toe d p t r u y e n song tren m a t nuoe bang v = A . f = 0,5.100 = 50 cm/s. A. Q u a t r i n h t r u y e n song k e m theo s y v a n c h u y e n vat chat theo p h u o n g => C h p n B t r u y e n song. Cau 8: M o t d i e m O tren m a t nuoe dao d p n g v o i tan so 20 H z , v g n toe t r u y e n B. Qua t r i n h t r u y e n song k e m theo s u t r u y e n nang l u p n g t u n g u o n t o i Song tren m a t n u o e thay d o i t u 0,8 m/s < v < Im/s. T r e n m a t n u o e h a i d i e m A n h i j n g cho t r o n g m o i t r u o n g ma song t r u y e n t a i . Va B each n h a u 10 c m tren p h u a n g t r u y e n song l u o n dao d p n g n g u p c pha C. Q u a t r i n h t r u y e n song la qua t r i n h t r u y e n pha dao d p n g . ' A a u . Buoc song tren m a t n u d e la: D. V ^ n to'c t r u y e n song la v a n toe t r u y e n pha dao d p n g . A. 4 cm. B. 16 c m . C. 25 cm. D. 5 cm. .,•
1^1 inuttigminnuTiHBins varu ii, rapi-Nguym nuangLac Huang dan Huang dan gidi: D p l§ch p h a d a o d p n g c u a h a i p h a n t u tren p h u o n g t r u y e n s o n g e a c h nhai. B u o c s o n g la A = y ; v k h o n g d o i n e n k h i t a n g f l e n 3 I a n thi b u o c s o n g g i a m d la: A(p = ^ | ^ ( r a d ) . T h e o g i a thiet 3 Ian. => C h p n B. 2 Ted g^ajH- P h a t b i e u n a o s a u d a y v e d a i l u p n g d a c t r u n g c u a s o n g c o h p c la 2d.f ^ ^, 2d.f m A(p = 71 + k.2n => = K + k.2.7t = : = l + 2k=> V = khong diing? l + 2k l + 2k g A. C h u k y ciia song chinh bang c h u ky dao dpng cua cac p h a n t u dao dpng. M a 0,8 m/s < V < 1 m/s -> 0,8 m/s < B. T a n so c i i a s o n g c h i n h b 3 n g tan so d a o d p n g c i i a c a c p h a n t u d a o d p n g . < l m / s ^ l , 5 < k < 2 ^ k = 2 l + 2lc C . T o e d p c u a s o n g c h i n h b a n g toc d p d a o d p n g c u a c a c p h a n t u d a o d p n g . ^ A = 0,04 m = 4 c m . ^ Chpn A D. B u o c s o n g l a q u a n g d u o n g s o n g t r u y e n d i d u p e t r o n g m p t c h u k y . C a u 9; M p t s o n g c o h p c I a n t r u y e n t r o n g m p t m o i t r u o n g v a t c h a t tai m p t diei Huang dan gidi: T r o n g c a c d a i l u p n g d a c t r u n g c u a s o n g thi: C h u k y c i i a s o n g c h i n h b a n ? each n g u o n x (m) co p h u o n g trinh song: u = 4cos c m . V a n toc 3 6 chu k y d a o d p n g c u a c a c p h a n t u d a o d p n g ; T a n so c i i a s o n g c h i n h b a n g tan so t r u y e n s o n g t r o n g m o i t r u o n g co g i a tri: dao d p n g c u a c a c p h a n t u d a o d p n g ; C o n toe d p t r u y e n s o n g l a toc d p t r u y e n A . 2 m/s. B. I m / s . C . 