Giàng A Sẩu
Giới thiệu về bản thân
x=Acos(5t+φ)(cm)
⇒
v
=
−
5
A
s
i
n
(
5
t
+
φ
)
=
5
A
c
o
s
(
π
2
+
5
t
+
φ
)
(
cm/s
)
⇒v=−5Asin(5t+φ)=5Acos(
2
π
+5t+φ)(cm/s)
Tại
t
=
0
:
t=0:
{
x
=
−
2
(
c
m
)
v
=
10
(
cm/s
)
{
x=−2(cm)
v=10(cm/s)
⇒
{
x
0
=
A
c
o
s
φ
=
−
2
(
c
m
)
v
0
=
5
A
c
o
s
(
π
2
+
φ
)
=
10
(
cm/s
)
⇒{
x
0
=Acosφ=−2(cm)
v
0
=5Acos(
2
π
+φ)=10(cm/s)
⇔
{
c
o
s
φ
=
−
2
A
(
1
)
5
A
(
c
o
s
π
2
.
c
o
s
φ
−
s
i
n
π
2
.
s
i
n
φ
)
=
10
⇔
⎩
⎨
⎧
cosφ=−
A
2
(1)
5A(cos
2
π
.cosφ−sin
2
π
.sinφ)=10
⇒
5
A
.
(
−
s
i
n
φ
)
=
10
⇔
s
i
n
φ
=
−
2
A
(
2
)
⇒5A.(−sinφ)=10⇔sinφ=
A
−2
(2)
Từ
(
1
)
;
(
2
)
⇒
φ
=
−
3
π
4
(
r
a
d
)
;
A
=
2
2
(
c
m
)
(1);(2)⇒φ=
4
−3π
(rad);A=2
2
(cm)
Vậy ta có ptdđ của vật:
x
=
2
2
c
o
s
(
5
t
−
3
π
4
)
(
c
m
)
x=2
2
cos(5t−
4
3π
)(cm)
b)
v
m
a
x
=
ω
A
=
5
A
=
10
2
(
cm/s
)
v
max
=ωA=5A=10
2
(cm/s)
a
m
a
x
=
ω
2
A
=
50
2
(
cm/s
2
)
a
max
=ω
2
A=50
2
(cm/s
2
)
c)
α
=
Δ
t
.
ω
=
1
,
4
π
.
5
=
7
π
(
r
a
d
)
=
6
π
+
π
(
r
a
d
)
α=Δt.ω=1,4π.5=7π(rad)=6π+π(rad)
⇒
S
=
3.4
A
+
2
2
−
2
+
2
2
+
2
=
12
A
+
4
2
=
28
2
(
c
m
)
⇒S=3.4A+2
2
−2+2
2
+2=12A+4
2
=28
2
(cm)
x=Acos(5t+φ)(cm)
⇒
v
=
−
5
A
s
i
n
(
5
t
+
φ
)
=
5
A
c
o
s
(
π
2
+
5
t
+
φ
)
(
cm/s
)
⇒v=−5Asin(5t+φ)=5Acos(
2
π
+5t+φ)(cm/s)
Tại
t
=
0
:
t=0:
{
x
=
−
2
(
c
m
)
v
=
10
(
cm/s
)
{
x=−2(cm)
v=10(cm/s)
⇒
{
x
0
=
A
c
o
s
φ
=
−
2
(
c
m
)
v
0
=
5
A
c
o
s
(
π
2
+
φ
)
=
10
(
cm/s
)
⇒{
x
0
=Acosφ=−2(cm)
v
0
=5Acos(
2
π
+φ)=10(cm/s)
⇔
{
c
o
s
φ
=
−
2
A
(
1
)
5
A
(
c
o
s
π
2
.
c
o
s
φ
−
s
i
n
π
2
.
s
i
n
φ
)
=
10
⇔
⎩
⎨
⎧
cosφ=−
A
2
(1)
5A(cos
2
π
.cosφ−sin
2
π
.sinφ)=10
⇒
5
A
.
(
−
s
i
n
φ
)
=
10
⇔
s
i
n
φ
=
−
2
A
(
2
)
⇒5A.(−sinφ)=10⇔sinφ=
A
−2
(2)
Từ
(
1
)
;
(
2
)
⇒
φ
=
−
3
π
4
(
r
a
d
)
;
A
=
2
2
(
c
m
)
(1);(2)⇒φ=
4
−3π
(rad);A=2
2
(cm)
Vậy ta có ptdđ của vật:
x
=
2
2
c
o
s
(
5
t
−
3
π
4
)
(
c
m
)
x=2
2
cos(5t−
4
3π
)(cm)
b)
v
m
a
x
=
ω
A
=
5
A
=
10
2
(
cm/s
)
v
max
=ωA=5A=10
2
(cm/s)
a
m
a
x
=
ω
2
A
=
50
2
(
cm/s
2
)
a
max
=ω
2
A=50
2
(cm/s
2
)
c)
α
=
Δ
t
.
