2012級下大學(xué)物理考試計算題總結(jié)

1、123大學(xué)物理答疑大學(xué)物理答疑42(/ )radsT cos()xAt 0cosxA 3 0sin03vA 0.12cos()3xtm ),/(2sradT 3 0.12cos()()3xtSI )cos( tAx3 oxyA5326t )(83. 0st 3 oxyA )(3cos(12. 0mtx 由振動表達(dá)式：由振動表達(dá)式：)sin( tA可可得得：(m/s)3-tsin(.120 )cos(2 tAa)/( )3cos(06. 022smt (m/s)18. 0 )/(03. 12sma )3cos(1202 t.dtda )3sin(120 t.dtdx 322A)3cos(1202

2、 t.dtda )3sin(120 t.dtdx )3(t3232 oxyA 325236t )(83. 0st 130 kbmg xbkmgFxk 1422tdxdmxk 022 xmktdxd即：即：bgmk 得：得：)tcos(Ax 設(shè)振動方程為：設(shè)振動方程為：,0t時時當(dāng)當(dāng) 0,bx00 xbkmgFkx 15)tcos(Ax 設(shè)振動方程為：設(shè)振動方程為：22020 xA 則則,b 或或0 xarctg00 )(cosSItbgbx 振動表達(dá)式為：振動表達(dá)式為：,0t時時當(dāng)當(dāng) 0,00 bx)SI(tmkcosx kmg或為：或為： ,0cos,0cos0Ax16,0 時時當(dāng)當(dāng) t0,

3、 000 x cos0A 2 0sin0 A2 0sin )m420cos(10. 0 tx0.10(m)A 110(Hz)2T 2020.1(s)T 4 18)m420cos(10. 0 txcos()xAt sin()vAt 2cos()aAt 2st 2(m)20 x 2 (m/s)v 2220 2(m/s )a 0sinklmg022 xmkdtxdkxxlkmgF )(sin0 20)(02xlkT tdxdmMxlkmg220)2()(sin tdxdMrraMrJrTrT2222121)21( tdxdmTmg221sin 0sinklmg0222 xmMktdxdcm521 A

4、As1 . 021 TTT 2 20 ,010 x010 v21 2cm)220cos(51 txcm)20cos(52 tx,520cmx 020 v55/cmx/st005. 01 . 012xO1M2M222100MMA cm25 45 cm)4520cos(25 txxO1M2MAcm)220cos(51 txcm)20cos(52 tx4523242)42(2cos50 xt.yTtAy(2cos )ox )(SI),2121(cos50 xt.y 25)m2cos(50 t.y0.5 sin()2dytdt 20.5cos()2datdt )(212xx 2)12(42 )2121

5、(cos50 xt.y 26)(2cos xTtAy2 mxty2)0 . 20 . 2(2cos)0 . 1( 波動方程為：波動方程為：0,0 tyy 處處時時00 xtyA0tm0 . 2m0 . 1As0 . 2Tm5 . 0 xs0.1 tmxty2)0 . 1(cos0 . 1 27)cos( tAyo2 0, 0 tyy 處處時時00 xtyAmtyo)2cos(0 . 1 0tm0 . 2m0 . 1As0 . 2Tm5 . 0 xs0.1 t28s0.1 ts0.1tm/ym/xs0.1tm)sin( 0 .1xm2cos0 .1xy1.0cos2()m2.02.02txy 2

6、9m5 . 0 xm5 . 0 xm/ys/tym5 . 0 x)mcos(0 . 1ty1.0cos2()m2.02.02txy 302 . 02 m4 . 0 T 2 u2 0cos()yAt52 4 . 0/08. 02 oyuPm2 .0m04. 0 x31mxty 208. 052cos04. 0 mty 252cos04. 00 2/ oyoyuPm2 .0m04. 0 x)(cos uxtAy32u,4m mA1 )/(4sradAP smTu/82 波速波速smP/4max A max0 P 又又o)(my)(mxP33)2(2 )(cos uxtAy 5 . 0時的相位時的相

