C\u00e2u 1:<\/b>\u00a0Ph\u01b0\u01a1ng tr\u00ecnh c\u1ee7a m\u1ed9t dao \u0111\u1ed9ng \u0111i\u1ec1u h\u00f2a l\u00e0:<\/p>\n
<\/p>\n
Ch\u1ecdn ph\u00e1t bi\u1ec3u \u0111\u00fang<\/p>\n
A. Bi\u00ean \u0111\u1ed9 A = -3 cm.\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0B. Pha ban \u0111\u1ea7u \u03c6 = \u03c0\/6 (rad).<\/p>\n
C. Chu k\u00ec T = 0,5 s.\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0D. Li \u0111\u1ed9 ban \u0111\u1ea7u xo<\/sub>\u00a0= 0,75 cm.<\/p>\n C\u00e2u 2:<\/b>\u00a0T\u00ecm ph\u00e1t bi\u1ec3u sai v\u1ec1 l\u1ef1c k\u00e9o v\u1ec1 t\u00e1c d\u1ee5ng l\u00ean v\u1eadt dao \u0111\u1ed9ng \u0111i\u1ec1u h\u00f2a.<\/p>\n A. Lu\u00f4n h\u01b0\u1edbng v\u1ec1 v\u1ecb tr\u00ed c\u00e2n b\u1eb1ng.<\/p>\n B. Lu\u00f4n ng\u01b0\u1ee3c pha v\u1edbi li \u0111\u1ed9.<\/p>\n C. Lu\u00f4n bi\u1ebfn thi\u00ean \u0111i\u1ec1u h\u00f2a theo th\u1eddi gian.<\/p>\n D. Lu\u00f4n ng\u01b0\u1ee3c chi\u1ec1u chuy\u1ec3n \u0111\u1ed9ng c\u1ee7a v\u1eadt.<\/p>\n C\u00e2u 3:<\/b>\u00a0M\u1ed9t v\u1eadt dao \u0111\u1ed9ng \u0111i\u1ec1u h\u00f2a theo ph\u01b0\u01a1ng tr\u00ecnh:<\/p>\n <\/p>\n Qu\u00e3ng \u0111\u01b0\u1eddng v\u1eadt \u0111i \u0111\u01b0\u1ee3c trong 0,125 s k\u1ec3 t\u1eeb th\u1eddi \u0111i\u1ec3m t = 0 l\u00e0:<\/p>\n A. 6 cm. \u00a0\u00a0\u00a0\u00a0B. 4,5 cm. \u00a0\u00a0\u00a0\u00a0 C. 7,5 cm. \u00a0\u00a0\u00a0\u00a0D. 9 cm.<\/p>\n C\u00e2u 4:<\/b>\u00a0Dao \u0111\u1ed9ng \u0111i\u1ec1u h\u00f2a c\u00f3 ph\u01b0\u01a1ng tr\u00ecnh:<\/p>\n <\/p>\n V\u1eadn t\u1ed1c c\u1ef1c \u0111\u1ea1i c\u1ee7a dao \u0111\u1ed9ng c\u00f3 gi\u00e1 tr\u1ecb<\/p>\n A. 20 cm\/s. \u00a0\u00a0\u00a0\u00a0B. 40 cm\/s. \u00a0\u00a0\u00a0\u00a0 C. 80 cm\/s. \u00a0\u00a0\u00a0\u00a0D. 100 cm\/s.<\/p>\n C\u00e2u 5:<\/b>\u00a0Khi dao \u0111\u1ed9ng \u0111i\u1ec1u h\u00f2a, v\u1eadt c\u00f3 chu k\u00ec \u03c0\/5 (s) v\u00e0 v\u1eadn t\u1ed1c c\u1ef1c \u0111\u1ea1i 15 cm\/s. T\u1eeb v\u1ecb tr\u00ed c\u00e2n b\u1eb1ng v\u1eadt c\u00f3 th\u1ec3 ra xa nh\u1ea5t m\u1ed9t \u0111o\u1ea1n<\/p>\n A. 1,5 cm. \u00a0\u00a0\u00a0\u00a0 B. 1,8 cm. \u00a0\u00a0\u00a0\u00a0 C. 2 cm. \u00a0\u00a0\u00a0\u00a0 D. 2,5 cm.