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7/31/2019 mt s bi tp sng c v sng m vt l 12
1/19
Mt s dng bi tp sng c-sng m
I,C s l thuytvt l.
+,sng c l s lan truyn dao ng c trong mi trng vt cht theo thi gian
+,phn loi sng c: -sng dc: truyn c trong mi trng rn lng kh
-sng ngang ch truyn trong mi trng rn ( mt cht lng l mt trng hp c bit)
+,tc truyn sng : tc truyn pha dao ng ,truyn nng lng,ph thuc c tnh mitrng,vrn>vlng>vkh.
+, bc sng: l khong cch 2 im gn nht dao ng cng pha
L qung ng m sng i c trong 1 chu k
f
vvT
+,nng lng sng: khi sng truyn th n s truyn nng lng-khi sng truyn theo 1 ng thng th nng lng khng i => A cng khng i
-khi sng truyn trn mt phng th nng lng sng t l nghch vi khong cch.
1
2
2
1
R
R
E
E
-khi sng truyn trong khng gian th nng lng sng t l nghch vi bnh phng khongcch:
7/31/2019 mt s bi tp sng c v sng m vt l 12
2/19
1
2
2
1
R
R
E
E
+, tng tcho bin sng :-khi truyn trn ng thng: A1=A2
-khi sng truyn trn mt phng :1
2
2
1
R
R
A
A
-khi sng truyn trong khng gian:1
2
2
1
R
R
A
A
+,khi quan st c n nh sng: S=(n-1)
Tnt ).1(
+ lch pha:
d2
-khi k2 =>d=K => 2 dao ng cng pha.
-khi ).2
1().12( kdk => 2 dao ng ngc pha
-khi2
).2
1(
2).12(
kdk => 2 dao ng vung pha
+giao thoa sng: cc bn nn thuc phng trnh sng 3 dng cng pha,ngc pha,vung pha
c thm sch gio khoa.
+qu tch cc im cc i cc tiu:-cng pha=> trung tm l cc i.
-ngc pha=> trung tm l cc tiu.
Cc phng php gii nhanh bi tp c bn:Dng 1: tm sim dao ng cc i ,cc tiu trn on AB.
Cch 1: p dng d2-d1= ?
V d2+d1=AB
p dng 0
7/31/2019 mt s bi tp sng c v sng m vt l 12
3/19
Cch 2: cch nhanh hn:da vo tnh cht qu tch cc im cc i cc tiu
-A,B cng pha=>xC=2
k
XCT=42
k
-A,B ngc pha th x cc i cc tiu ngc li vi cng pha
V hnh => -AB/2K => vhnh xc nh
7/31/2019 mt s bi tp sng c v sng m vt l 12
4/19
Dng 4: tm bin dao ng tng hp v pha ban u
+,Cng bin :trong sch gio khoa c dng ny da vo pt tng hp sng m thi.
+khc bin : a v tng hp dao ng
Tm A da vo nh l cosin
Cn tm pha dao ng ly hiu
Dng 5: tm sim dao ng vi bin Ao cho trc
+Cng bin : A=Ao
Giiphng trnh lng gic ri a v dng 1
Ch : n k ng trn lng gic mi lm c dng bi ny.+khc bin : da vo tng hp dao ng nh dng 4 khc bin .
Dng 6: tnh cht sng c:-hai ngun AB cng pha:uA=uB
-hai ngun AB ngc pha: uA=-uB
- hai ngun AB vung pha: Auu BA 22
Dng 7: bi ton qung ng thi gian trong sng c hc.
n k li 2 phng php trc thi gian v ng trn lng gic tm c t ,s
Ch cng thc:s
vt
Bi ton v sng dng:
)2
2cos(.