7、位處的質(zhì)點(diǎn)在處的質(zhì)點(diǎn)在stx5 . 00 225 . 045 . 0 2 mxty2)8(4cos1 oysT24 22 )cos(0 tAy kkt 822442,8,24 TsTsT,8 0cos()(m)82yAtcos()(m)8102xyAt10m/sxut tk 24)rr(12122 2 , 2)43(2 2 , )43(2)243(2x x42 2 )(21212rr xLcos()xyAtu cos()LLyAtu cos ()LLyAtu 反反cos (yAt 反反Lxu)Lu u入入u反反oxyLcos()0 xyAtu cos()2xLyAtuu 反反yyy 反反212

8、122cos()cos()22yAxt 222222cos()cos()22LLyAxt 222cos()cos()22AxLtL 2cos()12xL 2()2xLn 1()2 2xLn xL2cos()02xL 2()(21)22xLn 2xLn xL41mdDxxx11. 020201010 Nen )1(7)1( enN （暗）（暗）明明反反2)12()(2 kken24 . 422 menk449 . 3222 menk12l 22ln 121(1)2llln 112 ln ()41.7 10 rad0.0097 46 Rr102 nRr 102 2rrn21. 1 27. 140.

9、 12,nkRrk 47)(Rre2=20e空空氣氣玻玻璃璃 122121220kee)2(0ekRr )20 ek 48493sin(21)2ak 12axkf sintanxf72 sin(21)2ak 22+=sinfxx06624848cm2 5900 10210.m.10Lfa 261 105 10 (m)2000ab2sinsinadkakd 8=maxk, 3 . 8=90sin0maxdk)2, 1(, 2sinsinkkkadkkakd kdsin5312sinsind dkka 5444, 8dkkka ,sin kd ,sin ka )(105 . 2400016md ,

10、85. 390sin0 dkm kdsin3mk)2, 1(, 2sinsinkkkadkkakd kd)30sin(sin0max5k 77. 5)30sin90(sin0 dk1maxk 22+=sinfxx585001 bamm1023 kba sin)(,2, 1,0 kfx=tansin bafkx , 11bafxk bafxk 3, 33bafxxx 213)(12. 0m 59601sin0222adxaxAdx 21,2aA aA2 axAx sin axa 22sin2 21sin22/022/02dxaxadxaa 6102sin22 axadxd ,2, 1,0,2 k

11、kax axAx sin 62xanax sin2)( dxxPa240)( dxanaa402sin2 2sin2141 nn 2141 P%9 41 P1 n n63xanax22sin2=)(4sin2)(22aanax 4sin22 na 14sin2 n24 kn, 1 , 0 k24 kn4a,10, 6 , 2=:n即即6422201xA x edx 320 (0)( )2 (0)xxxxex 222223004xxAA x edxe 1234A 322A ) 0( ) 0( 0)(xAxexxx )(x )(x 652( )( )P xx 3220 (0)( )4 (0)xxP

12、 xx ex 0( )0 x xdP xdx 002232004(22)0 xxx ex e01x ) 0( ) 0( 0)(xAxexxx )(x )(x 2mm 0 0 Av dvvvdNNvvmvv2vdNf vNdv( )2Avm0vv 0mvvm2001vf v dvAv dv( )3m3/Avm0m3043vvvf v dvAvv dv( )m2242m0035vvv f v dvAv dvv( )2m35vv2mm 0 0 Av dvvvdNNvvmvv2vvo0v02vavNf v ( )0/Nf vav v( )Nf va( )0Nf v ( )0(0)vv00(2)vvv0(2)vvf v ( )0/av Nv/a N00(0)vv00(2)vvv0(2)vv000200vvvavdvadvvNv032=vNa00(21.5)Navv 0( )vvf v dv0002200vvvavdvavdvv22001 13()32vavavN13N01( )vNf v dv