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Dạng 1: Nhận biết phương trình dao động
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a.Xác định A, φ,
ω
………
g4+.K+9*"h(.2i+8.
g$$.DK+9h$<?SjSωkkk
b.Suy ra cách kích thích dao động :
g@>I.K+9
⇒⇒F.&"
x A cos( t )
v A sin( t )
= ω + ϕ
= − ω ω + ϕ
0
0
x
v
c.Chú ý:
g5+9h<=>?$-ωj/A
>gω?$-ωj/A>gω
/-#/ /-#/
/-#/ "/-#/
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Fi#%6*%"* n.6
S!$MSK!S."1
5. os(4. . )
6
x c t
π
π
= +
5. os( . )x c t
π
= −
5. os(2. . )
4
x c t
π
π
= − +
10.sin(5. . )
3
x t
π
π
= +
5. ( . ) 1x cos t
π
= − +
2
.a x
ω
= −
0; 0v a> >f
0; 0v a< <p p
,&'()#
&'
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E
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-#L$/
OvgviwM+8*D*"+b
eK"TI
I
1<
B<-#/
EM<-#L$/
ZM<-#L$
ω ω ϕ π π π π
= = + = + = +
5.sin(2. .5 ) 5.sin( ) 2,5( ).
6 6
x cm
π π
π
= + = =
3
10. . (2. .5 ) 10. . ( ) 10. . 5. 30
6 6 2
v cos cos
π π
π π π π
= + = = =
2 2
2 2
. 4. .2,5 100( ) 1( )
cm m
a x
s s
ω π
= − = − = − = −
0
5.sin120 2,5. 3x = =
0
10. . 120 5.v cos
π π
= = −
2 2
π π
= = − =
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5+9#"*1"<
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6o
4.sin(10. . )
4
x t
π
π
= +
Dạng 3: Vận tốc và gia tốc cực đại.
* !"#$#
EM"* A
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#
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x A t
v A cos t
ω ϕ
ω ω ϕ
= +
= +
.sin( )
. ( )
x A t
v
A cos t
ω ϕ
ω ϕ
ω
= +
⇒
= +
( )
( )
2 2
2
2 2 2 2
2
4 2
a v
A
ω ω
= +
,&'()#
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( )
10
T s
π
=
40
10
4 4
s
A cm= = =
2
v
A x v A x
ω
ω
= + ⇒ = ± −
2 2 2 2
20. 10 8 120( / )v A x cm s
ω
= − − = − − = −
2 2 2 2
. 20 .8 3200( / ) 32( / )a x cm s m s
ω
= − = − = − = −
&'%;"*67"I#JCtI"
}|St$ @9#MM&11
=>Wz#U*+D*E@FG
B(Z'
G6<?>AF&c<@>A
@!$M1<
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ZM<
10
5
2 2
l
cm= =
78,5
1,57
50
x t cm
π
π
= +
÷
3x cm= −
1
8x cm=
1
12 /v cm s=
2
16 /v cm s=
2
6x cm= −
2!"103$-45!(6'4'789)(4':;*95(<=4>,'?(@ABC-9D*DE
(
D
E
4
DF+GH!(I!
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0
thì :
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0
0
x ?
v ?
=
=
·
MOM '
0
T 360
t ?
→
= → ∆ϕ
t
d k2
t
− ϕ π
= +
ω ω
π − − ϕ π
= +
ω ω
b 0
b 0
− ϕ >
π − − ϕ >
b 0
b 0
− ϕ <
π − − ϕ <
= +
10.sin( . )
2
x t
π
π
= −
5 2
10.sin(10 . )x t
π
=
10.sin(5 . )
2
x t
π
π
= −
10.sin(10 . )x t
π
=
25 2.
π
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V
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I
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