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{"id":25263,"date":"2018-04-04T17:30:12","date_gmt":"2018-04-04T10:30:12","guid":{"rendered":"https:\/\/lop12.edu.vn\/?p=25263"},"modified":"2018-04-04T17:30:12","modified_gmt":"2018-04-04T10:30:12","slug":"chuong-2-bai-6-dao-dong-dieu-hoa","status":"publish","type":"post","link":"https:\/\/lop12.edu.vn\/chuong-2-bai-6-dao-dong-dieu-hoa\/","title":{"rendered":"Ch\u01b0\u01a1ng 2 – B\u00e0i 6: Dao \u0111\u1ed9ng \u0111i\u1ec1u h\u00f2a"},"content":{"rendered":"

C\u00c2U H\u1eceI<\/strong><\/p>\n

C\u00e2u 1 (trang 34 sgk V\u1eadt L\u00ed 12 n\u00e2ng cao):<\/b>\u00a0V\u1ebd \u0111\u1ed3 th\u1ecb li \u0111\u1ed9 c\u1ee7a dao \u0111\u1ed9ng \u0111i\u1ec1u h\u00f2a sau \u0111\u00e2y (c\u00f9ng d\u1ea1ng v\u1edbi \u0111\u01b0\u1eddng li\u1ec1n n\u00e9t (2) trong H\u00ecnh 2.3): x = 2cos(\u03c0t – \u03c0\/4) (cm)<\/span><\/p>\n

Ghi r\u00f5 t\u1ecda \u0111\u1ed9 giao \u0111i\u1ec3m giao c\u1ee7a \u0111\u01b0\u1eddng bi\u1ec3u di\u1ec5n v\u1edbi tr\u1ee5c tung (x) v\u00e0 tr\u1ee5c ho\u00e0nh (t).<\/p>\n

L\u1eddi gi\u1ea3i:<\/b><\/p>\n

H\u1ecdc sinh t\u1ef1 v\u1ebd theo d\u1ea1ng \u1edf H\u00ecnh 6.5 SGK<\/p>\n

C\u00e2u 2 (trang 34 sgk V\u1eadt L\u00ed 12 n\u00e2ng cao):<\/b>\u00a0X\u00e9t ba \u0111\u1ea1i l\u01b0\u1ee3ng \u0111\u1eb7c tr\u01b0ng A, \u03c6, \u03c9 cho dao \u0111\u1ed9ng \u0111i\u1ec1u h\u00f2a c\u1ee7a m\u1ed9t con l\u1eafc l\u00f2 xo \u0111\u00e3 cho. Nh\u1eefng \u0111\u1ea1i l\u01b0\u1ee3ng n\u00e0o c\u00f3 th\u1ec3 c\u00f3 nh\u1eefng gi\u00e1 tr\u1ecb kh\u00e1c nhau, t\u00f9y thu\u1ed9c c\u00e1ch k\u00edch th\u00edch dao \u0111\u1ed9ng? \u0110\u1ea1i l\u01b0\u1ee3ng n\u00e0o ch\u1ec9 c\u00f3 m\u1ed9t gi\u00e1 tr\u1ecb x\u00e1c \u0111\u1ecbnh \u0111\u1ed1i v\u1edbi con l\u1eafc l\u00f2 xo \u0111\u00e3 cho?<\/span><\/p>\n

L\u1eddi gi\u1ea3i:<\/b><\/p>\n

Trong ba \u0111\u1ea1i l\u01b0\u1ee3ng tr\u00ean th\u00ec bi\u00ean \u0111\u1ed9 A, pha ban \u0111\u1ea7u \u03c6 l\u00e0 c\u00e1c \u0111\u1ea1i l\u01b0\u1ee3ng c\u00f3 th\u1ec3 c\u00f3 nh\u1eefng gi\u00e1 tr\u1ecb kh\u00e1c nhau, t\u00f9y thu\u1ed9c c\u00e1ch k\u00edch th\u00edch dao \u0111\u1ed9ng. C\u00f2n t\u1ea7n s\u1ed1 g\u00f3c \u03c9 ch\u1ec9 c\u00f3 m\u1ed9t gi\u00e1 tr\u1ecb x\u00e1c \u0111\u1ecbnh \u0111\u1ed1i v\u1edbi con l\u1eafc l\u00f2 xo \u0111\u00e3 cho v\u00e0 lu\u00f4n b\u1eb1ng \u03c9 = \u221a(k\/m).<\/p>\n

