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R 7ω " !X U Y 1 HD NMA) " 7XW UY- " 89/05^! R 2 " 7X!"π U3+FG"()^6K " !"πSX!"'"n SK 1 R ^ K " IK?-Z 1RRk N m= )(./0 b Om kg= ;/dA "A cm= ~2 5Y9 m *B\pA>p'(`./0 R R'bm m= d Z)\) m 23*B\@Ai ( ) R m m+ -.A "R cm s ER E cm s "b cm s O b 1" cm s K }D p!>p!K7 " ! " )!•K7 " !Lπ7% }3 R d)\)';sfM /0$U R &V!!TV!Xπ7% }ri ( ) R m m+ -•V!"π E }]D NM(.=i) R 1 " ( ) R m m+ R " ! 1 " ( ) R m m+ V " U 1 " " !T R " !V " U R k m m+ " !T R !"R7% 8K(K- 1 !1.(M- " !X'1Aa-!L"bc72ri A)'%MA))<lZ23CK`B\ @A51'L?Alag8K;/dZ2 fg8!O'n7% " 2f/F<A)-FBg)`g) K21O'ncWR'"c M2bRcWXR'"c N2LRcWXRc P21"RcWnRc h€•€ U∆l! 1 7!R'R1bLn‚K Uf4F<A))]!c Uc _45!T([g t Uc6m!R!Tc !m6t !Tc! " 6$KS∆l&!EO'Onc Uc ABCg!T([D gg c 6t 6m " !R!Tc !m " Ut ! " U$KU∆l& !Tc !bO'Onc @K'?-Z!bRc7)(!bRR )A R A :;<j2ri( bRR E m g = A)M;/d(. R 1 7v m s= 2h+>5)) ))?)y8)Y9-)`g2^5( ))-)45)49,/0)1RR)nR2N " 1R 7g m s= 2M/5) R b 2A cm= 2 R 1R 2A cm= 2 R b " 2A cm = O R b EA cm = 2 K m " t c ] m " t c ] ∆l j K SK M K C [...]... nng m1 trc khi va chm a = - 2A, vi A l biờn dao ng ban u Tn s gúc = 2 = 1 (rad/s), Suy ra - 2cm/s2 = -A (cm/s2) -> A = 2cm T Gi A l biờn dao ng ca con lc sau va chm vi m2 Quóng ng vt nng i c sau va chm n khi i chiu s = A + A 2 2 Theo h thc c lõp: x0 =A, v0 = v -> A = A + v2 2 = 22 + (2 3 ) 2 =16 1 -> A = 4 (cm) > S = A + A = 6cm Chn ỏp ỏn C Cõu 24 : Mt con lc lũ xo dao ng iu ho trờn mt phng... biờn trong T/2 : A = 2àmg/k = 0,4à * Theo y/c ca : 0,04 + (0,04 - A) < S < 0,04 + 2(0,04 - A) => 0,08 0,4à < S < 0,12 0,8à 0 * Ti khi dng : kA2/2 = àmgS => S = 4.1 0-3 /à A -4 + 0,08 0,4à < 4.1 0-3 /à => ( - 0,1)2 > 0 + 4.1 0-3 /à < 0,12 0,8à => à2 0,15à + 0,005 < 0 => 0,05 < à < 0,1 Cõu 26: Mt con lc lũ xo cú cng k=100N/m, vt nng m=100g dao ng tt dn trờn mt phng nm ngang do ma sỏt, vi h s ma sỏt 0,1... àMgA0 -> = - àMgA0 = - 2àMgA0 2 2 2 2 2 2 2 Do ú v max = v0 - 4àgA0 > vmax = 0,4195 m/s = 0,42 m/s Cõu 30: Mt con lc lũ xo treo thng ng gm vt nng m=1kg, lũ xo nh cú cng k=100N/m t giỏ B nm ngang vt m lũ xo cú chiu di t nhiờn Cho giỏ B chuyn ng i xung vi gia tc a=2m/s2 khụng vn tc u Chn trc ta thng ng, chiu dng trờn xung, gc ta VTCB ca vt, gc thi gian lỳc vt ri giỏ B Phng trỡnh dao ng ca... xo gin: l0=mg/k=0,1m k Tn s dao ng: = =10rad/s m Vt m: P + N + Fdh = m a Chiu lờn trc Ox ó chn ta cú: mg-N-k l=ma Khi vt ri giỏ N=0, m( g a ) gia tc ca vt a=2m/s2( theo bi ra) Suy ra l = k at 2 Trong khong thi gian ú vt i c qung ng l c tớnh l= 2 Kt hp 2 biu thc ta cú: t=0,283(s) at 2 Qung ng vt i c n khi ri giỏ l: S= =0,08m 2 