chap19 Simple Harmonic MoNNNNtion1-1.ppt

GGGGGG

2. a  – x
a = – 
2
x
Negative sign:
Direction of a
is opposite to x
O-x
o
x
o
a a

3. 200
grams
Vibrating
Tuning fork
A weight on
a spring
A boy on
a swing

4. F  – x
F = – kx

5. O-x
o
x
o
Point of equilibrium:
a = 0
F = 0
a
x
Measured from the
point of equilibrium
Amplitude :
= x
o
=Maximum
distance

6. F = - kx
ma = - kx
a =
k
m
- x a = - 
2
xcompare

2
=
k
m
 = angular frequency
= 2f= 2
T

7. F
x
0 x
o-x
o
a
x
0 x
o-x
o

2
x
o
-
2
x
o
F = -kx a = - 
2
x

8. a = - 
2
x
d
2
x
dt
2
= - 
2
x
x = x
o
sin (t + )General solution
dx
dt
= x
o
cos (t + )v =
d
2
x
dt
2
= -
2
x
o
sin (t + )a =
= - 
2
x
x = x
o
sin (t + )
phase

Phase constant

9. x = x
o
sin (t + )
 depends on the
initial condition
when t = 0
O-x
o
x
o
At t = 0, x = 0
0 = x
o sin ((0) + )
sin (0 + ) = 0
sin  = 0
 = 0
Therefore :
x = x
o
sin t
v = x
o
cos t
a = - 
2
x
o
sin t
= - 
2
x

10. x
t
0
x = x
o
sin tx
o
-x
o
T 2T
v
T
0
-x
o
x
o
v = x
o
cos t
t
2T
a
t
0
a = -
2
x
o
sin t

2
x
o
-
2
x
o
T 2T
t=0,
v=x
o
t=0,
a=0

11. phase
= t + /2
phase = t
x
t
0
x
o
-x
o
T 2T
v
T
0
-x
o
x
o
t
2T
a
t
0

2
x
o
-
2
x
o
T 2T
x = x
o
sin t
v = x
o
cos t
To compare the phase :
They must all
be expressed
as same function
(either as sine
or cosine function)
Phase difference :
=(t+ ) - t

2
=

2
rad
= x
o
sin (t + /2)
v leading x

12. phase
= t + 
x
t
0
x
o
-x
o
T 2T
v
T
0
-x
o
x
o
t
2T
a
t
0

2
x
o
-
2
x
o
T 2T
x = x
o
sin t
phase = t
v = x
o
cos t
phase
= t + /2
a = -
2
x
o sin t= 
2
x
o
sin (t + )
Phase difference :
=(t+) - t = rad
a leading x
a and x are
antiphase

13. x
t
0
x
o
-x
o
T 2T
v
T
0
-x
o
x
o
t
2T
a
t
0

2
x
o
-
2
x
o
T 2T

14. x = x
o
sin (t + )
 depends on the
initial condition
when t = 0
Example 2 :
t=0 , x = x
o
:- find 
-x
o x
o0
t=0
x
t

15. -x
o
x
o
0
t=0
x
x = x
o
sin (t + )General solution
Substitute :
t=0 , x = x
o
to find 
x
o = x
o sin ((0) + )
sin  = 1
 =

2
Therefore ,
equation of x is:
x = x
o sin (t + /2)
t
= x
o
cos t

16. x = x
o
sin (t + )General solution
x = x
o
sin t
t=0 , x = x
o
x = x
o
sin (t + /2)
= x
o
cos t

17. The motion of a body in simple harmonic
motion is described by the equation :
x = 4.0 cos ( 2t + )

3
Where x is in metres, t is in seconds and
(2t + ) is in radians.

3
a)What is
i) the displacement
ii) the velocity
iii) the acceleration and
iv) the phase
When t = 2.0 s ?
b) Find
i) the frequency
ii) the period
of the motion.

18. x = 4.0 cos ( 2t + )

3
a) i)
Displacement = ?
t = 2.0 s ;x = 4.0 cos ( 2(2) + )

3
= 2.0 m
ii) velocity = ?
v =
dx
dt
=-2(4.0) sin ( 2t + )

3
v = -8.0 sin (4 + )t = 2.0 s ;

3
= -21.8 ms
-1

19. iii)acceleration = ?
a =
dv
dt
=-16.0
2
cos ( 2t + )

3
t = 2.0 s ;a =-16.0
2
cos ( 2(2) + )

3
= -79.0 ms
-2
iv) Phase = ?
x = 4.0 cos ( 2t + )

3
Phase = 2t +

3
t = 2.0 s ;Phase = 2(2) +

3
= 13.6 rad

20. The motion of a body in simple harmonic
motion is described by the equation :
x = 4.0 cos ( 2t + )

3
Where x is in metres, t is in seconds and
(2t + ) is in radians.

3
a)What is
i) the displacement
ii) the velocity
iii) the acceleration and
iv) the phase
When t = 2.0 s ?
b) Find
i) the frequency
ii) the period
of the motion.

21. x = 4.0 cos ( 2t + )

3
b) i) frequency = ?
x = x
o cos (t + )
General solution :
 = 2
2f = 2
f = 1.0 Hz
ii) Period = ?
T = 1
f
= 1.0 s

22. Figures (1) and (2) are the displacement-time
graph and acceleration-time graph respectively
of a body in simple harmonic motion.
x,m
3
-3
t,sT0
(1) a,ms
-2
12
-12
t,sT0
(2)
What is the frequency of the motion? Write an
expression to represent the variation of
displacement x with time t.

23. x,m
3
-3
t,sT0
(1) a,ms
-2
12
-12
t,sT0
(2)
frequency = ?
x
o
= 3 
2
x
o = 12

2
(3) = 12
substitute
 = 2 rad s
-1
2f = 2
f = 0.318 Hz

24. expression = ?
x = x
o
sin (t + )General solution
x
o
= 3 ,  = 2 ; x = 3 sin (2t + )
At t = 0, x = 3 3 = 3 sin (2(0) + )
sin  = 1
 = 
2
rad
Therefore, the
expression is
x = 3 sin (2t + )

2
x = 3 cos 2t