\n
\n

C\u00e2u 3 (trang 34 sgk V\u1eadt L\u00ed 12 n\u00e2ng cao):<\/b>\u00a0N\u00f3i r\u00f5 v\u1ec1 th\u1ee9 nguy\u00ean c\u1ee7a c\u00e1c l\u01b0\u1ee3ng A, \u03c6, \u03c9.<\/span><\/p>\n

L\u1eddi gi\u1ea3i:<\/b><\/p>\n

– Th\u1ee9 nguy\u00ean c\u1ee7a c\u00e1c \u0111\u1ea1i l\u01b0\u1ee3ng A l\u00e0 chi\u1ec1u d\u00e0i (m, cm …)<\/p>\n

– \u0110\u1ea1i l\u01b0\u1ee3ng \u03c6 l\u00e0 g\u00f3c n\u00ean kh\u00f4ng c\u00f3 th\u1ee9 nguy\u00ean.<\/p>\n

– Th\u1ee9 nguy\u00ean c\u1ee7a \u0111\u1ea1i l\u01b0\u1ee3ng [\u03c9] = 1\/T \u21d2 th\u1ee9 nguy\u00ean l\u00e0 ngh\u1ecbch \u0111\u1ea3o c\u1ee7a th\u1eddi gian (1\/s hay s(-1)<\/sup>)<\/p>\n

B\u00c0I T\u1eacP<\/strong><\/p>\n

B\u00e0i 1 (trang 34 sgk V\u1eadt L\u00ed 12 n\u00e2ng cao):<\/b>\u00a0T\u1ed1c \u0111\u1ed9 c\u1ee7a ch\u1ea5t \u0111i\u1ec3m giao \u0111\u1ed9ng \u0111i\u1ec1u h\u00f2a c\u1ef1c \u0111\u1ea1i khi:<\/span><\/p>\n

Li \u0111\u1ed9 c\u1ef1c \u0111\u1ea1i\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0B. gia t\u1ed1c c\u1ef1c \u0111\u1ea1i<\/p>\n

C. li \u0111\u1ed9 b\u1eb1ng 0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0D. pha b\u1eb1ng \u03c0\/4<\/p>\n

L\u1eddi gi\u1ea3i:<\/b><\/p>\n

Ch\u1ecdn C<\/p>\n

B\u00e0i 2 (trang 35 sgk V\u1eadt L\u00ed 12 n\u00e2ng cao):<\/b>\u00a0Gia t\u1ed1c c\u1ee7a ch\u1ea5t \u0111i\u1ec3m dao \u0111\u1ed9ng \u0111i\u1ec1u h\u00f2a b\u1eb1ng 0 khi:<\/span><\/p>\n

A. Li \u0111\u1ed9 c\u1ef1c \u0111\u1ea1i\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0B. li \u0111\u1ed9 c\u1ef1c ti\u1ec3u<\/p>\n

C. V\u1eadn t\u1ed1c c\u1ef1c \u0111\u1ea1i ho\u1eb7c c\u1ef1c ti\u1ec3u\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0D. v\u1eadn t\u1ed1c b\u1eb1ng 0.<\/p>\n

L\u1eddi gi\u1ea3i:<\/b><\/p>\n

Ch\u1ecdn C<\/p>\n

\n
\n

B\u00e0i 3 (trang 35 sgk V\u1eadt L\u00ed 12 n\u00e2ng cao):<\/b>\u00a0Dao \u0111\u1ed9ng c\u01a1 \u0111i\u1ec1u h\u00f2a \u0111\u1ed5i chi\u1ec1u khi:<\/span><\/p>\n

