Transcript No Slide Title
Chapter 14 - Simple Harmonic Motion
A PowerPoint Presentation by Paul E. Tippens, Professor of Physics Southern Polytechnic State University © 2007
Photo by Mark Tippens A TRAMPOLINE exerts a restoring force on the jumper that is directly proportional to the average force required to displace the mat. Such restoring forces provide the driving forces necessary for objects that oscillate with simple harmonic motion.
Objectives: After finishing this unit, you should be able to:
• Write and apply Hooke ’ s Law for objects moving with simple harmonic motion.
• Write and apply formulas for finding the frequency velocity f , period T , v , or acceleration terms of displacement x a or in time t .
• Describe the motion of pendulums and calculate the length required to produce a given frequency.
Periodic Motion
Simple periodic motion definite interval of time.
is that motion in which a body moves back and forth over a fixed path, returning to each position and velocity after a
f
1
T
Period , T, is the time for one complete oscillation. (seconds,s) Amplitude A 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?
x
F
T
15 s 30 cylces
0.50 s
Period: T = 0.500 s
f
1
T
1 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.
x m F F = -kx The spring constant k is a property of the spring given by: D F k = D x
Work Done in Stretching a Spring
Work done ON work BY the spring is positive ; spring is negative.
x From Hooke ’ s law the force F is: m F F (x) = kx F To stretch spring from x 1 to x 2 , work is:
Work
½
kx
2 2 ½
kx
1 2 x 1 x 2 (Review module on work)
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/s 2 ) = 39.2 N m F Now, from Hooke ’ s law, the force constant k of the spring is: D F D x 39.2 N 0.2 m k = 196 N/m
Example 2(cont.: potential energy The mass m is now stretched a distance of 8 cm and held. What is the ? (k = 196 N/m) The potential energy is equal to the work done in stretching the spring:
Work
½
kx
2 2 ½
kx
1 2 0 8 cm
U
½
kx
2 ½(196 N/m)(0.08 m) 2 m F U = 0.627 J
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 (-)
m
v (+) 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 negative, and negative when x is 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
a
kx 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 the instant the displacement is 12 cm and released. What is the acceleration at x = +7 cm ?
a
kx m a
(400 N/m)(+0.07 m) 2 kg
a = -
14.0 m/s 2
a
m Note: When the displacement is +7 cm (downward), the acceleration is -14.0 m/s 2 (upward) independent of motion direction.
+
x
Example 4: for the 2-kg What is the = 12 cm , k = 400 N/m ) maximum acceleration mass in the previous problem? ( A 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 x max =
A
+
x
m
a
kA
m
2 kg Maximum Acceleration: a max = ± m/s 2 24.0
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: ½
mv A 2 + ½kx A 2 =
½
mv B 2 + ½kx B 2
Energy of a Vibrating System:
a
A
x v
B
m
x = -A x = 0 x = +A
• At points A and B , the velocity is zero and the acceleration is a maximum. The total energy is: U + K = ½kA 2
x =
A and v = 0.
• At any other point: U + K = ½mv 2 + ½kx 2
Velocity as Function of Position.
a x v
m
x = -A x = 0 x = +A
1 2
mv
2 1 2
kx
2 1 2
kA
2
v
k m A
2
x
2
v max
when x = 0:
v
k m A
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 released. What is the velocity at the instant the displacement is x = +6 cm ?
10 cm and ½ mv 2 + ½kx 2 = ½kA 2
v
k m A
2
x
2
v
800 N/m 2 kg (0.1 m) 2 (0.06 m) 2 m +
x v
= ± 1.60 m/s
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: ½
mv 2
0
+ ½kx 2 = ½kA 2 v
k m A
800 N/m (0.1 m) 2 kg m +
x v
= ± 2.00 m/s
The Reference Circle
The reference circle compares the circular motion of an object with its horizontal projection.
x
A
cos w
t x
A
cos(2
ft
)
x = Horizontal displacement.
A = Amplitude (x max ).
