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 A? + A^ + A^ => A2 = Vai may FX570-ES ta bam n h u sau : [ 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 ,y Mat khac: - A ^ = V42 -2^ = 2V3 (cm) A^ = A^ + A^ + 2A^Aj cos(92 - 9]) |SHIFTJ|MOD£| § chuyen doi don vi goc la R (Rad): fi! |4>/3|^HIFTJ § g § ISHIFTJ [SHIFT] § '^^'^ ^^^^ H • '^^t • o j h li jisv'v' :jr 2^ =4^+(2V3)%2.4.2N/3COS Z 7t/2 cos Nghia la bien bang A = 4; pha ban dau cf) = n/2 => Chon A Cau 146 Mot vat thuc hi^n dong thoi hai dao dpng dieu hoa cung phuong c6 phuong trinh dao dong la X j = s i n ( t t + a) (cm), = 4\/3cos7ct (cm) Bien 371 D a = T: C.a = Huong dan gidi: Ta c6: x, = cos Ttt + a — (cm) va X j =4\/2cos(27tt) (cm) Bien dp dao dong tong hop xac djnh: +2AjA2COS a — De bien dao dong dat gia tri nho nha't thi cot-^ 3j cm; X2 = A2cos(o)t + (p2)cni iML.'''i^ \ B Tan so ciia goc dao dpnh tong hpp co = 27t rad/s , ; C Pha ban dau cua dao dong tong hpp = — D Phuong trinh cua dao dpng tong hpp x = 8cos cm Huong dan gidi: Taco: x, = V s i n i t ( c m ) va X j = 4\/2cos27tt (cm) A Bieu dien cac dao dong tren gian vec to, ta c6: OA = A i = 4cm; OB = A2; OC = A = 2cm dap an B d i i n g 2 A - A ^ + A ^ + A J A C O S ( - I ) = (4V2) + ( V ) + ( ) c o s ^ => 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 : A , sin(p, + A , sin(p, — = -1: (p = — dap an C diing jf' A^ C O S ( p j + A2 COS(p2 C sVScm; Huong dan gidi: BC2 = O C + OB2 A Bien dp cua dao dpng tong hpp la A = 8cm tgcp = — • Cap gia tri nao cua A2 va cp sau day la dung? T u gian vec to suy ra: 0ii_148 M o t vat thuc hien dong thoi dao dong c6 phuung trinh la Bien d p c i i a d a o d p n g tong hpp x a c d j n h : Phuong trinh dao dong tong hop x = 2cos((ot + (p)cm Trong cpj -(p = B 2V3cm; ' VP 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 Cau 147 Mot chat diem tham gia dong thoi hai dao dpng dieu hoa tren cung A sVScm; 2 tong hpp cOng co tan so goc (a = 2n rad/s cos a — = - l = > a - — = 7t=:i>a = — ^ Chpn A 2 xi = 4cos 3; Vay A2 = 2\/3(cm); (p = => Chpn D n A^ = A j + A j + A j A j cos((p2 -(pj) = A j + A j mot true Ox c6 phuong trinh: •cp = 6J I X j = 4^2 sin27it(cm); X2 = 4>/2 cos27it (cm) Ket luan nao sau day la sai? , > dao dong tong hop dat gia tri nho nha't A a = 571 71 n ^T^^-> D 2V3cm;0 V5 / Phuong trinh dao dpng dieu hoa tong hpp co bieu thuc: X = 8cos 27rt-i^ c m d a p a n D s a i -> C h p n D V o i may FX570-ES ta bam n h u sau : 7- P O D E ] ^ tren man hinh xuat hien chu: CMPLX | H I F T ] [ M O D E J § chuyen doi don vi goc la R (Rad): f72 SHIFT (-) -7t/2 SHIFT (-) 161 - SHIFT] y g y M a n h i n h hien t h i ket qua: Z - t / A=(V3B)^-4(52-25)>0 NghTa la bien d o bang A = 8; pha ban dau 4> = - H / 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 C O bien d o Ian lug-t la A i = 3cm va A = 4cm Bien dao d g n g t o n g h o p khorig the nhan gia t r i nao sau day: A 5,7cm B 1,0cm C 7,5cm Huong dan M p t lac 16 xo t h a m gia d o n g t h o i dao d p n g d i e u hoa c i i n g 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 dao dpng p h a n la A i = c m va A Biet d p I o n van toe ciia v a t tai t h o i d i e m D 5,0cm (Jpng nang bang the nang la 40( cm) Bien dp p h a n A bang gidi: 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 = cm B A2 = 4V3 l u o n phai thoa m a n : | A j - A j | < A < A j + A j Theo bai ra, ta c : c m < A < 7cm D ±407i cm/s (0 COS(A(P) = 75 => A = V c m a' V = ±4071 ( c m / s) gidi: • A = 8cm X = 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 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 = c o s ( t + (pj)cm; = c o s ( t + (p2)cm v o i < (P2 -(Pj ^ n Biet p h u o n g t r i n h dao d p n g tong x = 2cos hop la X = 5cos(cot + (p)(cm) Bien dp dao dpng B c6 gia tri cue dai k h i A bSng: B cm , C Syflcm Hwang dan D 2,5V2cm