A. L\u1ef1c t\u00e1c d\u1ee5ng \u0111\u1ed5i chi\u1ec1u\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0B. l\u1ee5c t\u00e1c d\u1ee5ng b\u1eb1ng 0<\/p>\n

C. l\u1ef1c t\u00e1c d\u1ee5ng c\u00f3 \u0111\u1ed9 l\u1edbn c\u1ef1c \u0111\u1ea1i\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0D. l\u1ef1c t\u00e1c d\u1ee5ng c\u00f3 \u0111\u1ed9 l\u1edbn c\u1ef1c ti\u1ec3u.<\/p>\n

L\u1eddi gi\u1ea3i:<\/b><\/p>\n

Ch\u1ecdn C<\/p>\n

B\u00e0i 4 (trang 35 sgk V\u1eadt L\u00ed 12 n\u00e2ng cao):<\/b>\u00a0a) Th\u1eed l\u1ea1i r\u1eb1ng:<\/span><\/p>\n

x = A1<\/sub>cos\u2061\u03c9t + A2<\/sub>sin\u2061\u03c9t (6.14)<\/p>\n

trong \u0111\u00f3 A1<\/sub>\u00a0v\u00e0 A2<\/sub>\u00a0l\u00e0 hai h\u1eb1ng s\u1ed1 b\u1ea5t k\u00ec c\u0169ng l\u00e0 nghi\u1ec7m c\u1ee7a ph\u01b0\u01a1ng tr\u00ecnh (6.3).<\/p>\n

b) Ch\u1ee9ng t\u1ecf r\u1eb1ng, n\u1ebfu ch\u1ecdn A1<\/sub>\u00a0v\u00e0 A2<\/sub>\u00a0trong bi\u1ec3u th\u1ee9c \u1edf v\u1ebf ph\u1ea3i c\u1ee7a (6.14) nh\u01b0 sau:<\/p>\n

A1<\/sub>\u00a0= A.cos\u2061\u03c6; A2<\/sub>\u00a0= -Asin \u03c6<\/p>\n

th\u00ec bi\u1ebfu th\u1ee9c \u1ea5y tr\u00f9ng v\u1edbi bi\u1ec3u th\u1ee9c \u1edf v\u1ebf ph\u1ea3i c\u1ee7a (6.4).<\/p>\n

L\u1eddi gi\u1ea3i:<\/b><\/p>\n

a) Th\u1eed l\u1ea1i r\u1eb1ng: x = A1<\/sub>\u00a0cos\u2061\u03c9t + A2<\/sub>sin\u2061\u03c9t (6.14)<\/p>\n

Trong \u0111\u00f3 A1<\/sub>; A2<\/sub>\u00a0l\u00e0 hai h\u1eb1ng s\u1ed1 b\u1ea5t k\u00ec c\u0169ng l\u00e0 nghi\u1ec7m c\u1ee7a Ph\u01b0\u01a1ng tr\u00ecnh (6.3)<\/p>\n

Ta c\u00f3 : x’= -\u03c9A1<\/sub>sin\u2061\u03c9t + \u03c9A2<\/sub>cos\u2061\u03c9t<\/p>\n

Suy ra : x”= -\u03c92<\/sup>A1<\/sub>cos\u2061\u03c9t – \u03c92<\/sup>A2<\/sub>sin\u2061\u03c9t (6.15)<\/p>\n

thay (6.15) v\u00e0 (6.14) v\u00e0o Ph\u01b0\u01a1ng tr\u00ecnh (6.3) x”+ \u03c92<\/sup>x = 0 ta th\u1ea5y nghi\u1ec7m \u0111\u00fang.<\/p>\n

b) N\u1ebfu ch\u1ecdn A1<\/sub>\u00a0v\u00e0 A2<\/sub>\u00a0trong bi\u1ec3u th\u1ee9c \u1edf v\u1ebf ph\u1ea3i c\u1ee7a (6.14) nh\u01b0 sau.<\/p>\n

