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{"id":26375,"date":"2018-04-10T17:37:30","date_gmt":"2018-04-10T10:37:30","guid":{"rendered":"https:\/\/lop12.edu.vn\/?p=26375"},"modified":"2018-04-10T17:37:30","modified_gmt":"2018-04-10T10:37:30","slug":"bai-tap-trac-nghiem-vat-li-12-con-lac-lo-xo-phan-2","status":"publish","type":"post","link":"https:\/\/lop12.edu.vn\/bai-tap-trac-nghiem-vat-li-12-con-lac-lo-xo-phan-2\/","title":{"rendered":"B\u00e0i t\u1eadp tr\u1eafc nghi\u1ec7m V\u1eadt L\u00ed 12: Con l\u1eafc l\u00f2 xo (ph\u1ea7n 2)"},"content":{"rendered":"

C\u00e2u 9:<\/b>\u00a0M\u1ed9t con l\u1eafc l\u00f2 xo treo th\u1eb3ng \u0111\u1ee9ng c\u00f3 \u0111\u1ed9 c\u1ee9ng c\u1ee7a l\u00f2 xo k = 100 N\/m, kh\u1ed1i l\u01b0\u1ee3ng v\u1eadt n\u1eb7ng m = 1kg. T\u1eeb v\u1ecb tr\u00ed c\u00e2n b\u1eb1ng k\u00e9o v\u1eadt n\u1eb7ng xu\u1ed1ng d\u01b0\u1edbi 6 cm r\u1ed3i bu\u00f4ng nh\u1eb9 cho v\u1eadt dao \u0111\u1ed9ng \u0111i\u1ec1u h\u00f2a. Ch\u1ecdn tr\u1ee5c t\u1ecda \u0111\u1ed9 c\u00f3 g\u1ed1c t\u1ea1i v\u1ecb tr\u00ed c\u00e2n b\u1eb1ng, chi\u1ec1u d\u01b0\u01a1ng h\u01b0\u1edbng th\u1eb3ng \u0111\u1ee9ng l\u00ean tr\u00ean. Ch\u1ecdn g\u1ed1c th\u1eddi gian khi bu\u00f4ng tay. Ph\u01b0\u01a1ng tr\u00ecnh dao \u0111\u1ed9ng \u0111i\u1ec1u h\u00f2a c\u1ee7a v\u1eadt l\u00e0<\/p>\n

A. x=3cos\u206110t (cm). \u00a0\u00a0\u00a0\u00a0\u00a0B. x=6cos10t (cm).<\/p>\n

C. x=6cos\u2061(10t+\u03c0\/2) (cm). \u00a0\u00a0\u00a0\u00a0\u00a0D. x=6cos\u2061(10t+\u03c0) (cm).<\/p>\n

C\u00e2u 10:<\/b>\u00a0M\u1ed9t con l\u1eafc l\u00f2 xo treo th\u1eb3ng \u0111\u1ee9ng c\u00f3 \u0111\u1ed9 c\u1ee9ng c\u1ee7a l\u00f2 xo k = 40 N\/m, kh\u1ed1i l\u01b0\u1ee3ng v\u1eadt n\u1eb7ng m = 400 g. \u0110\u01b0a v\u1eadt l\u00ean tr\u00ean v\u1ecb tr\u00ed c\u00e2n b\u1eb1ng 5 cm theo ph\u01b0\u01a1ng th\u1eb3ng \u0111\u1ee9ng r\u1ed3i truy\u1ec1n cho n\u00f3 v\u1eadn t\u1ed1c ban \u0111\u1ea7u 50 cm\/s h\u01b0\u1edbng xu\u1ed1ng d\u01b0\u1edbi. Ch\u1ecdn tr\u1ee5c t\u1ecda \u0111\u1ed9 c\u00f3 g\u1ed1c v\u1ecb tr\u00ed t\u1ea1i v\u1ecb tr\u00ed c\u00e2n b\u1eb1ng, chi\u1ec1u d\u01b0\u01a1ng h\u01b0\u1edbng th\u1eb3ng \u0111\u1ee9ng l\u00ean tr\u00ean. Ch\u1ecdn g\u1ed1c th\u1eddi gian khi b\u1eaft \u0111\u1ea7u dao \u0111\u1ed9ng \u0111i\u1ec1u h\u00f2a. Ph\u01b0\u01a1ng tr\u00ecnh dao \u0111\u1ed9ng \u0111i\u1ec1u h\u00f2a c\u1ee7a v\u1eadt l\u00e0<\/p>\n