Tir cong thiic tong h o p dao dpng ta c6: 571 c m Gia t r j ciia cpt la: 'ff': B 66 ' Huang dan gidi: Gian d o vecto n h u h i n h ve Ox la tri^c goc: gidi: 5^ = A ^ + B^ - A B C O S — hay A ^ - VSBA + B^ - = 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 : Al=A^-Al=8^-A^=>A2=4Scm => C h p n B Chpn D CO ^ 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 A 573cm = cm A ^ = A J + A ^ + 2A1A2 C O S ( A = A ^ + A ^ ) ^ CO A2 gidi: A p d y n g cong thuc: v2 D M a t khac ta co: A^ = A^ + A^ + 2A1A2 2v^ A^=2 Ta c6: co = 27:f = 871 (rad/s) ,2 = \/3 c m E = 2W^ = i k A = > - m v = i k A d 2 Van to'c ciia vat k h i no c6 gia toe 32>/2 (cm/s^) la Huong dan A2 K h i d p n g nang bang the nang, ta co: c i m g tan so' i = H z , c i i n g bien d o cm va c6 d o lech pha A(p = — Cho 7t^ = 10 C ± ^ cm/s 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, B ± ; : cm/s cm Huang dan Vay A k h o n g the bang 7,5cm => Chpn C A ±307c cm/s B T a m giac O A A can tai O => O A A = O A A O A la p h a n giac A j O A ( V i O A A A la h i n h thoi) Q 2- Thai gian chuyen dpng thling cua v|t m t u liic ban dau den v i tri 16 xo AOA2 = AOAj = OAA2 l^'hong bien dang la: => A O A , = O A A = O A A A A O A deu ^ ', A O A = - =i> AOA, A = - ' Theo bai: AOx = — => A^Ox = — =>(p, = - — ^ ^ kx = fjmg => X = |Limg/k = (cm) giay Biet dp Ian van toe ciia vat tai thoi diem dpng nang bang the nang ij ,^ B 2V3cm C 373cm D 4V3cm Thai gian chuyen' dpng thSng ciia vat m t u liic ban dau den v i tri 16 xo ' ^ ; rSu 157 * Mot lac 16 xo gom mot vat nho kho'i lupng 0,02kg va c6 dp cung Ion nhat vat nho dat dupe qua trinh dao dpng la A 1572cm/s • A = 8cm B 25V2em/s C 40^^ cm/s D 50v^em/s Huang dan gidi: Do hai dao dpng phan vuong pha nen Van toe ciia vat dat gia trj Ion nhat qua trinh dao dpng tat dan ehinl" A = ^A^ - A j = 78^-4^ = 4V3cm => Chpn D v$n toe Ian nhat ma vat dat dupe i chu ki dau tien Gpi x la v i tri l i Cau 155 Mot lac 16 xo dao dpng tat dan Cu sau moi chu ki bien dp ciia no giam 2,5% Phan nang lupng cua lac bj mat d i sau moi dao dpng toan phan la ' '^ •- bj nen 10 cm roi buong nh^ de lac dao dpng tat dan Lay g = lOm/s^ Toe dp 2imv2=ikA2 ^ "^'•> H? so ma sat trupt giiJa gia da va vat nho la 0,1 Ban dau g i u vat v i t r i 16 xo Khi dpng nang bang the nang ta c6: 2Ed = E A 35% T = t J ^ - 0,2n (s) N/m Vat nho dupe dat tren gia c6 dinh nam ngang dpe theo trvic 16 xo Huang dan gidi: k Chu ki dao dpng: khong bien dang la: t = j + - ^ = -^(s) => Chpn C 40cm/s Hay xac djnh bien dao dpng phan Aj ? , D ^- s 30 C.^s 15 Vi tri can bang ciia lac 16 xo each v i tri 16 xo khong bien dang x: Chpn B =4cos^5>/2t-njcm va X = A C O S 5\ll\ — c m , t tinh bang A VScm • Huong dan gidi: 7t Cau 154 Mgt vat tham gia dong thai hai dao dong dieu hoa cung phuang v6, Xj B.^s 20 25S B 45% C 55% D.95% Huang dan gidi: 2 Nang lupng chu ki dau la: W = - mco A Nang lupng chu ki tiep theo la: W ^mco^A'^ A/ = (0,975)^ * 0,95 = 95% =^ Chpn D Suy ra: w A Cau 156 * Mot IMc 16 xo c6 dp ciing k = 10 (N/m), khoi lupng vat nan? 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 lac va mat ban bang H * ma vat c6 toe dp Ion nhat Ap dung dinh luat bao toan nang lupng, nang lupng ban dau ciia h§ gom the nSng cue dai ban dau: ~ k A p , nang lupng ciia h§ tai thoi diem vat c6 van toe c i ; ?i gom dpng nang ^ m v ^ , the nang ^^x^ va dp Ion eong ciia lye ma sat: = ^mg Al = |img(Al + x) kx^ +)img(Al + x) -kAl^ =-mv^ + — ^ J i A l - A x - ^ g ( A l + x) m V / m Dat y = v = — A l ^ - — x - ^ g ( A l + x ) X e t d a o h a m y' = - — x - ^ i g m m m y' = » X = - ^ ^ = -0,02m = -2em ' ,, Ki thuatgilii iiluiiih BI'rN Vat Li 12, t&p - Nguyen Quang Lac Cfy TNHH K h a o sat svt bien thien ciia y ta tha'y v o l x = - 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 cong ciia l y c m a sat Suy v