Simple Harmonic Motion(SHM)
Harmonic Motion and Waves
• Vibration (oscillation)
• Equilibrium position – position of the natural
length of a spring
• Amplitude – maximum displacement
Period and Frequency
• Period (T) – Time for one complete cycle (back
to starting point)
• Frequency (Hz) – Cycles per second
F=1
T
T=1
f
Period and Frequency
A radio station has a frequency of 103.1 M Hz.
What is the period of the wave?
103.1 M Hz
1X106 Hz = 1.031 X 108 Hz
1M Hz
T = 1/f = 1/(1.031 X 108 Hz) = 9.700 X 10-9 s
1
Hooke’s Law
F = -kx
F = weight of an object
k = spring constant (N/m)
x = displacement when the object is placed on the
spring
Hooke’s Law: Example 1
What is the spring constant
if a 0.100 kg mass causes
the spring to stretch 6.0
cm?
(ANS: 16 N/m)
Special Note:
If a spring has a mass on it,
and then is stretched
further, equilibrium
position is the starting
length (with the mass on
it)
Hooke’s Law: Example 2a
How far will the car lower if a 300 kg family
borrows the car?
F = -kx
x = -F/k
x = -(300 kg)(9.8 m/s2)/ 6.5 X 104 N/m
x = 4.5 X 10-2 m = 4.5 cm
Hooke’s Law: Example 2
A family of four has a combined mass of 200 kg.
When they step in their 1200 kg car, the shocks
compress 3.0 cm. What is the spring constant of
the shocks?
F = -kx
k = -F/x
k = -(200 kg)(9.8 m/s2)/(-0.03 m)
k = 6.5 X 104 N/m
(note that we did not include the mass of the car)
Forces on a Spring
• Extreme Position (Amplitude)
– Force at maximum
– Velocity = 0
• Equilibrium position
– Force = 0
– Velocity at maximum
2
All PE
Energy and Springs
• KE = ½ mv2
• PE = ½ kx2
• Maximum PE = ½ kA2
All KE
All PE
Law of conservation of Energy
½ kA2 = ½ mv2+ ½ kx2
Spring Energy: Example 1
A 0.50 kg mass is connected to a light spring with a
spring constant of 20 N/m. Calculate the total
energy if the amplitude is 3.0 cm.
Maximum PE = ½ kA2
Maximum PE = ½ (20 N/m)(0.030 m2)
Maximum PE = 9 X 10-3 Nm (J)
Spring Energy: Example 1b
What is the potential energy and kinetic energy at x
= 2.0 cm?
Some KE and Some PE
Spring Energy: Example 1a
What is the maximum speed of the mass?
½ kA2 = ½ mv2+ ½ kx2
½ kA2 = ½ mv2
(x=0 at the origin)
-3
2
9 X 10 J = ½ (0.50 kg)v
v = 0.19 m/s
Spring Energy: Example 1c
At what position is the speed 0.10 m/s?
PE = ½ kx2
PE = ½ (20 N/m)(0.020 m2) = 4 X 10-3 J
½ kA2 = ½ mv2 + ½ kx2
½ mv2 = ½ kA2 - ½ kx2
KE = 9 X 10-3 J - 4 X 10-3 J = 5 X 10-3 J
(Ans: + 2.6 cm)
3
Spring Energy: Example 2a
Spring Energy: Example 2b
A spring stretches 0.150 m when a 0.300 kg mass is
suspended from it (diagrams a and b). Find the
spring constant.
The spring is now stretched an additional 0.100 m
and allowed to oscillate (diagram c). What is the
maximum velocity?
(Ans: 19.6 N/m)
The maximum velocity occurs through the origin:
½ kA2 = ½ mv2+ ½ kx2
½ kA2 = ½ mv2
(x=0 at the origin)
kA2 = mv2
v2 = kA2/m
v = \/kA2/m = \/(19.6 N/m)(0.100m)2/0.300kg
v = 0.808 m/s
Spring Energy: Example 2c
What is the velocity at x = 0.0500 m?
½ kA2 = ½ mv2+ ½ kx2
kA2 = mv2+ kx2
mv2 = kA2 - kx2
v2 = kA2 - kx2
m
v2 = 19.6 N/m(0.100m2 – 0.0500m2) = 0.49 m2/s2
0.300 kg
v = 0.700 m/s
Spring Energy: Example 2d
What is the maximum acceleration?
The force is a maximum at the amplitude
F = ma
and F = kx
ma = kx
a = kx/m = (19.6 N/m)(0.100 m)/(0.300 kg)
a = 6.53 m/s2
Trigonometry and SHM
• Ball rotates on a table
• Looks like a spring from the side
• One rev(diameter) = 2 A
T=2
m
k
f= 1
T
4
• Period depends only on mass and spring constant
• Amplitude does not affect period
vo = 2 Af or
vo = 2 A
T
vo is the initial (and maximum) velocity
Period: Example 2a
An insect (m=0.30 g) is caught in a spiderweb that
vibrates at 15 Hz. What is the spring constant of
the web?