\"\"<\/p>\n

C\u00e2u 11:<\/b>\u00a0Treo m\u1ed9t v\u1eadt c\u00f3 kh\u1ed1i l\u01b0\u1ee3ng 1 kg v\u00e0o m\u1ed9t l\u00f2 xo c\u00f3 \u0111\u1ed9 c\u1ee9ng k = 98 N\/m. K\u00e9o v\u1eadt ra kh\u1ecfi v\u1ecb tr\u00ed c\u00e2n b\u1eb1ng v\u1ec1 ph\u00eda d\u01b0\u1edbi, \u0111\u1ebfn v\u1ecb tr\u00ed x = 5 cm, r\u1ed3i th\u1ea3 nh\u1eb9. Gia t\u1ed1c c\u1ef1c \u0111\u1ea1i dao \u0111\u1ed9ng \u0111i\u1ec1u h\u00f2a c\u1ee7a v\u1eadt l\u00e0<\/p>\n

A. 2,45 m\/s2<\/sup>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 B. 0,05 m\/s2<\/sup>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 C. 0,1 m\/s2<\/sup>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 D. 4,9 m\/s2<\/sup><\/p>\n

C\u00e2u 12:<\/b>\u00a0M\u1ed9t con l\u1eafc l\u00f2 xo c\u00f3 qu\u1ea3 c\u1ea7u kh\u1ed1i l\u01b0\u1ee3ng m = 0,2 kg. K\u00edch th\u00edch cho qu\u1ea3 c\u1ea7u chuy\u1ec3n \u0111\u1ed9ng th\u00ec n\u00f3 dao \u0111\u1ed9ng v\u1edbi ph\u01b0\u01a1ng tr\u00ecnh x=5cos\u20614\u03c0t (cm). L\u1ea5y \u03c02<\/sup>\u00a0\u2248 10. N\u0103ng l\u01b0\u1ee3ng \u0111\u00e3 truy\u1ec1n cho qu\u1ea3 c\u1ea7u l\u00e0<\/p>\n

A. 2 J\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 B. 0,2 J\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 C. 0,02 J\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 D. 0,04 J<\/p>\n

C\u00e2u 13:<\/b>\u00a0M\u1ed9t v\u1eadt \u0111\u01b0\u1ee3c treo v\u00e0o m\u1ed9t l\u00f2 xo th\u1eb3ng \u0111\u1ee9ng c\u00f3 \u0111\u1ed9 c\u1ee9ng 40 N\/m. G\u1ecdi \u00d5 l\u00e0 tr\u1ee5c t\u1ecda \u0111\u1ed9 c\u00f3 ph\u01b0\u01a1ng tr\u00f9ng v\u1edbi ph\u01b0\u01a1ng dao \u0111\u1ed9ng c\u1ee7a v\u1eadt v\u00e0 c\u00f3 chi\u1ec1u h\u01b0\u1edbng l\u00ean tr\u00ean. L\u1ea5y g\u1ed1c t\u1ecda \u0111\u1ed9 tr\u00f9ng v\u1edbi v\u1ecb tr\u00ed c\u00e2n b\u1eb1ng c\u1ee7a v\u1eadt. Khi v\u1eadt dao \u0111\u1ed9ng t\u1ef1 do v\u1edbi bi\u00ean \u0111\u1ed9 5 cm, th\u00ec \u0111\u1ed9ng n\u0103ng c\u1ee7a v\u1eadt kh\u00ed n\u00f3 \u0111i qua v\u1ecb tr\u00ed x1<\/sub>\u00a0= 3 cm l\u00e0<\/p>\n