dat gia t r j eye dai tai v i t r i x = - c m Do do: A ^ ^ = Hmg.s = W => 0,02.0,1.10.5 = 0,5 => S = ( m ) =>ChpnB v ^ ^ = J — A l ^ - — x - n g ( A l + x) = , V m / s = 40N/2cm / s ' ' £^vi 161 M p t 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 Cau 159 M p t tarn v a n bac qua m p t m u o n g c6 tan so' dao d p n g rieng bang 0,5 H z M p t n g u 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 r u n g len m a n h nhat? A buoc B buoc C buoc 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 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 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 A 22,93em/s B 25,48em/s C 38,22cm/s d u o n g la chieu dan ciia 16 xo K h i vat chuyen d p n g theo chieu am: - k x + famg = ma = m x " -k r _^mg^ - m X \ lOOgphang 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 k = 0,02m = 2cm; co = j — = lOrad / s Vm X - = 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 = -> xo = c m => = Aeos cf) vo = han la: => = -lOasincp => cp = 0; a = em A 25 c m B.25m C 250 cm Huang dan gidi: C o nang ciia vat la: W = ^ k A ^ = ^.100.0,1^ = , ( j ) D.250m D 28,66cm/s 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: Cau 160 M p t 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 tien la: Huang dan gidi: D buoc p h a i bang tan so'dao d p n g rieng ciia t a m v a n N G p i N la so buoc chan: => — = , = > N = b u o c chan => C h p n B tu thoi d i e m tha den t h 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 => X - = 4cos(10t) (cm) K h i 16 xo k h o n g bien dang: x = =:> coslOt = -1/2 = cos27i/3 t = n/15 s Vtb = 7t/15 [jjen '^^ 90 a 28,66 cm/s 3,14 ila A 0,05 => Ch(?n D Cau 163 Mot lac 16 xo dao dong tat dan cham Cu sau moi chu ki, bien dao dong cua no giam 0,5% Nang lugng dao dong cua lac bi mat di sau moi dao dong toan phan la A 0,5% vau B 1% ca/.n: Ta c6: ^ A r C 1,5% ;r V Huangddngidi: = 0,5% => - — = 0,005 A ,;,„^_ C 0,005 Huang dan gidi: D 0,0025 Gpi A i , A la bien dp dao dpng tai hai thoi diem each nira dao dpng D 2% £)p giam CO nang cua lac: AW = ^ k ^ A j - A j , , , ,,,, Congcua lire ma sat: AA^^, = F ^ , { A J — = 0,995 A + A2) = nmg(Ai + A2) Do AW = A A ^ , = > A j - A = ^ ,!,v Nang chu k i tie'p theo: W = — mco A W B.0,025 f a tim bieu thue ciia dp giam bien dp sau moi dao dpng toan phan A2 Nang lugng chu k i dau: W = - mco2 A Suy ra: dpng toan phan va h$ so'ma sat giiia qua cau voi day kim Dp giam bien dp moi dao dpng: AA = ( A J - A ) = 4|img )' ^ ' Theo bai ra: 200AA = AQ = 2cm kAA = (0,995^ « , 9 = 99% Tudo: AA = r T r = 0,01cm va n = - ^ = 0,005 200 4mg A Chpn C 1< Vay nang lug-ng bi mat di la 1% => Chon B CSu 166 Mot lac 16 xo dat tren mat phSng nghieng goe ao = 60" so voi mat Cau 164 Mot lac don c6 dp dai lo = 16cm dupe treo toa tau a 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 van toe cua lupng m = 400 (g), lay g = 10 m/s^ Cong suat can cung cap cho lac de no doan tau bang 15 m/s Lay g = 10 m M 7t^ = 10, coi tau chuyen dpng thang dao dpng dieu h6a voi bien dp A = 5cm va tan so f = Hz la: deu, chieu dai moi ray bang A 12 m B 16 m A P = 180(mW) C 18 m C.P = 220(mW) D P = 240(mW) Huang dan gidi: D.24m Huang dan gidi: Sau 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 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) Nang lupng can cung cap sau moi chu ky la: Chu ki dao dpng rieng ciia lac: TQ = 27t J — VS Con lac dao dong manh nhat T = B.P = 200(mW) §iif4 27iJ— = - ==> = 2n,:^.v = 12m IA^, I =0,04 0) Cong suat can cung cap: P = ^ p=> => Chpn A AW = ,1 = AW.f = 0,04.6 = 0,24(w) = 240(mW) Chpn D Cau 165 Mot 16 xo c6 dp cung k = 600N/m, mot dau co'dinh, dau gan qun J67 Mot dao dpng tat dan c6 bien dp giam 2% sau moi chu ki Sau 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 khoi vj tri can bang 2cm roi tha cho qua cau