T = 1/f = 1/15 Hz = 0.0667 s
T=2 m
k
T2 = (2 )2m
k
k = (2 )2m = (2 )2(3.0 X 10-4 kg) = 2.7 N/m
T2
(0.0667)2
Period: Example 1
What is the period and frequency of a 1400 kg car
whose shocks have a k of 6.5 X 104 N/m after it
hits a bump?
m =
2 (1400 kg/6.5 X 104 N/m)1/2
k
T = 0.92 s
f = 1/T = 1/0.92 s = 1.09 Hz
T=2
Period: Example 2b
What would be the frequency for a lighter insect,
0.10 g? Would it be higher or lower?
T=2 m
k
T = 2 (m/k)1/2
T = 2 (1.0 X 10-4 kg/2.7 N/m)1/2 = 0.038 s
f = 1/T = 1/0.038 s = 26 Hz
Cosines and Sines
• Imagine placing a
pen on a vibrating
mass
• Draws a cosine
wave
x = A cos2 t
T
or
x = A cos2 ft
A = Amplitude
t = time
T = period
f = frequency
5
x = A cos2 ft
Cos: Example 1a
Velocity is the
derivative of
position
A loudspeaker vibrates at 262 Hz (middle C). The
amplitude of the cone of the speaker is 1.5 X 10-4
m. What is the equation to describe the position
of the cone over time?
v = -vosin2 ft
Acceleration is
the derivative of
velocity
x = A cos2 ft
x = (1.5 X 10-4 m) cos2 (262 s-1)t
x = (1.5 X 10-4 m) cos(1650 s-1)t
a = -aocos2 ft
Cos: Example 1b
Cos: Example 1c
What is the position at t = 1.00 ms (1 X 10 -3 s)
What is the maximum velocity and acceleration?
x = A cos2 ft
x = (1.5 X 10-4 m) cos2 (262 s-1) (1 X 10-3 s)
x = (1.5 X 10-4 m) cos(1.65 rad) = -1.2 X 10-5 m
vo = 2 Af
vo = 2 (1.5 X 10-4 m)(262 s-1) = 0.25 m/s
F = ma
kx = ma
a = kx/m
a=kx
m
T=2 m
k
T2 = (2 )2m
k
k = (2 )2 =
m T2
a=kx
m
a = (2 f)2x = (2 f)2A
a = [(2 )(262 Hz)]2(1.5 X 10-4 m) = 410 m/s2
But we don’t know k or m
Solve for k/m
(2 )2f2
6
Cos: Example 2a
x = A cos2 ft
x = (0.25 m)cos
t
8.0
Find the amplitude, frequency and period of motion
for an object vibrating at the end of a spring that
follows the equation:
Therefore A = 0.25 m
x = (0.25 m)cos
2 ft =
t
t
8.0
2f =
8.0
f = 1/16 Hz
8.0
Cos: Example 2b
Find the position of the object after 2.0 seconds.
x = (0.25 m)cos
t
8.0
x = (0.25 m)cos
T = 1/f = 16 s
The Pendulum
• Pendulums follow SHM only
for small angles (<15o)
• The restoring force is at a
maximum at the top of the
swing.
4.0
Fr = restoring Force
x = 0.18 m
Remember the circle (360o = 2 rad)
L
=x
L
x
Fr = mgsin
at small angles sin =
Fr = mg
mg
Fr = mg
Fr = mgx
L
(Look’s like Hook’s Law F = -kx)
k = mg
L
T=2 m
k
T = 2 mL
mg
Fr
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T=2
L
g
f=1 = 1 g
T 2 L
The Period and Frequency of a pendulum depends
only on its length
Example 1: Pendulum
What would be the period of a grandfather clock
with a 1.0 m long pendulum?
T=2
L
g
Ans: 2.0 s
Damped Harmonic Motion
•Most SHM systems slowly stop
•For car shocks, a fluid “dampens” the motion
Swings and the Pendulum
• To go fast, you need a high frequency
• Short length (tucking and extending your legs)
f= 1 g
2 L
decrease the denominator
Example 2: Pendulum
Estimate the length of the pendulum of a
grandfather clock that ticks once per second (T =
1.0 s).
