Chuyen de 1 dao dong dieu hoa

TT Luyện thi ĐH TÂN VIỆT - 54 Quốc Lộ 22 Quận 12 TP HCM - Chuyên  1 : u hòa http://thuvienvatly.com/u/32950 - GV: Đoàn Văn Lượng - Email: doanvluong@gmail.com Trang. 1 CHUYÊN  1:  1. Dao động điều hòa  x = Acos(t + ).   2. Các đại lượng đặc trưng của dao động điều hoà: x = Acos(t + ) thì: Các đại lượng đặc trưng   A x max = A >0 m, cm, mm (t + )     rad   rad/s. T  T = 2  = N t s ( giây) f   1 f T  Hz ( Héc) hay 1/s , T và f:  = T  2 = 2f=> Biên     3. Mối liên hệ giữa li độ , vận tốc và gia tốc của vật dao động điều hoà: Đại lượng Biểu thức So sánh, liên hệ  x = Acos(t + ):  :  2 x = 0   x max = A   2     v = x' = - Asin(t + ) v= Acos(t +  + 2  ) - A), v = 0. - max = A. -           2  so  - Khi v v trí biên v v trí cân bng thì vn t ln, khi v v trí cân bng v biên thì vn t ln gim dn.  -  2 Acos(t + ) a= -  2 x.    -   a max =  2 A. -  -  pha 2   -Khi v  v trí cân bn v trí biên, a c chiu vi v ( vt chuyng chm dn) -Khi v   v   n v trí cân bng, a cùng chiu vi v ( vt chuy ng nhanh dn).  F = ma = - kx       :luôn   F max = kA - Chuyng nhanh dn : a.v>0, vF    ; - ng chm dn a.v<0 , vF    ( F  là hp lc tác dng lên vt) [...]... Câu4 Câu5 Câu6 Câu7 Câu8 Câu9 Câu10 A A A A A B A D B A Câu 11 Câu12 Câu13 Câu14 Câu15 Câu16 Câu17 Câu18 Câu19 Câu20 D A C A A Câu 1: HD: Ta có: x  Acos  t+       v   A sin  t+   A sin  t+     Acos   t+      Acos   t+   2 2   1 v 2 2 2 Câu 2: HD: T1  ; v1  A 1  A ; T2  2 T1 ; v2  A2  A A  1  v1  2 v2 f T2 T1 2T1 2 2 2 2 3 Câu 3: HD: Chu kì:... t  ) cm C x  6 2 cos (10  t  ) cm D x  6 2 cos (10  t ) cm Câu 1 Câu2 Câu3 Câu4 Câu5 Câu6 Câu7 Câu8 Câu9 Câu10 B B A A B A B A B B Câu 11 Câu12 Câu13 Câu14 Câu15 Câu16 Câu17 Câu18 Câu19 Câu20 http://thuvienvatly.com/u/32950 - GV: Đoàn Văn Lượng - Email: doanvluong@gmail.com Trang 30 TT Luyện thi ĐH TÂN VIỆT - 54 Quốc Lộ 22 Quận 12 TP HCM - Chuyên đề 1 : Dao động điều hòa Câu 1: HD Giải:  = 2πf = π... 12 073  2 012    6 6 6 12 073 12 073 6 Suy ra: t     s  4 24   10 06.2  Ví dụ 12 : Một vật dao động điều hòa theo phương trình P   x  10 cos 10 t   2  (cm) Xác định thời điểm vật qua vị trí x = 5 cm lần thứ 2 014 Hướng dẫn giải: Ta có: 5  10 cos 10 t     cos 10 t     1  cos            2   10 t     10 t     k2   2 3 10 t    2  2  2 3  1. .. 