0,4m/s. D . M p t g i a tri khac. pha d a o d p n g v a k h o n g b a n g toc dp d a o d p n g c u a c a c p h a n t u d a o d p n g . . Chpn C. .a Huong dan gidi: C a u 14: M p t s o n g c o I a n t r u y e n tren m p t d u o n g t h a n g t u d i e m O d e n d i e m M Ta C O to = — ; T = 6s;A = 2 , 4 m ^ v = ^ = 0 , 4 m/s each O m p t d o a n d. Biet tan so f, b u o c s o n g X v a b i e n d p a c u a s o n g k h o n g doi 3 6 trong q u a t r i n h s o n g t r u y e n . N e u p h u o n g t r i n h d a o d p n g c u a p h a n tit v a t c •=> C h p n C tai d i e m M co d a n g U j ^ ( t ) = a sin27rft thi p h u o n g t r i n h d a o d p n g c u a p h a n ti'*- C a u 10: K h i m p t s o n g c o h p c t r u y e n t u k h o n g k h i v a o n u o c thi d a i l u p n g nao sau day khong thay doi: V9t c h a t tai O l a : A . V a n toc. B. T a n so'. C . B u o c song. D. N a n g lupng. B. U Q ( t ) = a s i n 7 t ft + - A. U Q ( t ) = a s i n 7 i Huang dan gidi: X V a n toc t r u y e n s o n g t r o n g c a c m o i t r u o n g k h a c n h a u l a k h a c n h a u ; d o vay C. Uo(t) = asin27t f t . ^ D. U Q ( t ) = asin27t k h i m p t s o n g c o t r u y e n t u m o i t r u o n g n a y s a n g m o i t r u o n g k h a c thi v a n toc X thay d o i , b u o c s o n g v a n a n g l u p n g c u n g t h a y d o i c h i co t a n so l a k h o n g thav Huong dan gidi: n o i . => C h p n B. P h u o n g t r i n h d a o d p n g c u a p h a n t u v a t c h a t tai O s o m p h a h o n tai M la C a u 11: M p t s o n g c o co tan s o f, t r u y e n t r o n g m o i t r u o n g v a t cha't d a n h o i , voi t= ^ v a n toe l a v thi b u o c s o n g la P h u o n g t r i n h d a o d p n g c i i a p h a n t i i v a t c h a t tai O l a A . ? . = vf. B.X = 2vf. D.A=--. V f U Q (t) = asin27tf t+ - = a s i n 271 = a s i n 271 ft + - Huang dan gidi: V V X T a CO buoc song A = j => C h p n D . => C h p n C . £|u_15: M p t n g u o n p h a t s o n g d a o d p n g v o i p h u o n g t r i n h u = a s i n 2 0 7 t t ( c m ) C a u 12: M p t s o n g c o h p c I a n t r u y e n trong m o i t r u o n g d a n h o i v o i v a n toc ^ o i t tinh b a n g g i a y . T r o n g k h o a n g t h o i g i a n 2s s o n g n a y t r u y e n d i d u p e q
Huang dan gidi: J a u 20: K h o a n g each g i i i a hai n g o n song lien tie'p tren m a t ho la 3 m . T r o n g to jyipt p h i i t song d a p vao b o 7 Ian. V a n {oc t r u y e n song la Tan so song la f = — = l O H z ; V a n toe song la v = A.f = lOX 271 A . 0,6 m/s. B. 0,2 m/s. C. 0,5 m/s. D . 0,3 m/s. ' Q u a n g d i r o n g la S = v.t = 10X.2 = 20X Chpn D . ci JW/fv Huang dan gidi: C a u 16: M p t n g u o i q u a n sat m p t chiec la tren m a t n u o c thay n o n h o cao 8 JaY, C h u k i eiia song la T = — = 10s; V a n to'c t r u y e n song v = — = 0 , 3 m / s trong 14 giay. K h o a n g caeh g i u a hai n g p n song lien tie'p la 2 m . V a n to'c truyen 6 T song la Chpn D . A. Im/s. B. 2m/s. C. 3m/s. D . 