I can . . .
Period, T�Amplitude, A�Position, x�Velocity, v�Centripetal Accel, ac��
from �from�from�from�from�
Frequency, f�Position, x�Angle, θ�Angular Velocity, ω�Acceleration, a
Simple Harmonic Motion� is that motion in which a body moves back and forth over a fixed path, returning to each position, velocity, and acceleration after a definite interval of time.
I can . . .
Period, T�Amplitude, A�Position, x�Velocity, v�Centripetal Accel, ac��
from �from�from�from�from�
Frequency, f�Position, x�Angle, θ�Angular Velocity, ω�Acceleration, a
Force (N)
Distance (m)
Hooke’s Law
Work = ½kx·x
When is the Forcespring, Velocity, and Acceleration �the greatest versus least? WHY?
When is the Velocity, and the Acceleration in the same direction versus opposite direction? WHY?
When is the Forcespring, and the Acceleration in the �same direction versus opposite direction? WHY?
Knowing:� the spring constant, k� the mass, m� and the Amplitude, A ��Find an equation that would predict how fast, v, the mass is traveling when it is at position, x.
Amplitude, A
Position, x
Conservation�of�Energy
½mv2 + ½kx2 = ½kA2
Knowing:� the spring constant, k� the mass, m� and the Amplitude, A �How fast, v, is it traveling when it is at position, x?
Amplitude, A
Position, x
Conservation�of�Energy
I can . . .
Period, T�Amplitude, A�Position, x�Velocity, v�Centripetal Accel, ac��
from �from�from�from�from�
Frequency, f�Position, x�Angle, θ�Angular Velocity, ω�Acceleration, a
Clock (seconds)
Estimate:� Period, T� Frequency, f� Amplitude, A
5 s�0.2 Hz�1.9 cm
(8.7 – 4.9)/2
Period, T� Time to make one complete cycle.
Frequency, f� Number of cycles made each second.��
Amplitude, A� Maximum displacement, x from equilibrium position.
(seconds/cycle)�
(cycles/second) (Hertz)
(meters)
Example 1: The suspended mass makes 30 complete oscillations in 15 s. What is the period and frequency of the motion?
x
F
Period: T = 0.500 s
Frequency: f = 2.00 Hz
Amplitude, A
Position, x
6.28 seconds
Angle (Radians) =
Arc Length (Meters)�Radius (Meters)
How many radians�are in ONE rotation?
2·π·R�R
= 2·π
An angle of ONE Radian is when�the arc length equals the radius.
ONE�Radian
(57.3°)
2
3
4
5
6
How many arc lengths (radii) fit into the circumference?
If your Angular Velocity is�1 radian per second, how long will it take to rotate once?
2 radians per second,
3.14 seconds
ω = 2πf
ω =
2π radians� cycle
cycles�second
θ
ω = θ/t
http://mathprecalculus.wikispaces.com/Chapter+5.1+Unit+Circle+Approach
Unit Circle
π
2π
3π/2
π/2
0
x = cos(θ)
θ
The Reference Circle
The reference circle compares the circular motion of an object with its horizontal projection.
x = Horizontal displacement.
A = Amplitude (xmax).
θ = Reference angle.
ω = 2πf
x = Acosωt
t = Time.
f = frequency
Learning Goal�for Simple Harmonic Motion
Period, T�Amplitude, A�Position, x�Velocity, v�Centripetal Accel, ac��
versus �versus�versus�versus�versus�
Frequency, f�Position, x�Angle, θ�Angular Velocity, ω�Acceleration, a
Does this equation�model reality?
Δx
Δt
$
y = ¼a · x + b
vertical�axis
horizontal�axis
constant�Slope
constant�Intercept
= · +
10
20
Slope = 2
y = 2 · x + 6
a = 8
y = 2a·x2
y
x
y
x2
Use a graph to find the value of “a”.
y = 2a·(x2)
= ·
vertical�axis
horizontal�axis
constant�Slope
Slope = 0.5
0.5 = 2a
0.25 = a
LINEARIZE THE DATA
100
50
What variables should be graphed in order to linearize the data?
Find the values of a and b in terms of slope and intercept.
LINEARIZING DATA
a = slope/(-5)
b = intercept
a = 3·slope
b = (intercept+2)2
a = 1/3·slope
b = 1/6·intercept
a = 1/2·slope
b = intercept
Simple average versus using a graph
Work and Kinetic Energy
A resultant force changes the velocity of an object and does work on that object.
m
vo
m
vf
x
F
F
Work Done in Stretching a Spring
F
x
m
Work done ON the spring is positive; work BY spring is negative.
From Hooke’s law the force F is:
F (x) = kx
x1
x2
F
To stretch spring from x1 to x2 , work is:
(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?