ω
=
1
,
4
π
.
5
=
7
π
(
r
a
d
)
=
6
π
+
π
(
r
a
d
)
α=Δt.ω=1,4π.5=7π(rad)=6π+π(rad)
⇒
S
=
3.4
A
+
2
2
−
2
+
2
2
+
2
=
12
A
+
4
2
=
28
2
(
c
m
)
⇒S=3.4A+2
2
−2+2
2
+2=12A+4
2
=28
2
(cm)
x=Acos(5t+φ)(cm)
⇒
v
=
−
5
A
s
i
n
(
5
t
+
φ
)
=
5
A
c
o
s
(
π
2
+
5
t
+
φ
)
(
cm/s
)
⇒v=−5Asin(5t+φ)=5Acos(
2
π
+5t+φ)(cm/s)
Tại
t
=
0
:
t=0:
{
x
=
−
2
(
c
m
)
v
=
10
(
cm/s
)
{
x=−2(cm)
v=10(cm/s)
⇒
{
x
0
=
A
c
o
s
φ
=
−
2
(
c
m
)
v
0
=
5
A
c
o
s
(
π
2
+
φ
)
=
10
(
cm/s
)
⇒{
x
0
=Acosφ=−2(cm)
v
0
=5Acos(
2
π
+φ)=10(cm/s)
⇔
{
c
o
s
φ
=
−
2
A
(
1
)
5
A
(
c
o
s
π
2
.
c
o
s
φ
−
s
i
n
π
2
.
s
i
n
φ
)
=
10
⇔
⎩
⎨
⎧
cosφ=−
A
2
(1)
5A(cos
2
π
.cosφ−sin
2
π
.sinφ)=10
⇒
5
A
.
(
−
s
i
n
φ
)
=
10
⇔
s
i
n
φ
=
−
2
A
(
2
)
⇒5A.(−sinφ)=10⇔sinφ=
A
−2
(2)
Từ
(
1
)
;
(
2
)
⇒
φ
=
−
3
π
4
(
r
a
d
)
;
A
=
2
2
(
c
m
)
(1);(2)⇒φ=
4
−3π
(rad);A=2
2
(cm)
Vậy ta có ptdđ của vật:
x
=
2
2
c
o
s
(
5
t
−
3
π
4
)
(
c
m
)
x=2
2
cos(5t−
4
3π
)(cm)
b)
v
m
a
x
=
ω
A
=
5
A
=
10
2
(
cm/s
)
v
max
=ωA=5A=10
2
(cm/s)
a
m
a
x
=
ω
2
A
=
50
2
(
cm/s
2
)
a
max
=ω
2
A=50
2
(cm/s
2
)
c)
α
=
Δ
t
.
ω
=
1
,
4
π
.
5
=
7
π
(
r
a
d
)
=
6
π
+
π
(
r
a
d
)
α=Δt.ω=1,4π.5=7π(rad)=6π+π(rad)
⇒
S
=
3.4
A
+
2
2
−
2
+
2
2
+
2
=
12
A
+
4
2
=
28
2
(
c
m
)
⇒S=3.4A+2
2
−2+2
2
+2=12A+4
2
=28
2
(cm)
x=Acos(5t+φ)(cm)
⇒
v
=
−
5
A
s
i
n
(
5
t
+
φ
)
=
5
A
c
o
s
(
π
2
+
5
t
+
φ
)
(
cm/s
)
⇒v=−5Asin(5t+φ)=5Acos(
2
π
+5t+φ)(cm/s)
Tại
t
=
0
:
t=0:
{
x
=
−
2
(
c
m
)
v
=
10
(
cm/s
)
{
x=−2(cm)
v=10(cm/s)
⇒
{
x
0
=
A
c
o
s
φ
=
−
2
(
c
m
)
v
0
=
5
A
c
o
s
(
π
2
+
φ
)
=
10
(
cm/s
)
⇒{
x
0
=Acosφ=−2(cm)
v
0
=5Acos(
2
π
+φ)=10(cm/s)
⇔
{
c
o
s
φ
=
−
2
A
(
1
)
5
A
(
c
o
s
π
2
.
c
o
s
φ
−
s
i
n
π
2
.
s
i
n
φ
)
=
10
⇔
⎩
⎨
⎧
cosφ=−
A
2
(1)
5A(cos
2
π
.cosφ−sin
2
π
.sinφ)=10
⇒
5
A
.