A1<\/sub>\u00a0= Acos\u03c6; A2<\/sub>\u00a0= -Asin\u03c6.<\/p>\n

Thay A2<\/sub>\u00a0v\u00e0 A1<\/sub>\u00a0b\u1eb1ng c\u00e1c gi\u00e1 tr\u1ecb \u0111\u00e3 ch\u1ecdn v\u00e0o (6.14) ta \u0111\u01b0\u1ee3c:<\/p>\n

x = A1<\/sub>\u00a0cos\u2061\u03c9t + A2<\/sub>sin\u2061\u03c9t<\/p>\n

= Acos\u2061\u03c6cos\u2061[\u03c9t] – Asin\u03c6sin\u2061\u03c9t = Acos(\u03c9t + \u03c6)<\/p>\n

B\u00e0i 5 (trang 35 sgk V\u1eadt L\u00ed 12 n\u00e2ng cao):<\/b>\u00a0Ph\u01b0\u01a1ng tr\u00ecnh dao \u0111\u1ed9ng c\u1ee7a m\u1ed9t v\u1eadt l\u00e0: x = 6 cos\u2061(4 \u03c0t + \u03c0\/6) (cm)<\/span><\/p>\n

X\u00e1c \u0111\u1ecbnh bi\u00ean \u0111\u1ed9, t\u1ea7n s\u1ed1 g\u00f3c, chu k\u00ec t\u1ea7n s\u1ed1 c\u1ee7a dao \u0111\u1ed9ng.<\/p>\n

X\u00e1c \u0111\u1ecbnh pha dao \u0111\u1ed9ng t\u1ea1i th\u1eddi \u0111i\u1ec3m t = 1\/4 s, t\u1eeb \u0111\u00f3 suy ra li \u0111\u1ed9 t\u1ea1i th\u1eddi \u0111i\u1ec3m \u1ea5y.<\/p>\n

V\u1ebd vect\u01a1 quay bi\u1ec3u di\u1ec5n dao \u0111\u1ed9ng v\u00e0o th\u1eddi \u0111i\u1ec3m t = 0.<\/p>\n

L\u1eddi gi\u1ea3i:<\/b><\/p>\n

<\/div>\n

Bi\u00ean \u0111\u1ed9 6 cm, t\u1ea7n s\u1ed1 g\u00f3c l\u00e0 4 \u03c0, chu k\u00ec T=(2 \u03c0)\/\u03c9 = 0,5s, t\u1ea7n s\u1ed1 2 Hz.<\/p>\n

Pha 4 \u03c0.1\/4 + \u03c0\/6 = 7\/2 \u03c0 rad, li \u0111\u1ed9 x = 6 cos\u2061[7\/6 \u03c0] = -3\u221a3 cm<\/p>\n

\u0110\u1ed9 d\u00e0i vectocm, g\u00f3c h\u1ee3p v\u1edbi Ox l\u00e0 \u03c0\/6<\/p>\n

\"\"<\/p>\n

B\u00e0i 6 (trang 35 sgk V\u1eadt L\u00ed 12 n\u00e2ng cao):<\/b>\u00a0M\u1ed9t v\u1eadt dao \u0111\u1ed9ng \u0111i\u1ec1u h\u00f2a v\u1edbi bi\u00ean \u0111\u1ed9 A = -4 cm v\u00e0 chu k\u00ec T = 2s.<\/span><\/p>\n

a) Vi\u1ebft ph\u01b0\u01a1ng tr\u00ecnh dao \u0111\u1ed9ng c\u1ee7a v\u1eadt, ch\u1ecdn g\u1ed1c th\u1eddi gian l\u00e0 l\u00fac n\u00f3 \u0111i qua v\u1ecb tr\u00ed c\u00e2n b\u1eb1ng theo chi\u1ec1u d\u01b0\u01a1ng.<\/p>\n

b) T\u00ednh li \u0111\u1ed9 c\u1ee7a v\u1eadt t\u1ea1i th\u01b0\u1eddi \u0111i\u1ec3m t = 5,5s.<\/p>\n