A. 4 mJ\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 B. 1,6 mJ\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 C. 32 mJ\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 D. 16mJ<\/p>\n

C\u00e2u14:<\/b>\u00a0M\u1ed9t con l\u1eafc l\u00f2 xo dao \u0111\u1ed9ng v\u1edbi c\u01a1 n\u0103ng l\u00e0 1,25.10-3<\/sup>\u00a0J th\u00ec bi\u00ean \u0111\u1ed9 dao \u0111\u1ed9ng l\u00e0 A1<\/sub>\u00a0= \u221a2 cm. N\u1ebfu c\u01a1 n\u0103ng c\u1ee7a con l\u1eafc c\u00f3 gi\u00e1 tr\u1ecb 1,8 mJ, bi\u00ean \u0111\u1ed9 dao \u0111\u1ed9ng c\u1ee7a con l\u1eafc ( A2<\/sub>) l\u00e0<\/p>\n

A. 1,4\u221a2 cm.\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 B. 1,5\u221a2 cm.\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 C. 1,1\u221a2 cm.\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 D. 1,2\u221a2 cm<\/p>\n

C\u00e2u 15:<\/b>\u00a0Con l\u1eafc l\u00f2 xo c\u00f3 kh\u1ed1i l\u01b0\u1ee3ng m = \u221a2 kg dao \u0111\u1ed9ng \u0111i\u1ec1u h\u00f2a theo ph\u01b0\u01a1ng n\u1eb1m ngang. V\u1eadn t\u1ed1c c\u1ee7a v\u1eadt c\u00f3 \u0111\u1ed9 l\u1edbn c\u1ef1c \u0111\u1ea1i b\u1eb1ng 0,6 m\/s. Ch\u1ecdn th\u1eddi \u0111i\u1ec3m t = 0 l\u00fac v\u1eadt \u0111i qua v\u1ecb tr\u00ed xo<\/sub>\u00a0= 3\u221a2 cm v\u00e0 t\u1ea1i \u0111\u00f3 th\u1ebf n\u0103ng b\u1eb1ng \u0111\u1ed9ng n\u0103ng. Chu k\u00ec dao \u0111\u1ed9ng c\u1ee7a con l\u1eafc v\u00e0 \u0111\u1ed9 l\u1edbn c\u1ee7a l\u1ef1c \u0111\u00e0n h\u1ed3i t\u1ea1i th\u1eddi \u0111i\u1ec3m t = \u03c0\/20 (s) l\u00e0<\/p>\n

A. T = 0,314 s; F = 3 N\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 B. T = 0,628 s; F = 6 N<\/p>\n

C. T = 0,628 s; F = 3 N\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 D. T = 0,314 s; F = 6 N.<\/p>\n

H\u01b0\u1edbng d\u1eabn gi\u1ea3i v\u00e0 \u0111\u00e1p \u00e1n<\/b><\/p>\n\n\n\n\n
C\u00e2u<\/td>\n9<\/td>\n10<\/td>\n11<\/td>\n12<\/td>\n13<\/td>\n14<\/td>\n15<\/td>\n<\/tr>\n
\u0110\u00e1p \u00e1n<\/td>\nD<\/td>\nD<\/td>\nD<\/td>\nD<\/td>\nC<\/td>\nD<\/td>\nB<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

C\u00e2u 9:<\/b>\u00a0D<\/p>\n

\"\"<\/p>\n

K\u00e9o v\u1eadt n\u1eb7ng xu\u1ed1ng d\u01b0\u1edbi 6 cm r\u1ed3i bu\u00f4ng nh\u1eb9 tay \u21d2 A = 6 cm<\/p>\n

