10. )cos( ϕω += tAx
Frequency频率
1
T
f =
Angular frequency角频率
fπω 2=
ϕω +t t 时刻的相位
Phase相位
ϕ t =0 时刻的相位 周相
初相位 退出返回
11. )cos( ϕω += tAx
CAI
Because SHM repeats after each period T and the
cosine function repeats after each 2 rad, one period
T represents a phase difference of 2 rad.π
π
退出返回
12. )cos( ϕω += tAx
)
4
cos(1
π
ω −= tAx
tAx ωcos2
=
12 ϕϕϕ∆ −=
4
)
4
(0
ππ
=−−=
相位差 phase difference
Increasing shifts the curve leftward along the t axisϕ
Decreasing shifts the curve rightward along the t axisϕ
退出返回
13. )cos( 1111 ϕω += tAx )cos( 2222 ϕω += tAx
相位差 )()( 1122 ϕωϕωϕ∆ +−+= tt
若 , 则 比 较早达到某一振动状态0>∆ϕ 1x2x
的相位比 的相位超前1x2x
T
1x
2x
x
1A
2A
O
2A−
1A−
t
的相位比 的相位落后1x 2x
返回 退出
14. )cos( 2222 ϕω += tAx)cos( 1111 ϕω += tAx
)()( 1122 ϕωϕωϕ∆ +−+= tt相位差
12 ϕϕϕ −=∆对两同频率的简谐振动 初相位差
)210(2 ...,,,kk =±= πϕ∆ 同相
)210()12( ...,,,kk =+±= πϕ∆ 反相
1x
2x
x
t
1A
2A
1A−
2A−
O
x
1A
2A
O
1A−
2A−
1x
2x t
退出返回
15. )cos( ϕω += tAx
)sin( ϕωω +−== tA
td
dx
v
)
2
cos(
π
ϕωω ++= tA
)cos(2
2
2
ϕωω +−== tA
td
xd
a
)cos(2
πϕωω ++= tA
退出返回
16. )cos( ϕω += tAx
)sin( ϕωω +−== tA
td
dx
v
的确定和ϕA
0=t 0xx = 0vv = 初始条件
0
0
2
2
02
0
x
v
tg
v
xA
ω
ϕ
ω
−=+=
简谐振动的能量 ( 以水平弹簧振子为例 )
动能 kinetic energy
)sin( ϕωω +−= tAv)cos( ϕω += tAx
退出返回
17. 2
2
1
mvK =
2
max
2
1
kAK =
∫
+
=〉〈
Tt
t
Kdt
T
K
1
势能 Potential energy
0min =K
)(sin
2
1 22
ϕω += tkA
m
k
=2
ω
)(sin)(
2
1 22
ϕωω +−= tAm
2
4
1
kA=
2
2
1
kxU = )(cos
2
1 22
ϕω += tkA
2
max
2
1
kAU = 0min
=U
∫
+
=〉〈
Tt
t
Udt
T
U
1 2
4
1
kA=
退出返回
18. 机械能 The mechanical energy
2
2
1
kA=UKE +=
The mechanical energy of a linear oscillator is constant and
independent of time
2
AE ∝简谐振动的机械能
退出返回