2sin.2
lt
dauM
Khi M l bng sng => bxkdd 4).12(12
Khi M l nt sng=> nxk
dd 2
12
Ch : khi M cch B m B li l nt =>
daAM
2sin.2
7/31/2019 mt s bi tp sng c v sng m vt l 12
5/19
Khi M cch B m B li l bng =>
daAM
2cos.2
iu kin trn dy c sng dng l
+, hai u cnh2kl
K=bng=> nt=k+1 ( v hnh ra nhnhanh hn thuc cng thc nha)
+,mt u cnh 1 u t do:42
kl
K=bng=nt
Cc dng bi tp v sng m:
Kin thc c s:-sng m l s lan truyn cc dao ng m trong mi trng rn lng kh
-tai ngi ch c thnghe c cc m c tn s t16Hz n 20000Hz
f>20kHz=> siu m thanh
f h m
-sng m ch truyn c trong cc mi trng rn lng khkhng truyn c trong chnkhng
-sng m khng truyn c qua cc cht nh bng len xp
- tc truyn m gim theo th t: khlnr vvv
Ph thuc vo : mi trng truyn m, nhit , D ring mi trng.
-cc c trng sinh l ca m:
+c trng sinh l: cao , to, m sc.
+,c trng vt l: tn sv bin
Ch : +, to l mt c trng m ph thuc vo tn s m
+ to ph thuc vo tn s m v mc cng m
Cng m l nng lng c sng m truyn qua mt n v din tch t vunggc vi phng truyn sng trong 1 n v thi gian
7/31/2019 mt s bi tp sng c v sng m vt l 12
6/19
)/(4 22 mWRP
S
PI
Chn m f=100Hz lm m chun ,Io=10-12 N/m2cng m chun.
Ta lun c : 2)(A
B
B
A
R
R
I
I
Mc cng m l i lng dng so snh to ca 1 m vi 1 m chunNhcng thc nhanh sau: 2)lg(.10lg.10
A
B
B
ABA
R
R
I
ILL
II, C s l thuyt ton hc v hypebol v ng dng vo gii bi ton vt l Kin thc c bn v hypelbol:
Phng trnh chnh tc: 12
2
2
2
b
y
a
x(ch ng c ln vi phng trnh elip du cng th kh
)
aMFMF 221
AB=2c
C2=a2+b2
Dng tm sim cc i cc tiu trn on bt k dng c phng php ny ngoi ta cn ccc bi ton giao nhau gia pt (H) vi cc ng thng cho trc trong cc bi tp s c.
II,Bi tpng dng
Bi tp hng dn khoa ly t thi th ln 1 ca anh Nguyn Vn Khnh v mt s bi trong kh ca khoa v bi tp ngoi.
Bi 1: Trn mt nc c 2 ngun sng ging ht nhau A v B cch nhau mt khong AB = 24cm. Cc sng c cng bc sng = 2,5 cm. Hai im M v N trn mt nc cng cch utrung im ca on AB mt on 16 cm v cng cch u 2 ngun sng v A v B. Simtrn on MN dao ng cng pha vi 2 ngun l
A. 7. B. 8. C. 6. D. 9.
A R = 4cm O B
7/31/2019 mt s bi tp sng c v sng m vt l 12
7/19
Hng dn
Phng trnh sng ti M do sng ti A truyn n l:
uAM = 3cos(40t +6
- 1
2 d
)
Phng trnh sng ti M do sng ti B truyn n l:
uBM = 4cos(40t +2
3
- 22 d
)
Phng trnh sng tng qut tng hp ti M l:
uM = uAM + uBM = 3cos(40t +
6
- 12 d
) + 4cos(40t +2
3
- 22 d
)
Bin sng tng hp ti M l: (p dng cng thc dao ng iu ha)
A = 2 2 2 12 2 2
3 4 2.3.4. os( ( ))3 6
d dc
= 2 22 1
23 4 2.3.4. os( ( ))
2c d d
Bin sng tng hp ti M bng 5 khi: 2 12os( ( ))
2c d d
= 0
Khi : 2 12
( )2
d d
2 12 (
2
d d
) =
2k
Do : d2d1 = k2
;
M - 8 d2d1 8 - 8 k 2
8 - 8 k 8
Tng t ti hai im M v N hai u bn knh l im dao ng vi bin bng 5cm
Nn sim dao ng vi bin 5cm l: n = 17x22 = 32
Bi 2: mt nc c hai ngun sng c A v B cch nhau 15 cm, dao ng iu ha cng tns, cng pha theo phng vung gc vi mt nc. im M nm trn AB, cch trung im O l
7/31/2019 mt s bi tp sng c v sng m vt l 12
8/19
1,5 cm, l im gn O nht lun dao ng vi bin cc i. Trn ng trn tm O, ngknh 20cm, nm mt nc c sim lun dao ng vi bin cc i l