L\u1eddi gi\u1ea3i:<\/b><\/p>\n

a) D\u1ea1ng t\u1ed5ng qu\u00e1t c\u1ee7a Ph\u01b0\u01a1ng tr\u00ecnh dao \u0111\u1ed9ng l\u00e0 x = Acos (\u03c9t + \u03c6)<\/p>\n

Bi\u00ean \u0111\u1ed9 A = -4cm.<\/p>\n

T\u1ea7n s\u1ed1 g\u00f3c \u03c9 = 2\u03c0f = \u03c0 (rad\/s)<\/p>\n

Pha ban \u0111\u1ea7u \u03c6 x\u00e1c \u0111\u1ecbnh b\u1edfi \u0111i\u1ec1u ki\u1ec7n ban \u0111\u1ea7u (ch\u1ecdn g\u1ed1c th\u1eddi gian):<\/p>\n

T = 0 th\u00ec x(0) = 0; v(0) = x\u2019(0) > 0<\/p>\n

Suy ra cos\u2061\u03c6 = 0; -sin\u2061\u03c6 > 0 t\u1ee9c l\u00e0 \u03c6 = -\u03c0\/2.<\/p>\n

V\u1eady ta \u0111\u01b0\u1ee3c: x = 4 cos\u2061(\u03c0t-\u03c0\/2) (cm )<\/p>\n

b) t = 5,5s th\u00ec x = 4 cos 5\u03c0 = -4cm<\/p>\n

B\u00e0i 7 (trang 35 sgk V\u1eadt L\u00ed 12 n\u00e2ng cao):<\/b>\u00a0M\u1ed9t v\u1eadt n\u1eb7ng treo v\u00e0o l\u00f2 xo l\u00e0m cho l\u00f2 xo d\u00e3n ra 0,8 cm. Cho v\u1eadt dao \u0111\u1ed9ng. T\u00ecm chu k\u00ec dao \u0111\u1ed9ng \u1ea5y. L\u1ea5y g = 10 m\/s2<\/sup>.<\/span><\/p>\n

L\u1eddi gi\u1ea3i:<\/b><\/p>\n

Chu k\u00ec dao \u0111\u1ed9ng T = 2 \u03c0 \u221a(m\/k), m l\u00e0 kh\u1ed1i l\u01b0\u1ee3ng c\u1ee7a v\u1eadt, k l\u00e0 \u0111\u1ed9 c\u1ee9ng c\u1ee7a l\u00f2 xo.<\/p>\n

X\u00e9t \u0111i\u1ec1u ki\u1ec7n c\u00e2n b\u1eb1ng c\u1ee7a v\u1eadt ta c\u00f3:<\/p>\n

mg = k\u2206l<\/p>\n

\u21d2m.10 = k.0,008<\/p>\n

V\u1eady m\/k = 0,008\/100 = [8.10](-4)<\/sup><\/p>\n

T = 2 \u03c0 \u221a([8.10](-4)<\/sup>\u00a0) = 2\u03c02 \u221a2.[10](-2)<\/sup>\u00a0\u2248 0,18s<\/p>\n