dao dpng Do c6 ma sat nho, dao dpng ch^m dan, sau 200 dao dpng thi qua cau dung l ^ i Lay g = 10m / Dp giam l-A 17,3% B 28.9% C 73,2% D 81,5% Huang dan gidi: 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 lai Wi: Q^aJTl: M p t 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 w 221 = (l-0,02) -1-0,04 A Is B 0,5s C 2,2s , „ , D 2s Huong dan gidi: = ( l - , ) W = 0,96W Chu k i dao d p n g cua eon lac T = 2nJ- Sau hai chu ky, ca nang lai: = 0,96Wi = {0,9ef = 2nP^ Vg W V = 2.1,1 = 2,2s 7t2 => C h p n C , Sau n a m chu k y , ca nang 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 = ( , f W = 0,815W = 81,5%W =^ Ch(?n D ' ' Cau 168 M o t lac d o n g o m vat c6 k h o i i u a n g m , day treo c6 chieu doi = I m Keo 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 nang la 0,18 J (moc the nang tai vj t r i can bang); lay A cm, v ti so d p n g nang va the nang la B C 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 , % t r o n g l u g n g ciia vat Q u a n g d u o n g ma =10 T a i l i d p 372 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 A 4m B m C m D m 2 Co nang W = => A = 0,06m = 6cm K h i d ^ = ^ = ^ 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: Wo = W + = i A«n W, W, x2 => Wo = Anwx = Fcan.Sm.nx => C h p n D ^ Smax = 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 mgl"° ^ = 500.1.ao = ( m ) => C h p n B mg u V / 1000 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 bang m p t nua d p Ian gia toe cue dai la A 0,083s B 0,125s Cau 169: M o t 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 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 l u p n g m ' t h i lac dao d p n g v a i chu k i 0,5s Gia t r i m2 bang A lOOg B 150g C.25g = 0,5Ti => c ^ | ^ = 0,5.27i =^ m ' = ^ Ta CO tai v j t r i l a i = , amax t h i 1x1= 0,5.A K h o a n g t h o i gian ngan nhat tir '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 = 75gam Chpn D •^Png eo bien d p bang A.7em B 11 em C.17em Huang dan gidi: d u a n g vat d i d u p e t r o n g 4s la: B 16 c m Huong dan gidi: £ | 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 Cau : 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 cm C 64 c m D 0,167s X = A den X = 0,5.A la t = T/6 = 0,5/6 = 1/12 = 0,083 ^ C h p n A D 75 g Huang dan gidi: Ta c6: C 0,104s 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 : D 23 em end A = A I + A | =17cm =>ChonC Dt - ' ' S O N G C O L Cau 175: H a i 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 lac dang 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 cho hai l i e dao I Song CO hpc va cac dac t r u n g ciia song d o n g dieu hoa v o i c i i n g bien goc, t r o n g hai m a t p h a n g song song v d i ^) 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 treo song song Gia t r i At gan gia t n nao nha't sau day? A 8,12s B 2,36s C 7,20s >.„ D 0,45s Huang dan gidi: PT dao d p n g xi, X : xi = A cos a),t ^ — ; tmin trtfong vat chat theo t h o i gian - 7t A /• = COS (0,t - C em D 12 cm chu k i 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 - X = cos x t + ^ (cm) (p = 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 CO song t r u y e n qua ^ = ~ D X = 5cos diem d o k h i song t r u y e n qua Thong t h u o n g cang xa t a m tao song t h i bien (cm) = n (rad/s) K h i t = vat d i qua can bang O theo chieu d u o n g : x = va v > => eoscp = va sintp < 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 B X = 5cos r t - ^ (em) 2j Huang dan gidi: Ta CO A = 5cm; co = 2n/T= lull 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 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 C Song ngang: la song c6 p h u o n g dao d p n g eua eae p h a n t u m o i t r u o n g 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 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 