A. 18. B. 16. C. 32. D. 17.
Hng dn
Sng ti M c bin cc i khi d2d1 = k
Ta c d1 = 15/2 + 1,5 = 9cm; d2 = 15/21,5 = 6cm
Khi d2d1 = 3.
Vi im M gn O nht chn k = 1. Khi ta c: = 3Sim dao ng vi bin cc i trn on AB l:
- S1S2 d2d1 S1S2
Hay -15 k 15 -5 k 5
Vy sim dao ng vi bin cc i trn ng trn tm O bn knh 20cm l
n = 10x22 = 18 cc i (y t A v B l hai cc i do ch c 8 ng cc i ct ngtrn ti 2 im, 2 cc i ti A v B tip xc vi ng trn)
Bi 3: Hai mi nhn S1, S2 cch nhau 9cm, gn u mt cu rung c tn sf = 100Hz c tcho chm nh vo mt mt cht lng. Vn tc truyn sng trn mt cht lng l v = 0,8 m/s. Gnh cho cn rung th 2 im S1,S2dao ng theo phng thng ng vi phng trnh dng: u =acos2ft. im M trn mt cht lng cch u v dao ng cng pha S1 , S2 gn S1S2 nht cphng trnh dao ng.
Hng dn
Phng trnh sng tng qut tng hp ti M l:
uM = 2acos(2 1d d
)cos(20t - 2 1
d d
)
d1
d2
A S1 O S2
7/31/2019 mt s bi tp sng c v sng m vt l 12
9/19
Vi M cch u S1, S2 nn d1 = d2. Khi d2d1 = 0 cos(2 1d d
) = 1 A = 2a
M dao ng cng pha vi S1, S2 th: 2 1d d
= 2k
suy ra: 2 1 2d d k 1 2 2
d dk
v d1 = d2 = k
Gi x l khong cch tM n AB: d1 = d2 =
2
2
2
ABx
=k
Suy ra 2
2
2
ABx k
= 20,64 9k ; ( = v/f = 0,8 cm)
Biu thc trong cn c ngha khi 20,64 9k 0 k 3,75
Vi x 0 v khong cch l nh nht nn ta chn k = 4;
Khi 1 2 2 8d d
k
Vy phng trnh sng ti M l:
uM = 2acos(200t - 8) = uM = 2acos(200t)
Bi 4: Hai ngun sng kt hp trn mt nc cch nhau mt on S1S2= 9 pht ra dao ng
u=cos(t). Trn on S1S2, s im c bin cc i cng pha vi nhau v ngc pha vingun (khng k hai ngun) l:
A. 8. B. 9 C. 17. D. 16.
Hng dn
Phng trnh sng tng qut tng hp ti M l:
uM = 2cos(2 1d d
)cos(20t - 2 1
d d
)
S1 O x
d1
7/31/2019 mt s bi tp sng c v sng m vt l 12
10/19
Vi d1 + d2 = S1S2= 9
Khi : Phng trnh sng tng qut tng hp ti M l:
uM = 2cos(2 1d d
)cos(20t - 9) = 2cos( 2 1
d d
)cos(20t - ) = -
2cos( 2 1d d
)cos(20t)
Vy sng ti M ngc pha vi ngun khi cos( 2 1d d
) = 1
2 1d d
= k2 d1 - d2 = 2k
Vi - S1S2 d1 - d2 S1S2 -9 2k 9 4,5 k 4,5
Suy ra k = 0; 1, 2; 3; 4. C 9 gi tr (c 9 cc i) Chn p n B
Bi 5 Trn mt nc c hai ngun kt hp S1, S2 cch nhau 6 2 cm dao ng theo phng trnhtau 20cos (mm). Bit tc truyn sng trn mt nc l 0,4 m/s v bin sng khng i
trong qu trnh truyn. im gn nht ngc pha vi cc ngun nm trn ng trung trc ca
S1S2 cch S1S2 mt on: A. 6 cm. B. 2 cm. C. 3 2 cmD. 18 cm.