(
−
s
i
n
φ
)
=
10
⇔
s
i
n
φ
=
−
2
A
(
2
)
⇒5A.(−sinφ)=10⇔sinφ=
A
−2
(2)
Từ
(
1
)
;
(
2
)
⇒
φ
=
−
3
π
4
(
r
a
d
)
;
A
=
2
2
(
c
m
)
(1);(2)⇒φ=
4
−3π
(rad);A=2
2
(cm)
Vậy ta có ptdđ của vật:
x
=
2
2
c
o
s
(
5
t
−
3
π
4
)
(
c
m
)
x=2
2
cos(5t−
4
3π
)(cm)
b)
v
m
a
x
=
ω
A
=
5
A
=
10
2
(
cm/s
)
v
max
=ωA=5A=10
2
(cm/s)
a
m
a
x
=
ω
2
A
=
50
2
(
cm/s
2
)
a
max
=ω
2
A=50
2
(cm/s
2
)
c)
α
=
Δ
t
.
ω
=
1
,
4
π
.
5
=
7
π
(
r
a
d
)
=
6
π
+
π
(
r
a
d
)
α=Δt.ω=1,4π.5=7π(rad)=6π+π(rad)
⇒
S
=
3.4
A
+
2
2
−
2
+
2
2
+
2
=
12
A
+
4
2
=
28
2
(
c
m
)
⇒S=3.4A+2
2
−2+2
2
+2=12A+4
2
=28
2
(cm)
x=Acos(5t+φ)(cm)
⇒
v
=
−
5
A
s
i
n
(
5
t
+
φ
)
=
5
A
c
o
s
(
π
2
+
5
t
+
φ
)
(
cm/s
)
⇒v=−5Asin(5t+φ)=5Acos(
2
π
+5t+φ)(cm/s)
Tại
t
=
0
:
t=0:
{
x
=
−
2
(
c
m
)
v
=
10
(
cm/s
)
{
x=−2(cm)
v=10(cm/s)
⇒
{
x
0
=
A
c
o
s
φ
=
−
2
(
c
m
)
v
0
=
5
A
c
o
s
(
π
2
+
φ
)
=
10
(
cm/s
)
⇒{
x
0
=Acosφ=−2(cm)
v
0
=5Acos(
2
π
+φ)=10(cm/s)
⇔
{
c
o
s
φ
=
−
2
A
(
1
)
5
A
(
c
o
s
π
2
.
c
o
s
φ
−
s
i
n
π
2
.
s
i
n
φ
)
=
10
⇔
⎩
⎨
⎧
cosφ=−
A
2
(1)
5A(cos
2
π
.cosφ−sin
2
π
.sinφ)=10
⇒
5
A
.
(
−
s
i
n
φ
)
=
10
⇔
s
i
n
φ
=
−
2
A
(
2
)
⇒5A.(−sinφ)=10⇔sinφ=
A
−2
(2)
Từ
(
1
)
;
(
2
)
⇒
φ
=
−
3
π
4
(
r
a
d
)
;
A
=
2
2
(
c
m
)
(1);(2)⇒φ=
4
−3π
(rad);A=2
2
(cm)
Vậy ta có ptdđ của vật:
x
=
2
2
c
o
s
(
5
t
−
3
π
4
)
(
c
m
)
x=2
2
cos(5t−
4
3π
)(cm)
b)
v
m
a
x
=
ω
A
=
5
A
=
10
2
(
cm/s
)
v
max
=ωA=5A=10
2
(cm/s)
a
m
a
x
=
ω
2
A
=
50
2
(
cm/s
2
)
a
max
=ω
2
A=50
2
(cm/s
2
)
c)
α
=
Δ
t
.
ω
=
1
,
4
π
.
5
=
7
π
(
r
a
d
)
=
6
π
+
π
(
r
a
d
)
α=Δt.ω=1,4π.5=7π(rad)=6π+π(rad)
⇒
S
=
3.4
A
+
2
2
−
2
+
2
2
+
2
=
12
A
+
4
2
=
28
2
(
c
m
)
⇒S=3.4A+2
2
−2+2
2
+2=12A+4
2
=28
2
(cm)
x=Acos(5t+φ)(cm)
⇒
v
=
−
5
A
s
i
n
(
5
t
+
φ
)
=
5
A
c
o
s
(
π
2
+
5
t
+
φ
)
(
cm/s
)
⇒v=−5Asin(5t+φ)=5Acos(
2
π
+5t+φ)(cm/s)
Tại
t
=
0
:
t=0:
{
x
=
−
2
(
c
m
)
v
=
10
(
cm/s
)
{
x=−2(cm)
v=10(cm/s)
⇒
{
x
0
=
A
c
o
s
φ
=
−
2
(
c
m
)
v
0
=
5
A
c
o
s
(
π
2
+
φ
)
=
10
(
cm/s
)
⇒{
x
0
=Acosφ=−2(cm)
v
0
=5Acos(
2
π
+φ)=10(cm/s)
⇔
{
c
o
s
φ
=
−
2
A
(
1
)
5
A
(
c
o
s
π
2
.
c
o
s
φ
−
s
i
n
π
2
.
s
i
n
φ
)
=
10
⇔
⎩
⎨
⎧
cosφ=−
A
2
(1)
5A(cos
2
π
.cosφ−sin
2
π
.sinφ)=10
⇒
5
A
.