= Reference angle.
w
2
f
Velocity in SHM
The velocity (v) of an oscillating body at any instant is the horizontal component of its tangential velocity (v T ).
v T =
w
R =
w
A;
w 2
f v = -v T
sin
;
=
w
t v = -
w
A
sin w
t v = -
2
f A
sin 2
f t
Acceleration Reference Circle
The acceleration (
a
) of an oscillating body at any instant is the horizontal component of its centripetal acceleration (
a c
).
a = -a c
cos = -
a c
cos( w t)
a c
v
2 w 2
R
2 ;
a c R R a = -
w 2 A cos( w t) w 2
R R = A a
4 2 2
f A
cos(2
ft
)
a
4 2 2
f x
The Period and Frequency as a Function of
a
and
x
.
For any body undergoing simple harmonic motion : Since
a = -4
2
f 2 x and T = 1/f f
1 2
a x T
2
x a
The frequency displacement and the period and acceleration that the signs of
a
and x can be found if the are known. Note will always be opposite.
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 :
f
1 2
k m T
2
m k
The the frequency f and the spring constant k period and mass T m can be found if of the vibrating body are known. Use consistent SI units.
Example 6: below has a The frictionless system shown 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 x = -0.2
m
f
1 2
k m
1 2
x = +0.2
m 400 N/m 2 kg
f =
2.25 Hz
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 = -0.2
m
x = 0 x = +0.2
m Acceleration is a maximum when x = A
a
4 2 2
f x
4 2 2
a
= 40 m/s 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 v = -
2
f A
sin 2
f t
m
v
(Note: in rads )
x = -0.2
v
m
x = 0 x = +0.2
m
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 x
A
cos(2
ft
) m cos(2
ft
) 2
x A
x = -0.2
m 0.12 m ; (2 0.20 m
x = 0 ft
)
ft
2.214 rad;
t
2.214 rad
x = +0.2
m 1
t =
0.157 s
The Simple Pendulum
The period of a simple pendulum is given by:
L
T
2
L g
For small angles .
f
1 2
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)?
T
2
L g
L
T
2 4 2
L
; L =
g
2
T g
4 2
L
2 2 (2 s) (9.8 m/s ) 4 2 L = 0.993 m
The Torsion Pendulum
The period pendulum T of a torsion is given by:
T
2
I k
' Where k ’ is a torsion constant that depends on the material from which the rod is made; I is the rotational inertia of the vibrating system.
Example 9: A 160 g solid disk is attached to the end of a wire, then twisted at 0.8 rad and released. The torsion constant k ’ is 0.025 N m/rad . Find the period.
(Neglect the torsion in the wire) For Disk:
I = ½mR 2
I = ½(0.16 kg)(0.12 m) 2 = 0.00115 kg m 2
T
2
I k
' 2 0.00115 kg m 2 0.025 N m/rad
T =
1.35 s Note: Period is independent of angular displacement.
Summary
Simple harmonic motion (SHM) after a definite interval of time.
is that motion in which a body moves back and forth over a fixed path, returning to each position and velocity x m F The frequency (rev/s) is the reciprocal of the period (time for one revolution).
f
1
T
Summary (Cont.)
Hooke force ’ s Law: In a spring, there is a restoring that is proportional to the displacement .
x m F
F
kx
The spring constant k is defined by:
k
D
F
D
x
x = -A
Summary (SHM)
a x v
m
x = 0 F
ma
kx x = +A a
kx m
Conservation of Energy: ½
mv A 2 + ½kx A 2 =
½
mv B 2 + ½kx B 2
Summary (SHM)
1 2
mv
2 1 2
kx
2 1 2
kA
2
v
k m A
2
x
2
x
A
cos(2
ft
)
v
0
k m A a
4
2 2
f x v
2
fA
sin(2
ft
)
Summary: Period and Frequency for Vibrating Spring.
a x v
m
x = -A x = 0 x = +A f
1 2
a x T
2
x a f
1 2
k m T
2
m k
Summary: Simple Pendulum and Torsion Pendulum
f
1 2
g L
L
T
2
L g T
2
I k
'