Ta ban u ca vt l x0=0,0 8-0 ,1 =-0 ,02m =-2 cm Vn tc ca vt khi ri giỏ... thớch cho vt iu hũa theo phng thng ng vi biờn 3cm Ly g = 10m/s2 Cụng ca lc n hi khi vt di chuyn theo chiu dng t v trớ cú ta x1 = 1cm n v trớ x2 = 3cm B A - 4 J B - 0,04 J C - 0,06 J D 6 J Gii: p dng nh lý ng nng: AFh = W = 2 mv 2 mv12 2 2 2 mv 2 mv12 k ( A 2 x12 ) AFh = W = == - 50 (32 1).1 0-4 = - 0,04 J 2 2 2 O mv12 kA 2 kx12 Vi v2 = 0 (vt v trớ biờn), v = 2 2 2 N M Chn ỏp ỏn B Cõu 16 Gn mt... sau va chm: v = 2cm/s; v = -1 cm/s Theo nh lut bo ton ng lng v ng nng ta cú: mv = m0v0 + mv (1) > m0v0 = m(v v) (1) 2 mv 2 m0 v 0 mv '2 (2) > m0v02 = m(v2 v2) (2) = + 2 2 2 T (1) v (2) ta cú v0 = v + v = 2 1 = 1cm/s Gia tc vt nng trc khi va chm a = - 2A, vi A l biờn dao ng ban u Tn s gúc = 2 = 1 (rad/s), Suy ra - 2cm/s2 = -Acm/s2 -> A = 2cm T Gi A l biờn dao ng ca con lc sau va chm... x0=0,0 8-0 ,1 =-0 ,02m =-2 cm Vn tc ca vt khi ri giỏ cú giỏ tr: v0=at=40 2 cm/s v2 =6cm Ti t=0 thỡ 6 cos =-2 = 1,91rad 2 Phng trỡnh dao ng :x=6cos(10t-1,91)(cm) P N D Biờn dao ng l:A= x 2 + Cõu 31: con lc lũ xo cú cng k=100N/m qu cu khi lng m dao ng iu ho vi biờn A=5cm Khi qu cu n v trớ thp nht ta nh nhng gn thờm vt M=300g sau ú 2 vt cựng dao ng iu ho vi biờn l ỏp ỏn 3cm Gii: O V trớ cõn bng c l O Khi o gión ca lũ xo... con lc n gm dõy treo cú chiu di 1m, vt nng cú khi lng 100g, dao ng nh ti ni cú gia tc trng trng g = 10 m/s2 Cho con lc dao ng vi biờn gúc 0,2 rad trong mụi trng cú lc cn khụng i thỡ nú ch dao ng c 150s ri dng hn Ngi ta duy trỡ dao ng bng cỏch dựng h thng lờn dõy cút, bit rng 70% nng lng dựng thng lc ma sỏt do h thng cỏc bỏnh rng Ly 2 =10 Cụng cn thit lờn dõy cút duy trỡ con lc dao ng trong 2 tun vi... 0,01.Từ vị trí lò xo không biến dạng truyền cho vật vận tốc ban đầu 1m/s thì thấy con lắc dao động tắt dần trong giới hạn đàn hồi của lò xo.độ lớn của lực đàn hồi cực đại của lò xo trong quá trình dao động là: A 19,8N B.1,5N C.2,2N D.1,98N Gii: Gi A l biờn cc i ca dao ng Khi ú lc n hi cc i ca lũ xo trong quỏ trỡnh dao ụng: Fhmax = kA tỡm A t da vo L bo ton nng lng: mv 2 kA2 kA2 = + Fms A = + àmgA 2... thay pin: s 23 ngy 4 FC = 4,4.10 3 rad P 5 4,4.10 3 = 0.0828rad -Sau 1 chi kỡ biờn cũn li l: 1 = 0 = 180 1 1 2 2 -Sau 1 chu kỡ c nng gim: W = mgl 0 mgl 1 = 3,759.10 3 J 2 2 - gim biờn sau 1 chu kỡ: = -Nng lng do pin cung cp l:W=0,25.Q.E -sau thi gian T Cn cung cp nng lng W -sau thi gian t cung cp nng lng W t = T W = 46ngy W Cõu 8: Mt CLLX nm ngang gm lũ xo cú cng k=20N/m va vt nng m=100g T . 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