Transcript +x - SeyedAhmad.com

Simple Harmonic Motion
Periodic Motion
Simple periodic motion is that motion in which a
body moves back and forth over a fixed path,
returning to each position and velocity after a
definite interval of time.
1
f 
T
Amplitude
A
Period, T, is the time
for one complete
oscillation. (seconds,s)
Frequency, f, is the
number of complete
oscillations per
second. Hertz (s-1)
Example 1: The suspended mass makes 30
complete oscillations in 15 s. What is the
period and frequency of the motion?
15 s
T
 0.50 s
30 cylces
x
F
Period: T = 0.500 s
1
1
f  
T 0.500 s
Frequency: f = 2.00 Hz
Simple Harmonic Motion, SHM
Simple harmonic motion is periodic motion in
the absence of friction and produced by a
restoring force that is directly proportional to
the displacement and oppositely directed.
x
F
A restoring force, F, acts in
the direction opposite the
displacement of the
oscillating body.
F = -kx
Hooke’s Law
When a spring is stretched, there is a restoring
force that is proportional to the displacement.
F = -kx
x
m
F
The spring constant k is a
property of the spring given by:
k=
DF
Dx
The period and frequency of a
wave
•
the period T of a wave is the amount of time it takes
to go through 1 cycle
•
the frequency f is the number of cycles per second
– the unit of a cycle-per-second is commonly referred to as
a hertz (Hz), after Heinrich Hertz (1847-1894), who
discovered radio waves.
•
frequency and period are related as follows:
1
f 
T
•
Since a cycle is 2p radians, the relationship between
frequency and angular frequency is:
  2pf
t
T
The Reference Circle
The reference circle compares
the circular motion of an object
with its horizontal projection.
x  A cos   t
x  A cos(2p ft )
x = Horizontal displacement.
A = Amplitude (xmax).
 = Reference angle.
  2pf
Velocity in SHM
The velocity (v) of an
oscillating body at any
instant is the horizontal
component of its
tangential velocity (vT).
vT = R = A;   2pf
v = -vT sin  ;
 = t
v = - A sin  t
v = -2pf A sin 2pf t
Acceleration Reference Circle
The acceleration (a) of an
oscillating body at any instant is
the horizontal component of its
centripetal acceleration (ac).
a = -ac cos  = -ac cos(t)
v2  2 R2
ac  
; ac   2 R
R
R
a = -2A cos(t)
R=A
a  4p 2 f 2 A cos(2p ft )
a  4p f x
2
2
The Period and Frequency as a
Function of a and x.
For any body undergoing simple harmonic motion:
Since a = -4p2f2x
1
f 
2p
a
x
and
T = 1/f
x
T  2p
a
The frequency and the period can be found if the
displacement and acceleration are known. Note
that the signs of a and x will always be opposite.
Springs and SHM
•
Attach an object of mass m to the end of a spring, pull it out
to a distance A, and let it go from rest. The object will then
undergo simple harmonic motion:
x (t )  A cos(t )
v (t )   A sin( t )
a(t )   A 2 cos(t )
– Use Newton’s 2nd law, together with Hooke’s law,
and the above description of the acceleration to
find:
k

m
Period and Frequency as a Function
of Mass and Spring Constant.
For a vibrating body with an elastic restoring force:
Recall that F = ma = -kx:
1
f 
2p
k
m
m
T  2p
k
The frequency f and the period T can be found if
the spring constant k and mass m of the vibrating
body are known. Use consistent SI units.
Example 6: The frictionless system shown
below has a 2-kg mass attached to a spring
(k = 400 N/m). The mass is displaced a
distance of 20 cm to the right and released.
What is the frequency of the motion?
x
a
v
m
x = -0.2 m
1
f 
2p
x=0
k
1