Hng dn
Gi M l im dao ng ngc pha vi ngun
Phng trnh sng tng hp ti M l: uM = 2acos(2 1d d
)cos(20t - 2 1
d d
)
M dao ng ngc pha vi S1, S2 th: 2 1d d
= (2k + 1) suy ra: 2 1 2 1d d k
Vi d1 = d2 ta c: 2 1 2 12
d d k
7/31/2019 mt s bi tp sng c v sng m vt l 12
11/19
Gi x l khong cch tM n AB: d1 = d2 =
2
2 1 2
2
S Sx
= 2 1
2k
Suy ra
2 2
1 2(2 1)2 2
S Sx k
= 24(2 1) 18k ; Vi = v/f = 4cm
Biu thc trong cn c ngha khi 24(2 1) 18k 0 k 0,56
Vi x 0 v khong cch l nh nht nn ta chn k = 1 suy ra x = 3 2 cm; Chn p nC
Bi 6: Trn mt nc c 2 ngun sng ging ht nhau A v B cch nhau mt khong AB =24cm. Cc sng c cng bc sng = 2,5 cm. Hai im M v N trn mt nc cng cch utrung im ca on AB mt on 16 cm v cng cch u 2 ngun sng v A v B. Simtrn on MN dao ng cng pha vi 2 ngun l: A. 7. B. 8.C. 6. D. 9.
Hng dn
Gi M l im dao ng cng pha vi ngun
Phng trnh sng tng hp ti M l: uM = 2acos(2 1d d
)cos(20t - 2 1
d d
)
M dao ng ngc pha vi S1 th: 2 1d d
= 2k suy ra: 2 1 2d d k
Vi d1 = d2 ta c: 2 1d d k ; Gi x l khong cch tM n AB: d1 = d2 =
2
2
2
ABx
= k
Suy ra 2
2
2
ABx k
= 26,25 144k ;
Vi 0 x 16 4,8 k 8 k = 5, 6, 7, 8.
7/31/2019 mt s bi tp sng c v sng m vt l 12
12/19
Vy trn on MN c 2x4 = 8 imdao ng cng pha vi hai ngun Chnp n B
Bi 7: Ngun m ti O c cng sut khng i. Trn cng ng thng qua O c ba im A, B,C cng nm v mt pha ca O v theo th t xa c khong cch ti ngun tng dn. Mc cng
m ti B km mc cng m ti A l a (dB), mc cng m ti B hn mc cng m ti C l 3a (dB). Bit OA =
2
3OB. Tnh t s
OC
OA
A.81
16B.
9
4C.
27
8D.
32
27
Hng dn :
Cng thc lin hcng m v cng sut ngun pht :2
PI
4d
Ta cn tnh : C
A
dOC
OA d
- Mc cng m ti B km mc cng m ti A l a (dB)
a
A B A A 10A B
0 0 B B
I I I IaL L a 10lg 10lg a lg 10
I I I 10 I . (1)
- Mc cng m ti B hn mc cng m ti C l 3a (dB)
3a
CB B B 10B C
0 0 C C
II I I3aL L 3a 10lg 10lg 3a lg 10
I I I 10 I . (2)
- Theo gi thit : B
A
d2 3OA OB
3 d 2 .
- T(1)
2a a a
A B10 10 10
B A
I d 9: 10 10 10
I d 4
.