<\/p>\n C\u00e2u 6:<\/b>\u00a0M\u1ed9t v\u1eadt dao \u0111\u1ed9ng \u0111i\u1ec1u h\u00f2a v\u1edbi bi\u00ean \u0111\u1ed9 6 cm, chu k\u00ec 2 s. Ch\u1ecdn g\u1ed1c th\u1eddi gian l\u00e0 l\u00fac v\u1eadt \u0111\u1ea1t li \u0111\u1ed9 c\u1ef1c \u0111\u1ea1i, \u1edf ph\u00eda chi\u1ec1u d\u01b0\u01a1ng c\u1ee7a tr\u1ee5c t\u1ecda \u0111\u1ed9. Ph\u01b0\u01a1ng tr\u00ecnh dao \u0111\u1ed9ng c\u1ee7a v\u1eadt l\u00e0:<\/p>\n <\/p>\n C\u00e2u 7:<\/b>\u00a0M\u1ed9t v\u1eadt dao \u0111\u1ed9ng \u0111i\u1ec1u h\u00f2a v\u1edbi t\u1ea7n s\u1ed1 g\u00f3c b\u1eb1ng 10 rad\/s. \u1ede li \u0111\u1ed9 3 cm, v\u1eadn t\u1ed1c dao \u0111\u1ed9ng c\u1ee7a v\u1eadt b\u1eb1ng n\u1eeda v\u1eadn t\u1ed1c c\u1ef1c \u0111\u1ea1i. V\u1eadn t\u1ed1c c\u1ef1c \u0111\u1ea1i c\u1ee7a v\u1eadt c\u00f3 gi\u00e1 tr\u1ecb<\/p>\n A. 15\u221a3 cm\/s.\u00a0\u00a0\u00a0\u00a0B. 20\u221a3 cm\/s.<\/p>\n C. 60\/\u03c0 cm\/s.\u00a0\u00a0\u00a0\u00a0D. 30 cm\/s.<\/p>\n C\u00e2u 8:<\/b>\u00a0Ch\u1ecdn m\u1ed9t ch\u1ea5t \u0111i\u1ec3m dao \u0111\u1ed9ng \u0111i\u1ec1u h\u00f2a tr\u00ean \u0111o\u1ea1n th\u1eb3ng MN d\u00e0i 6 cm v\u1edbi t\u1ea7n s\u1ed1 2 Hz. Ch\u1ecdn g\u1ed1c th\u1eddi gian l\u00e0 l\u00fac ch\u1ea5t \u0111i\u1ec3m c\u00f3 li \u0111\u1ed9 (3\u221a3)\/2 cm v\u00e0 chuy\u1ec3n \u0111\u1ed9ng ng\u01b0\u1ee3c chi\u1ec1u v\u1edbi chi\u1ec1u d\u01b0\u01a1ng \u0111\u00e3 ch\u1ecdn. Ph\u01b0\u01a1ng tr\u00ecnh dao \u0111\u1ed9ng c\u1ee7a ch\u1ea5t \u0111i\u1ec3m l\u00e0<\/p>\n <\/p>\n C\u00e2u 9:<\/b>\u00a0M\u1ed9t v\u1eadt dao \u0111\u1ed9ng \u0111i\u1ec1u h\u00f2a theo ph\u01b0\u01a1ng tr\u00ecnh x=2cos\u206120\u03c0t (cm). V\u1eadn t\u1ed9c trung b\u00ecnh c\u1ee7a v\u1eadt khi \u0111i t\u1eeb v\u1ecb tr\u00ed c\u00e2n b\u1eb1ng \u0111\u1ebfn v\u1ecb tr\u00ed c\u00f3 li \u0111\u1ed9 l\u00e0 1 cm l\u00e0<\/p>\n A. 1,2 m\/s\u00a0\u00a0\u00a0\u00a0 B. 1,6 m\/s<\/p>\n C. 2,4 m\/s\u00a0\u00a0\u00a0\u00a0 D. 2,8 m\/s<\/p>\n C\u00e2u 10:<\/b>\u00a0V\u1eadt giao \u0111\u1ed9ng \u0111i\u1ec1u h\u00f2a theo ph\u01b0\u01a1ng tr\u00ecnh x=Acos\u2061\u03c9t (cm). Sau khi dao \u0111\u1ed9ng \u0111\u01b0\u1ee3c 