m 2p
f = 2.25 Hz
x = +0.2 m
400 N/m
2 kg
Example 6 (Cont.): Suppose the 2-kg mass
of the previous problem is displaced 20 cm
and released (k = 400 N/m). What is the
maximum acceleration? (f = 2.25 Hz)
x
a
v
m
x=0
x = -0.2 m
x = +0.2 m
Acceleration is a maximum when x =  A
a  4p f x  4p (2.25 Hz) (0.2 m)
2
2
2
a =  40 m/s2
2
Example 6: The 2-kg mass of the previous
example is displaced initially at x = 20 cm
and released. What is the velocity 2.69 s
after release? (Recall that f = 2.25 Hz.)
x
a
v
m
v = -2pf A sin 2pf t
x = -0.2 m x = 0
x = +0.2 m
v  2p (2.25 Hz)(0.2 m)sin 2p (2.25 Hz)(2.69 s)
(Note:  in rads)
v  2p (2.25 Hz)(0.2 m)(0.324)
v = -0.916 m/s
The minus sign means it
is moving to the left.
Example 7: At what time will the 2-kg mass
be located 12 cm to the left of x = 0?
(A = 20 cm, f = 2.25 Hz) -0.12 m
x
a
v
m
x  A cos(2p ft )
x = -0.2 m x = 0
x 0.12 m
cos(2p ft )  
;
A 0.20 m
2p ft  2.214 rad;
x = +0.2 m
(2p ft )  cos 1 ( 0.60)
2.214 rad
t
2p (2.25 Hz)
t = 0.157 s
To Be Continued…
Work Done in Stretching a Spring
Work done ON the spring is positive;
work BY spring is negative.
x
From Hooke’s law the force F is:
F (x) = kx
F
F
m
To stretch spring from
x1 to x2 , work is:
Work  ½kx  ½kx
2
2
x1
x2
2
1
Example 2: A 4-kg mass suspended from a
spring produces a displacement of 20 cm.
What is the spring constant?
The stretching force is the weight
(W = mg) of the 4-kg mass: 20 cm
F = (4 kg)(9.8 m/s2) = 39.2 N
F
m
Now, from Hooke’s law, the force
constant k of the spring is:
k=
DF
Dx
=
39.2 N
0.2 m
k = 196 N/m
Example 2(cont.: The mass m is now stretched
a distance of 8 cm and held. What is the
potential energy? (k = 196 N/m)
The potential energy is equal to
the work done in stretching the
spring:
Work  ½kx  ½kx
2
2
8 cm
0
m
2
1
U  ½kx  ½(196 N/m)(0.08 m)
2
U = 0.627 J
F
2
Displacement in SHM
x
m
x = -A
x=0
x = +A
• Displacement is positive when the position is
to the right of the equilibrium position (x = 0)
and negative when located to the left.
• The maximum displacement is called the
amplitude A.
Velocity in SHM
v (-)
v (+)
m
x = -A
x=0
x = +A
• Velocity is positive when moving to the right
and negative when moving to the left.
• It is zero at the end points and a maximum
at the midpoint in either direction (+ or -).
Acceleration in SHM
+a
-x
+x
-a
m
x = -A
x=0
x = +A
• Acceleration is in the direction of the
restoring force. (a is positive when x is
negative, and negative when x is positive.)
F  ma  kx
• Acceleration is a maximum at the end points
and it is zero at the center of oscillation.
Acceleration vs. Displacement
a
v
x
m
x = -A
x=0
x = +A
Given the spring constant, the displacement, and
the mass, the acceleration can be found from:
F  ma  kx or
 kx
a
m
Note: Acceleration is always opposite to displacement.
Example 3: A 2-kg mass hangs at the end
of a spring whose constant is k = 400 N/m.
The mass is displaced a distance of 12 cm
and released. What is the acceleration at
the instant the displacement is x = +7 cm?
 kx
a
m
(400 N/m)(+0.07 m)
a
2 kg
a = -14.0 m/s2
a
m
Note: When the displacement is +7 cm
(downward), the acceleration is -14.0 m/s2
(upward) independent of motion direction.
+x
Example 4: What is the maximum acceleration
for the 2-kg mass in the previous problem? (A
= 12 cm, k = 400 N/m)
The maximum acceleration occurs
when the restoring force is a
maximum; i.e., when the stretch or
compression of the spring is largest.
F = ma = -kx
xmax =  A
kA 400 N(  0.12 m)
a

m
2 kg
Maximum
Acceleration:
m
amax = ± 24.0 m/s2
+x
Conservation of Energy
The total mechanical energy (U + K) of a
vibrating system is constant; i.e., it is the
same at any point in the oscillating path.
a
v
x
m
x = -A
x=0
x = +A
For any two points A and B, we may write:
½mvA2 + ½kxA 2 = ½mvB2 + ½kxB 2
Energy of a Vibrating System:
A
x
a
v
m
x = -A
x=0
B
x = +A
• At points A and B, the velocity is zero and the
acceleration is a maximum. The total energy is:
U + K = ½kA2 x =  A and v = 0.
• At any other point: U + K = ½mv2 + ½kx2
Velocity as Function of Position.
x
a
v
m
x = -A
1
2
x=0
k
v
A2  x 2
m
mv  kx  kA
2
1
2
2
1
2
vmax when
x = 0:
x = +A
2
v
k
A
m
Example 5: A 2-kg mass hangs at the end of
a spring whose constant is k = 800 N/m. The
mass is displaced a distance of 10 cm and
released. What is the velocity at the instant
the displacement is x = +6 cm?
½mv2 + ½kx 2 = ½kA2
k
v
A2  x 2
m
800 N/m
v
(0.1 m) 2  (0.06 m) 2
2 kg
v = ±1.60 m/s
m
+x
Example 5 (Cont.): What is the maximum
velocity for the previous problem? (A = 10
cm, k = 800 N/m, m = 2 kg.)
The velocity is maximum when x = 0:
0
½mv2 + ½kx 2 = ½kA2
v
k
800 N/m
A
(0.1 m)
m
2 kg
v = ± 2.00 m/s
m
+x
The Simple Pendulum
The period of a simple
pendulum is given by:
L
T  2p
g
L
For small angles .
1
f 
2p
g
L
mg
Example 8. What must be the length of a
simple pendulum for a clock which has a period
of two seconds (tick-tock)?
L
T  2p
g
2
2 L
T  4p
;
g
(2 s) 2 (9.8 m/s 2 )
L
2
4p
L
T 2g
L=
2
4p
L = 0.993 m
To Be Continued…