7/31/2019 mt s bi tp sng c v sng m vt l 12
13/19
7/31/2019 mt s bi tp sng c v sng m vt l 12
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d1+ d2 = AB = 20 cm
d1 = 10 +1,5k
0 d1= 10 +1,5k 20
- 6 k 6 Trn ng trn c 26 im dao ng vi bin cc i
im gn ng thng AB nht ng vi k = 6. im M thuc cc i th 6
d1d2 = 6 = 18 cm; d2 = d118 = 2018 = 2cm
Xt tam gic AMB; h MH = h vung gc vi AB. t HB = x
h2 = d12AH2 = 202(20x)2
h2 = d22BH2 = 22x2
202(20x)2 = 22x2 x = 0,1 cm = 1mm
h = mmxd 97,193991202222 . Chn p n C
Cch khc:
v3
fcm ; AM = AB = 20cm
AM - BM = kBM = 20 - 3k
AB ABk 6,7
kmax = 6BMmin = 2cm
AMB cn: AM = AB = 200mm; BM = 20mm.
Khong cch tM n AB l ng cao MH ca AMB:
h = p p a p b p c a b c2 ; p 21cm
a 2
2 21.1.1.19h 1,997cm 19,97mm
20
Bi 10: Ti mt im trn mt phng cht lng c mt ngun dao ng to ra sng n nh trnmt cht lng. Coi mi trng tuyt i n hi. M v N l 2 im trn mt cht lng, cchngun ln lt l R1 v R2. Bit bin dao ng ca phn t ti M gp 4 ln ti N. T s
2
1
R
Rbng
A. 1/4 B. 1/16 C. 1/2 D. 1/8
Hng dn:
7/31/2019 mt s bi tp sng c v sng m vt l 12
15/19
Nng lng sng c t l vi bnh phng bin , ti mt im trn mt phng cht lng c mtngun
dao ng to ra sng n nh trn mt cht lng th nng lng sng truyn i sc phn bu cho ng trn (tm ti ngun sng) Cng sut t ngun truyn n cho 1 n
v di vng trn tm O bn knh R l R
E
2
0
Suy ra1
2
0
0
2
2
2
2
R
R
R
R
R
E
R
E
A
A
E
E
M
N
N
M
N
M
N
M
Vy16
1164
2
12
2
2
1
2 R
R
A
A
R
R
N
M
Bi 11: Cng sut m thanh cc i ca mt my nghe nhc gia nh l 10W. Cho rng c truyntrn khong cch 1m, nng lng m b gim 5% so vi ln u do s hp th ca mi trngtruyn m. Bit
I0 = 10-12W/m2. Nu mto ht cth mc cng m khong cch 6m l:
A. 102 dB B. 107 dB C. 98 dB D. 89 dB
Hng dn:
Cng m pht i t ngun im c xc nh l:2d4
P
S
PI
Nng lng m gim nn cng sut gim theo quan h: P = E/t, c 1m th gim 5% hay
6066
0
6
0
1
0
10 95,0.PP95,0E
E95,0
E
E05,0
E
EE
Vy mc cng m ti v tr cch ngun m 6m l:
dB102I.d4
95,0.Plog10L
02
6
0
Bi 12:Mt si dy n hi cng ngang, ang c sng dng n nh. Trn dy, A l mt im
nt, B l im bng gn A nht vi AB = 18 cm, M l mt im trn dy cch B mt khong 12cm. Bit rng trong mt chu k sng, khong thi gian m ln vn tc dao ng ca phn tB nhhn vn tc cc i ca phn t M l 0,1s. Tc truyn sng trn dy l:
A. 3,2 m/s. B. 5,6 m/s. C. 4,8 m/s. D. 2,4 m/s.
Hng dn:
B MA
7/31/2019 mt s bi tp sng c v sng m vt l 12
16/19
+ A l nt; B l im bng gn A nht Khong cch:
AB =4
= 18cm, = 4.18 = 72cm
+ Bin sng dng ti mt im M bt k trn dy:2
2 | sin |MMd
A a
(Vi dM l khong cch tB n M; a l bin ca sng ti v sng phn x)
Vi dM = MB = 12cm =6
2 .122 | sin |
72MA a
= 2a. sin
3
= 2a.