em, (em) \nrn' C h p n A ;j'.) - ^ = -^ 2r 1^2 n + ' < Buoc song dai nhat Amax = 21 k h i k = (chi co bo song) xen ke n h a u Ne'u hai n g u o n dao d p n g cung pha t h i d u o n g t r u n g t r y c eiia 51-^' 17A 177 - D a y d a i /co d i n h rriQt dau, m p t d a u t y do: I = ( k + ) — (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 + S6'nut:k + d) Do van toe truyeh - N e u d u n g d o n v i la dexiben (dB) t h i : L = l O l o g — f'j^L- IQ song ^) Cac dac trtmg sinh U cua dm 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 > ^ p 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 d u p e h i e u qua s u t r a m hay b o n g cua am - a) Song + dm Song a m la song co hpc ma tai n g u o i c6 the cam n h a n d u p e Song am 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 phat song a m - 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) - Cac a m co tan s o d u o i 16 H z gpi la am (tai cho cam t h u dugfc) b) M6i trubng truyen dm vd van toe truyen - dm 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, 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 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 l i f u each am ' - 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 {pxbng d p a m va tan so am) Song 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 Gia t r i c u o n g d p am I be nha't ma tai n g u o i 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' + thi gpi + 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 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' 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 nghe dupe M i e n r p n g hep p h u thupc tan so' - 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 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 dpng cua a m + A m CO ban va hoa am: M p t nhac cu phat m p t a m co tan so fo (am co ban hay hoa a m t h u nha't) t h i bao g i o cung phat d o n g t h o i cac hoa am t h i i 2, 3, CO tan s6'2fo, 3fo, Do hien t u p n g do, am phat 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 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 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 + ran bao g o m ca song dpc va i o n g ngang c) Cac dac tntng vdt ly cua dm - T a n so am: f t u l H z d e ' n 20.000Hz - C u o n g d p am va m u c c u o n g dp am: + E diem I = — = S-t + ^ khae n h a u phat sac thai am t h a n h ma ta nghe khac la d o so' hpa a m toan khac + 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! 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' E P — 47tR^t u n g v o i tan so' f = OOOHz) ' ^ o n g t u a n hoan ^guon - 47iR^ 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 A m d u p e phat t u m p t tieng no, tieng go vao t a m k i m loai, g p i la 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 D o n v j : W / m ^ (oat tren met v u o n g ) M u c c u o n g d p am L la dai l u p n g bang loga thap p h a n cua ty s^' 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 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 ^ nhac dm vd hgp cong huong N g u o n nhac a m la n h i r n g n g u o n phat nhac a m , tuc la phat n h u n g CO tan so xae d j n h , m o i nhac cu la m p t n g u o n nhac am - 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 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 ... + A ^ ) ^ CO A2 gidi: A p d y n g cong thuc: v2 D M a t khac ta co: A^ = A^ + A^ + 2A1A2 2v^ A^ =2 Ta c6: co = 27 :f = 871 (rad/s) ,2 = /3 c m E = 2W^ = i k A = > - m v = i k A d 2 Van to'c ciia... = 7t + 2k7r = 27 td => V = 2df _ fm^ 2k + l ~ 2k + l l s Theo bai 1,6 m/s < v < 2, 9 m/s => 1,6 < — ^ — < 2, 9 2k+ Huang dan => k = gidi: pao dpng tai M luon ciing pha voi dao dpng tai N 27 td nen... each Si,S2 Ian l u p t la 12cm va 9cm C 13 ; 12 Huang dan gidi: 'AD-BD