Bộ tài liệu bài tập tham khảo môn vật lý bậc đại học bằng tiếng anh Chapter 13 | a T f x 10 3s œ T 2nf x 103 rad s. b 10 x 10 3 s œ 2nf x 103 rad s. a Since the glider is released form rest its initial displacement m is the amplitude. b The glider will return to its original position after another s so the period is s. c The frequency is the reciprocal of the period Eq. f ik Hz. 1 I h ic 0 50s 1 1 1 x 1 3 c onr ffi onm11 qv iz xt -i c The period is 440 10 s and the angular frequency is a x 103 rad s. a From the graph of its motion the object completes one full cycle in s its period is thus s and its frequency 1 period s . b The displacement varies from - m to m so the amplitude is m. c s see part a This displacement is of a period. T 1 f s so t s. The period will be twice the time given as being between the times at which the glider is at the equilibrium position see Fig. k2n Y 2n Y . k ok m I I m I-------- I kg N m. T 2 s J a T f s. b œ 2nf rad s. c m kg. Solving Eq. for k k 2n V k 2n V k m I kg I 2n I I T I x 103 Nm. g f T 1 T-1 1 1 1 T-1 1 1 A rT1 O I kg C J 1 Z Z TT From Eq. and Eq. T 2nJ 140 m s j y Hz œ 2nf rad s. a ax dy- -œ2A sin œt fi -œ2x so x t is a solution to Eq. if œ2 a 2Aœ a constant so Eq. is not satisfied. c vx d iœ œt fi a zœ 2 Aé œt fi œ2x so x t is a solution to Eri 13 4 if œ2 k I m x dt j -ZT. e x s x x j is a. s xi i i q. . j a a. rx y ni a x mm cos 2n 440Hz t b 10 3 m 2n 440 Hz s 2 2 440Hz 2 x 104m s2. c j t x 107 m s3 sin 2n 440Hz t jmax x 107 ms3. max a From Eq. A v0 Jk m m. b Equation is indeterminant but from Eq. 0 y and from Eq. sin 0 so y. c cos ol n 2 - sin ol so x m sin rad s t . With the same value for o Eq. gives A m and Eq. gives and x