4m/s. Cau 21: Song tai m p t d i e m M dao d p n g v o i chu k i T = 2s, v a n to'c t r u y e n song la -» Huang dan gidi: V = 0,5 m/s. K h o a n g each giiJa hai d i e m ngupe pha nhau gan nha't la Chiec la n h o cao 8 Ian, tuc la thuc h i ^ n d u p e 7 chu k i t r o n g 14 giay A . 1,5m. B. I m . C. 2 m . D . 0,5 m . => C h u k i dao d p n g eiia chiec la la T = 2s; Huang dan gidi: V a n to'c t r u y e n song la v = A/T = 1 m/s . => C h p n A . Buoc song A, = v T = l m C a u 17: M p t song co hpe t r u y e n t r o n g m o i t r u a n g d a n h o i v o i v a n toe 3m/s va Khoang each giij-a hai d i e m dao dpng ngupe pha gan nhau nha't ^ = 0,5m chu k i 0,5s. Buoc song cua no la A . 1,5m. B. 3 m . C. I m . D . 2m. => C h p n D Huang dan gidi: Cau 22: M p t spi day dan h o i n a m ngang co m p t d a u co' d i n h , m p t d a u dao Ta C O b u o c song la A. = v T = 1,5m dpng v o i bien d p a = 10cm va chu k i T = I s . C h p n go'e t h o i gian t = 0 k h i vat d i => C h p n A . qua v j t r i co l i d p bang 0 theo chieu d u o n g . P h u o n g t r i n h dao d p n g eiia song la C a u 18: P h u o n g t r i n h n g u o n song tai O la UQ = 4COS lOOTtt-- (cm). Bie't van A . u = 10eos(27it - 7i/2) (cm). B. u = 10cos(27:t - n/2) ( m ) . 2 C. u = 10cos(27it + n/2) (em). D . u = 10cos(27it + 7i) ( m ) . to'c t r u y e n song la v = 5m/s. P h u o n g t r i n h song tai M each O m p t d o ^ n 5cm la , Huang dan gidi: A. Uj^ = 4 cos l O O T i t - - B. U j ^ = 4 c o s ( l 0 0 r t t - 7 i ) T a n s o a) = - ^ - 2 7 i ( r a d / s ) 2 C. u ^ = 4 cos 1007tt + - D . u ^ =4cosl007it Khi t = 0 : U(3 = 0; Vj) > 0 . Ta CO (p = - | 2 P h u o n g t r i n h dao d p n g eiia song la u = 1 0 c o s ( 2 7 r t - T c / 2 ) ( e m ) . C h p n A . Huang dan gidi: P h u o n g t r i n h song tai M la £ a u 23: T r e n mat m p t chat l o n g tai O co m p t n g u o n song co dao d p n g v o i tan 1 3n so f = 30Hz, van toe t r u y e n song la m p t gia t r j t r o n g k h o a n g l , 6 m / s < v < u ^ = 4cosl007i t - - = 4 cos lOOTlt- = 4 cos 1007tt + - cm 200 500 2) 2,9m/s. Bie't tai d i e m M each O m p t k h o a n g lOem, song tai d o l u o n dao d p n g nguoc pha v o i dao d p n g tai O . Gia t r i van toe do la , => C h p n C. A . 2m/s. B. 4m/s. C. I m / s . D . l,2m/s. C a u 19: M p t song c6 p h u o n g t r i n h u = 3eos(57tt + 0,047:x) ( e m ) , t r o n g do > Huong dan gidi: (i d u p e t i n h b a n g c m , t d u p e t i n h bang giay. Buoc song eiia song do la - Dao d p n g tai M l u o n n g u p e pha v o i dao d p n g tai O nen A . 70cm. B. 40em. C. 60cm. D . 50cm. 27td 2df _ 6 fm^ Huang dan gidi: A(p = 7t + 2k7r = => V = 2k + l ~ 2k + l l s 2nx T a c o 0,04Ttx = - X = 50cm Chpn D .