1\/6 chu k\u00ec v\u1eadt c\u00f3 li \u0111\u1ed9 \u221a3\/2 cm. Bi\u00ean \u0111\u1ed9 dao \u0111\u1ed9ng c\u1ee7a v\u1eadt l\u00e0<\/p>\n A. 2\u221a2 cm\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 B. \u221a3 cm<\/p>\n C. 2 cm\u00a0\u00a0\u00a0\u00a0 D. 4\u221a2 cm<\/p>\n H\u01b0\u1edbng d\u1eabn gi\u1ea3i v\u00e0 \u0111\u00e1p \u00e1n<\/b><\/p>\n C\u00e2u 1:<\/b>\u00a0C<\/p>\n <\/p>\n C\u00e2u 3:<\/b>\u00a0A<\/p>\n <\/p>\n V\u1eadt dao \u0111\u1ed9ng v\u1edbi chu k\u00ec T = 0,25 s v\u00e0 bi\u00ean \u0111\u1ed9 3 cm.<\/p>\n V\u00ec 0,125 s = T\/2 n\u00ean qu\u00e3ng \u0111\u01b0\u1eddng v\u1eadt \u0111i \u0111\u01b0\u1ee3c l\u00e0 s = 2.3 = 6 cm<\/p>\n C\u00e2u 4:<\/b>\u00a0C<\/p>\n V\u1eadn t\u1ed1c c\u1ef1c \u0111\u1ea1i c\u1ee7a c\u1ee7a dao \u0111\u1ed9ng \u0111i\u1ec1u h\u00f2a c\u00f3 gi\u00e1 tr\u1ecb vmax<\/sub>=\u03c9A<\/p>\n V\u1edbi bi\u00ean \u0111\u1ed9 A = 4 cm, v\u00e0 t\u1ea7n s\u1ed1 g\u00f3c \u03c9 = 20 rad\/s \u0111\u00e3 cho \u1edf ph\u01b0\u01a1ng tr\u00ecnh dao \u0111\u1ed9ng ta c\u00f3 vmax<\/sub>=20.4=80 m\/s<\/p>\n C\u00e2u 5:<\/b>\u00a0A<\/p>\n <\/p>\n V\u1edbi vmax<\/sub>=15 cm\/s ; \u03c9 = 10 rad\/s th\u00ec A = 1,5 cm<\/p>\n \u0110\u00f3 l\u00e0 \u0111\u1ed9 d\u1eddi xa nh\u1ea5t c\u1ee7a v\u1eadt v\u1edbi v\u1ecb tr\u00ed c\u00e2n b\u1eb1ng khi dao \u0111\u1ed9ng<\/p>\n C\u00e2u 6:<\/b>\u00a0A<\/p>\n Ph\u01b0\u01a1ng tr\u00ecnh dao \u0111\u1ed9ng c\u00f3 d\u1ea1ng : x = Acos(\u03c9t + \u03c6)<\/p>\n \u21d2 v = \u2013 \u03c9Asin(\u03c9t + \u03c6), trong \u0111\u00f3 A = 6 cm, \u03c9 = 2\u03c0\/T = \u03c0 (rad\/s)v<\/p>\n Ch\u1ecdn g\u1ed1c th\u1eddi gian t = 0 v\u00e0o l\u00fac x = + A v\u00e0 v = 0<\/p>\n <\/p>\n Ta ch\u1ecdn : \u03c6 = 0. Ph\u01b0\u01a1ng tr\u00ecnh dao \u0111\u1ed9ng c\u1ee7a v\u1eadt l\u00e0 : x = 6 cos\u03c0t (cm)<\/p>\n C\u00e2u 7:<\/b>\u00a0B<\/p>\n B. \u1ede v\u1ecb tr\u00ed c\u00e2n b\u1eb1ng x = 0 c\u00f3 v\u1eadn t\u1ed1c c\u1ef1c \u0111\u1ea1i vmax<\/sub>\u00a0.