How to compare the phases of two shos with e same a =Xml cos(at+)4p=(at+2)-(at+, x=x cos(at+ 2)A0=(2= 超前 △O=0同相△9=士兀反相Ao为其它 落后 x O
= 0 x t o 同相 How to compare the phases of two SHOs with same ? cos( ) 1 = 1 +1 x x t m cos( ) 2 = 2 +2 x x t m ( ) ( ) = + 2 − +1 t t = 2 −1 x t o 为其它 超前 落后 t x o = π 反相
Fig 17-6 shows several simple harmonic motions Fg17-6 0.5 6 -0.5 a same xX different: o (b)same: 0,o different (c)same: xm, o different: a
(c) same: , different: 1 2 3 4 5 6 t -1 -0.5 0.5 1 x t 2 4 6 8 t -1 -0.5 0.5 1 x t a 1 2 3 4 5 6 t -1 -0.5 0.5 1 x t Fig 17-6 shows several simple harmonic motions. (a) (b) (c) (a) same: , different: m x (b) same: , different: m x Fig 17-6 m x
d) Displacement, velocity, and acceleration Displacement x=xm coS(at +o) Velocity v dx =-Oxm sin( ot+o)=@xm cos(@t+o+o Acceleration ax =-ox cos(at axm cos(at +I+o) (17-11) When the displacement is a maximum in either direction, the speed is zero, because the velocity must now change its direction
d). Displacement, velocity, and acceleration Displacement Velocity Acceleration When the displacement is a maximum in either direction, the speed is zero, because the velocity must now change its direction. x = x cos(t +) m = = −x sin(t + ) dt dx vx m cos( ) 2 2 2 = = − x t + dt d x ax m ) 2 cos( =xm t + + cos( ) 2 = xm t + + (17-11)
xx一t图 .x=xm cos(at+o) 21 T 取Q=0 x 7【7=1图 X 0 au=-xm@sin( at+p) O T x.coslot+(0+ a=-x@ cos(ot +/a Q图 xm,ucos(ot+o+T)O
x − t 图 v − t 图 a − t 图 T xm− xm 2 xm 2 − xm xva ttt mxm − xooo TT x = x cos(t +) m 取 = 0 2π T = ) 2π = x m cos( t + + = − x sin( t + ) v m cos( π ) 2 = x m t + + cos( ) 2 a = − x m t +
17-4 Energy in simple harmonic motion 1. The potential energy U=-k kxm coS(at+o) 2 2 p=0 (17-12) U(t) →2. The kinetic energy K moxa sin (at+o) K(t) kxm sin(at+o (17-13) T/2 Fig17-8(a) U=-xmn @sin( at +p)
17-4 Energy in simple harmonic motion 1.The potential energy (17-12) 2.The kinetic energy 1 2 3 4 5 6 0.2 0.4 0.6 0.8 1 cos ( ) 2 1 2 1 2 2 2 U = k x = k xm t + U(t) K(t) T/2 T sin ( ) 2 1 sin ( ) 2 1 2 1 2 2 2 2 2 2 = + = = + k x t K m v m x t m m = −x sin(t +) v m = 0 (17-13) Fig 17-8(a)