3
2= a 3
+. Tc cc i ti M: vMmax = AM. = a 3
+. Tc ca phn t ti B (bng sng) khi c li xB = AM l: vB = xB = a 3 = vMmax
* Phn t ti bng sng: Cng ra bin tc cng gim Thi gian m ln vn tc dao ngca phn t B nhhn vn tc cc i ca phn t M (ng vi lc phn t ca bng sng qua vtr c li M ra bin v trv M)
+ Cos =3
2
a
a=
3
2 =
6
+ Trong 1 chu k: Thi gian m ln vn tc dao ng ca phn t Bnhhn vn tc cc i ca phn t M l
4t
= 4.
.
6.2
T
= 3
T
= 0,1s
T = 3.0,1 = 0,3s
* Tc truyn sng c: v =T
=
72
0,3= 240 cm/s = 2,4m/s
* Lu : M trong on AB hay M ngoi on AB u ng.
p n D.
Bi 13 ( ca 1 bn hi khoa v ca anh Nguyn Vn Khnh) :Hai im M, N cng nm trn mt phng truyn sng cch nhau x = /3, sng c bin A,chu k T. Ti thi im t1 = 0, c uM = +3cm v uN = -3cm. thi im t2 lin sau c uM = +A,bit sng truyn tN n M. Bin sng A v thi im t2 l
A. cm32 v12
11T B. cm23 v
12
11TC. cm32 v
12
22TD.
cm23 v12
22T
Hng dn:
3a
M0
7/31/2019 mt s bi tp sng c v sng m vt l 12
17/19
Ta c lch pha gia M v N l:
3
22
x
6
,
T hnh v, ta c thxc nh bin
sng l: A = 32cos
Mu (cm)
thi im t1, li ca im M l :
uM = +3cm, ang gim. n thi im t2lin sau , li ti M l : uM = +A.
Ta c
/
12
ttt vi
T
2;
6
112/
1211
2.
61112 TTttt
Vy: 121112 Tttt
Bi 14 : sng (A, B cng pha so vi S v AB = 100m). im M l trung im AB v cch S 70m c mc cng m 40dB. Bit vn tc m trong khng kh l 340m/s v cho rng mitrng khng hp thm (cng m chun Io = 10
-12W/m2). Nng lng ca sng m trongkhng gian gii hn bi hai mt cu tm S qua A v B l
A. J9,207 B. 207,9 mJ C. 20,7mJ D. 2,07J
Hng dn:
Sng truyn trong khng gian. Nng lng sng t l nghch vi bnh phng khong cch.Nng lng sng bng g? y cho mc cng m ti im M l trung im AB, nghal sxc nh c cng m ti M. Cn csuy ra cng m ti A v B. Cng m tiA v B t l nghch vi bnh phng khong cch n v l W/m 2 Nng lng sng ti ccmt cu tm (S, SA) v (S, SB). Ly hiu th c nng lng trong vng gii hn.
Theo gi thit:
2
2
ABrr
ABrr
MB
MA
. Cng m ti 1 im l nng lng i qua mt n v din
tch tnh
trong 1 n v thi gian. T gi thit suy ra cng sut ngun S l P= 24. MM rI
Nng lng trong hnh cu tm (S, SA) v (S, SB) l: :
Jrrv
rI
v
rP
v
rP AB
MMBA
9,207)100(340
75.4.10)(
4.WWW.W;.W
282
ABBA
M
M2
M1
u(cm)
N
A
3
-3
-A
7/31/2019 mt s bi tp sng c v sng m vt l 12
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Bi 15: Sng dng xut hin trn dy vi tn s f = 5 Hz. Gi th tcc im thuc dy ln lt
l O, M, N, P sao cho O l im nt sng, P l im bng sng nm gn O nht( M, N thuc
on OP). Khong thi gian gia 2 ln lin tip gi trli ca im P bng bin dao ng
ca im M v N ln lt l 1/20 s v 1/15 s. Bit khong cch gia 2 im M, N l 0,2cm. Bc
sng ca si dy l:A 3 B 4,8 C.1,2 D.4,8
Hng dn:
+ Thi gia i tli c ln bng bin ti M(tm gi lido M) l 1/20 ==> Thi gian i t
lido M v VTCB ht
+ L lun Tng t ta c N cch nt
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Chc cc bn hc tp tht tt.
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