X\u00e9t \u0111i\u1ec1u ki\u1ec7n ban \u0111\u1ea7u t = 0 : xo<\/sub>\u00a0= 6cos\u03c6 = \u2013 6 cm<\/p>\n

\u21d2 cos\u03c6 = \u20131 \u21d2 \u03c6 = \u03c0 (rad)<\/p>\n

C\u00e2u 10:<\/b>\u00a0D<\/p>\n

\"\"<\/p>\n

C\u00e2u 11:<\/b>\u00a0D<\/p>\n

\"\"<\/p>\n

C\u00e2u 12:<\/b>\u00a0D<\/p>\n

T\u1eeb ph\u01b0\u01a1ng tr\u00ecnh dao \u0111\u1ed9ng x = 5cos4\u03c0t (cm), ta c\u00f3 :<\/p>\n

A=5 cm=5.10-2<\/sup>\u00a0m ; \u03c9= 4\u03c0 rad\/s<\/p>\n

N\u0103ng l\u01b0\u1ee3ng \u0111\u00e3 truy\u1ec1n cho v\u1eadt ch\u00ednh l\u00e0 c\u01a1 n\u0103ng c\u1ee7a con l\u1eafc, n\u00ean ta c\u00f3 :<\/p>\n

\"\"<\/p>\n

C\u00e2u 13:<\/b>\u00a0C<\/p>\n

\"\"<\/p>\n

C\u00e2u 14:<\/b>\u00a0D<\/p>\n

\"B\u00e0i<\/p>\n

C\u00e2u 15:<\/b>\u00a0B<\/p>\n

\"\"<\/p>\n","protected":false},"excerpt":{"rendered":"

C\u00e2u 9:\u00a0M\u1ed9t con l\u1eafc l\u00f2 xo treo th\u1eb3ng \u0111\u1ee9ng c\u00f3 \u0111\u1ed9 c\u1ee9ng c\u1ee7a l\u00f2 xo k = 100 N\/m, kh\u1ed1i l\u01b0\u1ee3ng v\u1eadt n\u1eb7ng m = 1kg. T\u1eeb v\u1ecb tr\u00ed c\u00e2n b\u1eb1ng k\u00e9o v\u1eadt n\u1eb7ng xu\u1ed1ng d\u01b0\u1edbi 6 cm r\u1ed3i bu\u00f4ng nh\u1eb9 cho v\u1eadt dao \u0111\u1ed9ng \u0111i\u1ec1u h\u00f2a. Ch\u1ecdn tr\u1ee5c t\u1ecda \u0111\u1ed9 c\u00f3 g\u1ed1c t\u1ea1i v\u1ecb […]<\/p>\n","protected":false},"author":3,"featured_media":26376,"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":"\nB\u00e0i t\u1eadp tr\u1eafc nghi\u1ec7m V\u1eadt L\u00ed 12: Con l\u1eafc l\u00f2 xo (ph\u1ea7n 2)<\/title>\n<meta name=\"robots\" content=\"index, follow, max-snippet:-1, max-image-preview:large, max-video-preview:-1\" \/>\n<link rel=\"canonical\" href=\"https:\/\/lop12.edu.vn\/bai-tap-trac-nghiem-vat-li-12-con-lac-lo-xo-phan-2\/\" \/>\n<meta property=\"og:locale\" content=\"en_US\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"B\u00e0i t\u1eadp tr\u1eafc nghi\u1ec7m V\u1eadt L\u00ed 12: Con l\u1eafc l\u00f2 xo (ph\u1ea7n 2)\" \/>\n<meta property=\"og:description\" content=\"C\u00e2u 9:\u00a0M\u1ed9t con l\u1eafc l\u00f2 xo treo th\u1eb3ng \u0111\u1ee9ng c\u00f3 \u0111\u1ed9 c\u1ee9ng c\u1ee7a l\u00f2 xo k = 100 N\/m, kh\u1ed1i l\u01b0\u1ee3ng v\u1eadt n\u1eb7ng m = 1kg. 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