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

C\u00c2U H\u1eceI C\u00e2u 1 (trang 34 sgk V\u1eadt L\u00ed 12 n\u00e2ng cao):\u00a0V\u1ebd \u0111\u1ed3 th\u1ecb li \u0111\u1ed9 c\u1ee7a dao \u0111\u1ed9ng \u0111i\u1ec1u h\u00f2a sau \u0111\u00e2y (c\u00f9ng d\u1ea1ng v\u1edbi \u0111\u01b0\u1eddng li\u1ec1n n\u00e9t (2) trong H\u00ecnh 2.3): x = 2cos(\u03c0t – \u03c0\/4) (cm) Ghi r\u00f5 t\u1ecda \u0111\u1ed9 giao \u0111i\u1ec3m giao c\u1ee7a \u0111\u01b0\u1eddng bi\u1ec3u di\u1ec5n v\u1edbi tr\u1ee5c tung (x) v\u00e0 […]<\/p>\n","protected":false},"author":3,"featured_media":25264,"comment_status":"closed","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"_mi_skip_tracking":false,"tdm_status":"","tdm_grid_status":""},"categories":[1304],"tags":[1432,1431],"yoast_head":"\nCh\u01b0\u01a1ng 2 - B\u00e0i 6: Dao \u0111\u1ed9ng \u0111i\u1ec1u h\u00f2a<\/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\/chuong-2-bai-6-dao-dong-dieu-hoa\/\" \/>\n<meta property=\"og:locale\" content=\"en_US\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"Ch\u01b0\u01a1ng 2 - B\u00e0i 6: Dao \u0111\u1ed9ng \u0111i\u1ec1u h\u00f2a\" \/>\n<meta property=\"og:description\" content=\"C\u00c2U H\u1eceI C\u00e2u 1 (trang 34 sgk V\u1eadt L\u00ed 12 n\u00e2ng cao):\u00a0V\u1ebd \u0111\u1ed3 th\u1ecb li \u0111\u1ed9 c\u1ee7a dao \u0111\u1ed9ng \u0111i\u1ec1u h\u00f2a sau \u0111\u00e2y (c\u00f9ng d\u1ea1ng v\u1edbi \u0111\u01b0\u1eddng li\u1ec1n n\u00e9t (2) trong H\u00ecnh 2.3): x = 2cos(\u03c0t – \u03c0\/4) (cm) Ghi r\u00f5 t\u1ecda \u0111\u1ed9 giao \u0111i\u1ec3m giao c\u1ee7a \u0111\u01b0\u1eddng bi\u1ec3u di\u1ec5n v\u1edbi tr\u1ee5c tung (x) v\u00e0 […]\" \/>\n<meta property=\"og:url\" content=\"https:\/\/lop12.edu.vn\/chuong-2-bai-6-dao-dong-dieu-hoa\/\" \/>\n<meta property=\"og:site_name\" content=\"Lop12.edu.vn - C\u1ed9ng \u0111\u1ed3ng h\u1ecdc sinh l\u1edbp 12 l\u1edbn nh\u1ea5t t\u1ea1i Vi\u1ec7t Nam\" \/>\n<meta property=\"article:published_time\" content=\"2018-04-04T10:30:12+00:00\" \/>\n<meta property=\"og:image\" content=\"https:\/\/lop12.edu.vn\/wp-content\/uploads\/2018\/04\/1-54.png\" \/>\n\t<meta property=\"og:image:width\" content=\"170\" \/>\n\t<meta property=\"og:image:height\" content=\"152\" \/>\n\t<meta property=\"og:image:type\" content=\"image\/png\" \/>\n<meta name=\"author\" content=\"H\u00e0 Trang\" \/>\n<meta name=\"twitter:card\" content=\"summary_large_image\" \/>\n<meta name=\"twitter:label1\" content=\"Written by\" \/>\n\t<meta name=\"twitter:data1\" content=\"H\u00e0 Trang\" \/>\n\t<meta name=\"twitter:label2\" content=\"Est. reading time\" \/>\n\t<meta name=\"twitter:data2\" content=\"6 minutes\" \/>\n<script type=\"application\/ld+json\" class=\"yoast-schema-graph\">{\"@context\":\"https:\/\/schema.org\",\"@graph\":[{\"@type\":\"WebPage\",\"@id\":\"https:\/\/lop12.edu.vn\/chuong-2-bai-6-dao-dong-dieu-hoa\/\",\"url\":\"https:\/\/lop12.edu.vn\/chuong-2-bai-6-dao-dong-dieu-hoa\/\",\"name\":\"Ch\u01b0\u01a1ng 2 - 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