22 Quận 12 TP HCM - Chuyên đề 1 : Dao động điều hòa 2  10  t1  3  2k ; k  Z  =>  2 10  t    2 k ; k  Z 2  3  2 k 1  t1  30  15 ; k  0  t1min  15  Chon B  2 k 2 t    ; k  3  t2min  2 30 15 15  Hướng dẫn giải 2: Dùng vòng tròn lượng giác : Vẽ hình với góc quay 2π/3 => Thời gian là T/3 =1/ 15s   Ví dụ 4: Một vật dao động điều hòa với phương trình x  20cos 10 t ... Khi đó: vmax  A  4.5  20  cm/s  Ví dụ 3: Một chất điểm dao động điều hòa theo phương trình:x = 5cos 10 t (cm) Thời điểm chất điểm qua điểm M1 có li độ x1 = - 2,5 cm lần thứ nhất là: A 1 s 60 B 1 s 15 1 C s 6 D 11 s 60 Hướng dẫn giải 1: Thế li độ x1 = - 2,5 cm vào phương trình dao động ta có:  2,5  5 cos10t n  cos10t n   1 2 http://thuvienvatly.com/u/32950 - GV: Đoàn Văn Lượng - Email:... - 1 6035 12 07 )T= 0,2 = (s) Chọn B 12 12 12 http://thuvienvatly.com/u/32950 - GV: Đoàn Văn Lượng - Email: doanvluong@gmail.com Trang 18 TT Luyện thi ĐH TÂN VIỆT - 54 Quốc Lộ 22 Quận 12 TP HCM - Chuyên đề 1 : Dao động điều hòa Câu 7: Giải: Giả sử phương trình dao động của vật có dạng x = Acos 2 t (cm) T x1 = Acos 2 t1 (cm) T x2 = Acos 2 2 2 2 T  t2 = Acos (t1+ ) = Acos( t1 + ) (cm) = - Asin t1... A2 2  A2 v  A2 x2 nếu v là tốc độ thì v x2 v + 1 A2 x2 + Với hai thời điểm t 1, t2 vật có các c p giá trị x1, v1 và x2, v2 thì ta có hệ thức sau: x1 A 2 v1 A 2 x2 A 2 v2 A 2 2 x1 2 x2 2 v1 A2 2 v1 2 x1 2 v2 2 A2 T 2 v2 2 2 x2 2 x1 2 v1 2 x2 v2 2 b) Từ các phương trình của vận tốc và gia tốc ta có v A sin a A 2 t cos a 2 A t 2 v A 2 1 a2 4 2 A v2 2 2 A 1, 2 (2) được gọi là hệ thức liên hệ của a, A,... (1)  v  Asin  t     v 2  2 A 2 sin 2  t        2   a2  2 2 2 a  2 Acos  t    a  4 A 2cos 2  t      2   A cos  t    (2)    Lấy (1) cộng (2), ta được: v2  a2  2 A 2  a 2  2  2 A 2  v 2   a   2A 2  v 2  10  10 0 2 2  10 00 2   a  10  2000  10 00  10 10 10  10 02  10 00  cm/s2   10  m/s2  Ví dụ 5: Một một có khối lượng m = 10 0... 10 0 3  cm/s  - Li độ: - Lực hồi phục: Fhp  m2 x  0 ,1. 40.2,5 3 .10 2  0 ,1 3  N    12 00       t    2 6    - Li độ: x  5cos     5sin  2,5  cm  6 2 6    - Vận tốc: v  2.5sin     10 cos  5 3  cm/s  6 2 6 2 2 - Gia tốc: a   x  40. 2,5  10 0  cm/s  b Khi pha dao động 12 00 - Lực hồi phục: Fhp  m2 x  10 1. 4 .10 . 2,5 .10 2   0 ,1 ... 6: Cho một vật dao động điều hoà với phương trình x = 10 cos (10 t) cm Vận tốc của vật có độ lớn 50cm/s lần thứ 2 012 tại thời điểm A 6209 s 60 B 12 07 s 12 C 12 05 s 12 D 60 31 s 60 Câu 7: Một vật nhỏ dao động điều hòa với chu k T=1s Tại thời điểm t1 nào đó, li độ của vật là -2cm Tại thời điểm t2 = t1+0.25s,vận tốc của vật có giá trị : A: 4 cm/s B:-2 m/s C:2cm/s D:- 4m/s Câu 8: Một vật dao động điều . 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