F
20 cm
m
The stretching force is the weight (W = mg) of the 4-kg mass:
F = (4 kg)(9.8 m/s2) = 39.2 N
Now, from Hooke’s law, the force constant k of the spring is:
k = =
ΔF
Δx
39.2 Ν
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)
F
8 cm
m
U = 0.627 J
The potential energy is equal to the work done in stretching the spring:
0
Displacement in SHM
m
x = 0
x = +A
x = -A
x
Velocity in SHM
m
x = 0
x = +A
x = -A
v (+)
v (-)
Acceleration in SHM
m
x = 0
x = +A
x = -A
+x
-a
-x
+a
Maximum Force�Maximum Acceleration
Maximum Force�Maximum Acceleration
Net Force = Zero�Maximum Velocity�
m·g
m·g
m·g
Fspring
Fspring
Fspring
FNet
FNet
Fnet = 0
Acceleration vs. Displacement
m
x = 0
x = +A
x = -A
x
v
a
Given the spring constant, the displacement, and the mass, the acceleration can be found from:
or
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?
m
+x
a = -14.0 m/s2
a
Note: When the displacement is +7 cm (downward), the acceleration is -14.0 m/s2 (upward) independent of motion direction.
Example 4: What is the maximum acceleration for the 2-kg mass in the previous problem? (A = 12 cm, k = 400 N/m)
m
+x
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
amax = ± 24.0 m/s2
Maximum Acceleration:
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.
m
x = 0
x = +A
x = -A
x
v
a
For any two points A and B, we may write:
½mvA2 + ½kxA 2 = ½mvB2 + ½kxB 2
KA + UA = KB + UB
Energy of a Vibrating System:
m
x = 0
x = +A
x = -A
x
v
a
U + K = ½kA2 x = ± A and v = 0.
A
B
Velocity as Function of Position.
m
x = 0
x = +A
x = -A
x
v
a
vmax when x = 0:
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?
m
+x
½mv2 + ½kx 2 = ½kA2
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.)
m
+x
½mv2 + ½kx 2 = ½kA2
v = ± 2.00 m/s
0
The velocity is maximum when x = 0:
Velocity in SHM
The velocity (v) of an oscillating body at any instant is the horizontal component of its tangential velocity (vT).
vT = ωA;
v = -vT sin θ
v = -2πf A sin 2πf t
-
vT = 2πf A
θ = ωt = 2πf t
ω = 2πf
Acceleration Reference Circle
ac =
vt2�R
The acceleration (a) of an oscillating body at any instant is the horizontal component of its centripetal acceleration (ac).
a = -ac cos θ
R = A
a = -ω2Α cos(ωt)
θ = ωt = 2πft
a = - 4π2f 2Α cos(2πft)
= ω2A
(ωR)2R
=
= ω2R
a = - 4π2f 2 x
The Period and Frequency as a�Function of a and x.
T = 1/f
a = - 4π2f 2 x
Recall that F = ma = -kx
-a�x
=
k�m
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?
m
x = 0
x = +0.2 m
x
v
a
x = -0.2 m
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)
m
x = 0
x = +0.2 m
x
v
a
x = -0.2 m
Acceleration is a maximum when x = ± A
a = ± 40 m/s2
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.)
m
x = 0
x = +0.2 m
x
v
a
x = -0.2 m
v = -0.916 m/s
v = -2πf A sin 2πf t
(Note: θ in rads)
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)
m
x = 0
x = +0.2 m
x
v
a
x = -0.2 m
t = 0.157 s
-0.12 m
The Simple Pendulum
The period of a simple pendulum is given by:
mg
L
For small angles θ.
What must be the length of a simple pendulum for a clock which has a period of two seconds (tick-tock)?
L
L = 0.993 m
http://pirt.asu.edu/news%20Pendulum%20Wave.asp?index=0
p
I can . . .
Period, T�Amplitude, A�Position, x�Velocity, v�Centripetal Accel, ac��
from �from�from�from�from�
Frequency, f�Position, x�Angle, θ�Angular Velocity, ω�Acceleration, a
Summary
Simple harmonic motion (SHM) 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.
F
x
m
The frequency (rev/s) is the reciprocal of the period (time for one revolution).
Summary (Cont.)
F
x
m
Hooke’s Law: In a spring, there is a restoring force that is proportional to the displacement.
The spring constant k is defined by:
Summary (SHM)
m
x = 0
x = +A
x = -A
x
v
a
½mvA2 + ½kxA 2 = ½mvB2 + ½kxB 2
Conservation of Energy:
Summary (SHM)
Summary: Period and Frequency for Vibrating Spring.
m
x = 0
x = +A
x = -A
x
v
a
Summary: Simple Pendulum and Torsion Pendulum
L