Cdu t. Mpt yAt ktxii lao'ng m : I0g clao d6ng theo phuong trinh x = \A stnA,25n{2t + 1) (c;rr.). Tim gid tri cvc dqi cua
lryc tdc d4ng l\n vQt vd cila rh€ ndug dao d6ng. Ldy tr2 =10.
Gi6i: E0 ldn lu'c cgc d4i tdc dqng l€n v4t: F,,,,,* : rnol2A : 10-?.10,5n;':.0, I * 2,5.10-r N.
Th{l ndng cr;'c dai bang thi5 ning dao dQng: W = 11/2)mo2A2:0,5.10-2.(0,5r;2.0,12 - 12,5.10-5 J.
C6u 2. Quan sdt mQt dao cldng diiu hda cuu ntdt.t,dt nh6 c6 kh6i lrong 100 gant, ngtdi ta thay ctr sau 0,5 giay thivit lai
cdchv! tricdnbdngnhiutgioqnbdngl,Scnt.Bi6tbi1nd0daodQngcuavqtldA> l,5cm,tinhndngluo.'ngdaodQngcila
vQt:
Giiii: DotinhfuAnhoirn(c[r'sau0,5gi6yvdtlai c6 lidQbang+l,5hoic-l,5crn)vdbi6nd6A>1,5cm
= v! tri tr€n ung v6i li d0 , = x+ vd thbi gian ngan nh6t Ae Al gita hai vi tri ffAn litTl4.
^12
= A: 1,5J, ,T:2s, o = n rad/s = W: (112)ma2A2 :0,5.0,1.n2.2.(1,5.10-2)2 :2,25.104 J.
Cdu 3. MQt l)At .nho tham gia ding thd i hai clao d6ng diiu hda cirng phaong, cilng tin s6, v6'i cac biAn dQ tao'ng irng biing 3
,':
cm vd 5 cm. K,il qud ld dao dQng t6ng h.o'p c6 bi2n d0 7 cnt. D0 l€ch pha giira hai dao dpng thdnh phdn bdng
A:-A:-A: j1 -\2-5' I tr
Gi6i: A' = Ai + Ai +2A,A" cos Ap =cos Ag = " "t "z
2 A1A2 2.3.5 2 3
Cfru 4, Hai nguin sdng kdt hqp S 1 vd. 52, ndnt tuAn nfit chdt ldng tl'u.rc hi)n cac dao dQng diiu hda theo plnrong vu6ng gdc v6i m{t chtit l6ng t6'i hi€u s6 pha ban ddu bing rp. Xdc dinh rp biet rhtg n'En dr.rdng n6i hai ngudn ta thtiy n'ong sd nhitrtg
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