(
−
s
i
n
φ
)
=
10
⇔
s
i
n
φ
=
−
2
A
(
2
)
⇒5A.(−sinφ)=10⇔sinφ=
A
−2
(2)
Từ
(
1
)
;
(
2
)
⇒
φ
=
−
3
π
4
(
r
a
d
)
;
A
=
2
2
(
c
m
)
(1);(2)⇒φ=
4
−3π
(rad);A=2
2
(cm)
Vậy ta có ptdđ của vật:
x
=
2
2
c
o
s
(
5
t
−
3
π
4
)
(
c
m
)
x=2
2
cos(5t−
4
3π
)(cm)
b)
v
m
a
x
=
ω
A
=
5
A
=
10
2
(
cm/s
)
v
max
=ωA=5A=10
2
(cm/s)
a
m
a
x
=
ω
2
A
=
50
2
(
cm/s
2
)
a
max
=ω
2
A=50
2
(cm/s
2
)
c)
α
=
Δ
t
.
ω
=
1
,
4
π
.
5
=
7
π
(
r
a
d
)
=
6
π
+
π
(
r
a
d
)
α=Δt.ω=1,4π.5=7π(rad)=6π+π(rad)
⇒
S
=
3.4
A
+
2
2
−
2
+
2
2
+
2
=
12
A
+
4
2
=
28
2
(
c
m
)
⇒S=3.4A+2
2
−2+2
2
+2=12A+4
2
=28
2
(cm)
x=Acos(5t+φ)(cm)
⇒
v
=
−
5
A
s
i
n
(
5
t
+
φ
)
=
5
A
c
o
s
(
π
2
+
5
t
+
φ
)
(
cm/s
)
⇒v=−5Asin(5t+φ)=5Acos(
2
π
+5t+φ)(cm/s)
Tại
t
=
0
:
t=0:
{
x
=
−
2
(
c
m
)
v
=
10
(
cm/s
)
{
x=−2(cm)
v=10(cm/s)
⇒
{
x
0
=
A
c
o
s
φ
=
−
2
(
c
m
)
v
0
=
5
A
c
o
s
(
π
2
+
φ
)
=
10
(
cm/s
)
⇒{
x
0
=Acosφ=−2(cm)
v
0
=5Acos(
2
π
+φ)=10(cm/s)
⇔
{
c
o
s
φ
=
−
2
A
(
1
)
5
A
(
c
o
s
π
2
.
c
o
s
φ
−
s
i
n
π
2
.
s
i
n
φ
)
=
10
⇔
⎩
⎨
⎧
cosφ=−
A
2
(1)
5A(cos
2
π
.cosφ−sin
2
π
.sinφ)=10
⇒
5
A
.
(
−
s
i
n
φ
)
=
10
⇔
s
i
n
φ
=
−
2
A
(
2
)
⇒5A.(−sinφ)=10⇔sinφ=
A
−2
(2)
Từ
(
1
)
;
(
2
)
⇒
φ
=
−
3
π
4
(
r
a
d
)
;
A
=
2
2
(
c
m
)
(1);(2)⇒φ=
4
−3π
(rad);A=2
2
(cm)
Vậy ta có ptdđ của vật:
x
=
2
2
c
o
s
(
5
t
−
3
π
4
)
(
c
m
)
x=2
2
cos(5t−
4
3π
)(cm)
b)
v
m
a
x
=
ω
A
=
5
A
=
10
2
(
cm/s
)
v
max
=ωA=5A=10
2
(cm/s)
a
m
a
x
=
ω
2
A
=
50
2
(
cm/s
2
)
a
max
=ω
2
A=50
2
(cm/s
2
)
c)
α
=
Δ
t
.
ω
=
1
,
4
π
.
5
=
7
π
(
r
a
d
)
=
6
π
+
π
(
r
a
d
)
α=Δt.ω=1,4π.5=7π(rad)=6π+π(rad)
⇒
S
=
3.4
A
+
2
2
−
2
+
2
2
+
2
=
12
A
+
4
2
=
28
2
(
c
m
)
⇒S=3.4A+2
2
−2+2
2
+2=12A+4
2
=28
2
(cm)