T=2 L
g
Ans: 0.25 m
Resonance: Forced Vibrations
• Can manually move a spring (sitting on a car and
bouncing it)
• Natural or Resonant frequency (fo)
• When the driving frequency f = fo, maximum
amplitude results
– Tacoma Narrows Bridge
– 1989 freeway collapse
– Shattering a glass by singing
8
Wave Medium
• Mechanical Waves
– Require a medium
– Water waves
– Sound waves
– Medium moves up and down but wave moves
sideways
• Electromagnetic Waves
– Do not require a medium
– EM waves can travel through the vacuum of space
Parts of a wave
•
•
•
•
•
•
Crest
Trough
Amplitude
Wavelength
Frequency (cycles/s or Hertz (Hz))
Velocity
v= f
9
Velocity of Waves in a String
Example 1: Strings
• Depends on:
• Tension (FT) [tighter string, faster wave]
• Mass per unit length (m/L) [heavier string, more
inertia]
A wave of wavelength 0.30 m is travelling down a
300 m long wire whose total mass is 15 kg. If the
wire has a tension of 1000 N, what is the velocity
and frequency?
v = FT
m/L
v=
1000 N
15 kg/300 m
v= f
f = v/
= 140 m/s
= 140 m/s/0.30 m = 470 Hz
Transverse and Longitudinal Waves
• Transverse Wave – Medium vibrates
perpendicular to the direction of wave
– EM waves
– Water waves
– Guitar String
• Longitudinal Wave – Medium vibrates in the
same direction as the wave
– Sound
Longitudinal Waves: Velocity
• Wave moving along a long solid rod
– Wire
– Train track
vlong=
E
Elastic modulus
• Wave moving through a liquid or gas
vlong = B
Bulk modulus
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Ex. 1: Longitudinal Waves: Velocity
How fast would the sound of a train travel down a
steel track? How long would it take the sound to
travel 1.0 km?
vlong=
E = (2.0 X 1011/7800 kg/m3)1/2
vlong = 5100 m/s (much fast than in air)
v = x/t
t = x/v
1000m/5100m/s = 0.20 s
Energy Transported by Waves
Intensity = Power transported across a unit area
perpendicular to the wave’s direction
I = Power =
Area
P
4 r2
Comparing two distances:
I1r12 = I2r22
Earthquakes
• Both Transverse and Longitudinal waves are
produced
• S(Shear) –Transverse
• P(Pressure) – Longitudinal
• In a fluid, only p waves pass
• Center of earth is liquid iron
Intensity: Example 1
The intensity of an earthquake wave is 1.0 X 106
W/m2 at a distance of 100 km from the source.
What is the intensity 400 km from the source?
I1r12 = I2r22
I2 = I1r12/r22
I2 = (1.0 X 106 W/m2)(100 km)2/(400 km)2
I2 = 6.2 X 104 W/m2
Reflection of a Wave
•Hard boundary inverts the
wave
•Exerts an equal and
opposite force
•Continue in same direction if
using another rope boundary
•Loose rope returns in same
direction
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Constructive and Destructive
Interference
Constructive and Destructive
Interference : Phases
Waves “in phase”
Destructive
Constructive
Interference
Interference
Resonance
Standing Wave – a wave that doesn’t
appear to move
Node – Point of destructive interference
Antinode – Point of constructive
interference (think “Antinode,Amplitude)
“out of phase”
in between
Resonance: Harmonics
Fundamental
•Lowest possible frequency
•“first harmonic”
•L = ½
First overtone (Second Harmonic)
Second overtone (Third Harmonic)
“Standing waves are produced only at the
natural (resonant) frequencies.”
Resonance: Equations
L=n
n
2
f = nv = nf1
2L
v= f
v = FT
m/L
n = 1, 2, 3…..
Example 1: Resonance
A piano string is 1.10 m long and has a mass of
9.00 g. How much tension must the string be
under to vibrate at 131 Hz (fund. freq.)?
L=n n
2
2L
= 2.20 m
1=
1
v = f = (2.20 m)(131 Hz) = 288 m/s
12
v = FT
m/L
What are the frequencies of the first four harmonics
of this string?
v2 = FT
m/L
f1 = 131 Hz
f2 = 262 Hz
f3 = 393 Hz
f4 = 524 Hz
FT = v2m = (288 m/s)2(0.009 kg) = 676 N
L
(1.10 m)
•Velocity of a wave changes when crossing between substances
• Both reflection and refraction occur
• Angle of incidence = angle of reflection
2
1
=
1st Overtone
2nd Overtone
3rd Overtone
Refraction
Hitting a Boundary
1
1st Harmonic
2nd Harmonic
3rd Harmonic
4th Harmonic
•Soldiers slow down marching into mud
sin
1
= v1
sin
2
= v2
2
Reflected
wave
air
water
Refracted
wave
Example 1: Refraction
An earthquake p-wave crosses a rock boundary
where its speed changes from 6.5 km/s to 8.0
km/s. If it strikes the boundary at 30o, what is the
angle of refraction?
sin
sin
2
= v1
2 = v2
1
= 38o
Example 2: Refraction
A sound wave travels through air at 343 m/s and
strikes water at an angle of 50. If the refracted
angle is 21.4o, what is the speed of sound in
water?
sin 30o = 6.5 km/s
sin 2
8.0 km/s
(Ans: 1440 m/s)
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Diffraction
Note bending of wave into “shadow region”
Diffraction
• Bending of waves around an object
• Only waves diffract, not particles
• The smaller the obstacle, the more diffraction in
the shadow region
14