<\/p>\n \u1ede li \u0111\u1ed9 x = 3 cm c\u00f3 v\u1eadn t\u1ed1c vmax<\/sub>\/2<\/p>\n <\/p>\n C\u00e2u 8:<\/b>\u00a0B<\/p>\n Ph\u01b0\u01a1ng tr\u00ecnh dao \u0111\u1ed9ng c\u00f3 d\u1ea1ng : x = Acos(\u03c9t + \u03c6)<\/p>\n \u21d2 V\u1eadn t\u1ed1c v = \u2013 \u03c9Asin(\u03c9t + \u03c6), trong \u0111\u00f3 A = 3 cm, \u03c9 = 2\u03c0f = 4\u03c0 (rad\/s).<\/p>\n <\/p>\n C\u00e2u 9:<\/b>\u00a0A<\/p>\n T\u1ea1i t = 0 : V\u1eadt \u0111ang \u1edf v\u1ecb tr\u00ed c\u00e2n b\u1eb1ng (VTCB) theo chi\u1ec1u (+).<\/p>\n Th\u1eddi gian v\u1eadt \u0111i \u0111\u1ebfn li \u0111\u1ed9 1 cm l\u00e0 t = T\/12<\/p>\n \u21d2 V\u1eadn t\u1ed1c c\u1ee7a v\u1eadt khi t\u1eeb VTCB \u0111\u1ebfn x = 1 cm l\u00e0 :<\/p>\n <\/p>\n C\u00e2u 10:<\/b>\u00a0B<\/p>\n x = Acos\u03c9t<\/p>\n Khi dao \u0111\u1ed9ng \u0111\u01b0\u1ee3c 1\/6 chu k\u00ec th\u00ec v\u1eadt \u0111\u00f3 c\u00f3 li \u0111\u1ed9 l\u00e0 A\/2=\u221a3\/2 \u21d2 A = \u221a3 cm<\/p>\n C\u00e2u 1:\u00a0Ph\u01b0\u01a1ng tr\u00ecnh c\u1ee7a m\u1ed9t dao \u0111\u1ed9ng \u0111i\u1ec1u h\u00f2a l\u00e0: Ch\u1ecdn ph\u00e1t bi\u1ec3u \u0111\u00fang A. Bi\u00ean \u0111\u1ed9 A = -3 cm.\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0B. Pha ban \u0111\u1ea7u \u03c6 = \u03c0\/6 (rad). C. Chu k\u00ec T = 0,5 s.\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0D. Li \u0111\u1ed9 ban \u0111\u1ea7u xo\u00a0= 0,75 cm. C\u00e2u 2:\u00a0T\u00ecm ph\u00e1t bi\u1ec3u sai v\u1ec1 l\u1ef1c k\u00e9o v\u1ec1 t\u00e1c d\u1ee5ng l\u00ean v\u1eadt […]<\/p>\n","protected":false},"author":3,"featured_media":26405,"comment_status":"closed","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"_mi_skip_tracking":false,"tdm_status":"","tdm_grid_status":""},"categories":[158],"tags":[1445,1444],"yoast_head":"\n\n\n
\n C\u00e2u<\/td>\n 1<\/td>\n 2<\/td>\n 3<\/td>\n 4<\/td>\n 5<\/td>\n 6<\/td>\n 7<\/td>\n 8<\/td>\n 9<\/td>\n 10<\/td>\n<\/tr>\n \n \u0110\u00e1p \u00e1n<\/td>\n C<\/td>\n D<\/td>\n A<\/td>\n C<\/td>\n A<\/td>\n A<\/td>\n B<\/td>\n B<\/td>\n A<\/td>\n B<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n XEM TH\u00caM: B\u00e0i t\u1eadp tr\u1eafc nghi\u1ec7m V\u1eadt L\u00ed 12: Dao \u0111\u1ed9ng \u0111i\u1ec1u h\u00f2a (ph\u1ea7n 2)<\/a><\/h1>\n","protected":false},"excerpt":{"rendered":"