Harmonic Motion and Waves

Simple Harmonic Motion(SHM)
• 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 1X10 6 Hz = 1.031 X 10 8 Hz
1M Hz
T = 1/f = 1/(1.031 X 10 8 Hz) = 9.700 X 10 -9 s

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 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/s 2 )/(-0.03 m) k = 6.5 X 10 4 N/m
(note that we did not include the mass of the car)

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/s 2 )/ 6.5 X 10 4 N/m x = 4.5 X 10 -2 m = 4.5 cm

Forces on a Spring
• Extreme Position (Amplitude)
– Force at maximum
– Velocity = 0
• Equilibrium position
– Force = 0
– Velocity at maximum

Energy and Springs
• KE = ½ mv 2
• PE = ½ kx 2
• Maximum PE = ½ kA 2
Law of conservation of Energy
½ kA 2 = ½ mv 2 + ½ kx 2

All PE
All KE
All PE
Some KE and Some PE

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 kA 2
Maximum PE = ½ (20 N/m)(0.030 m 2 )
Maximum PE = 9 X 10 -3 Nm (J)

Spring Energy: Example 1a
What is the maximum speed of the mass?
½ kA 2 = ½ mv 2 + ½ kx 2
½ kA 2 = ½ mv 2
9 X 10 -3 J = ½ (0.50 kg)v 2 v = 0.19 m/s
(x=0 at the origin)

Spring Energy: Example 1b
What is the potential energy and kinetic energy at x
= 2.0 cm?
PE = ½ kx 2
PE = ½ (20 N/m)(0.020 m 2 ) = 4 X 10 -3 J
½ kA 2 = ½ mv 2 + ½ kx 2
½ mv 2 = ½ kA 2 - ½ kx 2
KE = 9 X 10 -3 J - 4 X 10 -3 J = 5 X 10 -3 J

Spring Energy: Example 1c
At what position is the speed 0.10 m/s?
(Ans: + 2.6 cm)

Spring Energy: Example 2a
A spring stretches 0.150 m when a 0.300 kg mass is suspended from it (diagrams a and b). Find the spring constant.
(Ans: 19.6 N/m)

Spring Energy: Example 2b
The spring is now stretched an additional 0.100 m and allowed to oscillate (diagram c). What is the maximum velocity?

The maximum velocity occurs through the origin:
½ kA 2 = ½ mv 2 + ½ kx 2
½ kA 2 = ½ mv 2 kA 2 = mv 2 v 2 = kA 2 /m
(x=0 at the origin) v = \/kA 2 /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 kA 2 = ½ mv 2 + ½ kx 2 kA 2 = mv 2 + kx 2 mv 2 = kA 2 - kx 2 v 2 = kA 2 - kx 2 m v 2 = 19.6 N/m(0.100m
2 – 0.0500m
2 ) = 0.49 m 2 /s 2
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/s 2

Trigonometry and SHM
• Ball rotates on a table
• Looks like a spring from the side
• One rev(diameter) = 2 p
A
T = 2 p m k f = 1
T

• Period depends only on mass and spring constant
• Amplitude does not affect period v o
= 2 p
Af or v o
= 2 p
A
T v o is the initial (and maximum) velocity

Period: Example 1
What is the period and frequency of a 1400 kg car whose shocks have a k of 6.5 X 10 4 N/m after it hits a bump?
T = 2 p m = 2 p
(1400 kg/6.5 X 10 4 N/m) 1/2 k
T = 0.92 s f = 1/T = 1/0.92 s = 1.09 Hz

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 p m k
T 2 = (2 p
) 2 m k k = (2 p
) 2 m = (2 p
) 2 (3.0 X 10 -4 kg) = 2.7 N/m
T 2 (0.0667) 2

Period: Example 2b
What would be the frequency for a lighter insect,
0.10 g? Would it be higher or lower?
T = 2 p m k
T = 2 p
(m/k) 1/2
T = 2 p
(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 p t or x = A cos2 p ft
T
A = Amplitude t = time
T = period f = frequency

x = A cos2 p ft
Velocity is the derivative of position v = -v o sin2 p ft
Acceleration is the derivative of velocity a = -a o cos2 p ft

Cos: Example 1a
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?
x = A cos2 p ft x = (1.5 X 10 -4 m) cos2 p
(262 s -1 )t x = (1.5 X 10 -4 m) cos(1650 s -1 )t

Cos: Example 1b
What is the position at t = 1.00 ms (1 X 10 -3 s) x = A cos2 p ft x = (1.5 X 10 -4 m) cos2 p
(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

Cos: Example 1c
What is the maximum velocity and acceleration?
v o v o
= 2 p
Af
= 2 p
(1.5 X 10 -4 m)(262 s -1 ) = 0.25 m/s

F = ma kx = ma a = kx/m a = k x
But we don’t know k or m
Solve for k/m m
T = 2 p m k
T 2 = (2 p
) 2 m k k = (2 p
) 2 = (2 p
) 2 f 2 m T 2

a = k x m a = (2 p f) 2 x = (2 p f) 2 A a = [(2 p
)(262 Hz)] 2 (1.5 X 10 -4 m) = 410 m/s 2

Cos: Example 2a
Find the amplitude, frequency and period of motion for an object vibrating at the end of a spring that follows the equation: x = (0.25 m)cos p t
8.0

x = A cos2 p ft x = (0.25 m)cos p t
8.0
Therefore A = 0.25 m
2 p ft = p t
2f =
1
8.0
8.0
f = 1/16 Hz T = 1/f = 16 s

Cos: Example 2b
Find the position of the object after 2.0 seconds.
x = (0.25 m)cos p t
8.0
x = (0.25 m)cos p
4.0
x = 0.18 m

The Pendulum
• Pendulums follow SHM only for small angles (<15 o )
• The restoring force is at a maximum at the top of the swing.
q
F r
= restoring Force

Remember the circle (360 o = 2 p rad) q
L q
= x
L x q F r
= mgsin q at small angles sin q
= q
F r
= mg q mg
F r

F r
= mg q
F r
= mgx
L k = mg
L
T = 2 p m k
T = 2 p mL mg
(Look’s like Hook’s Law F = -kx)

T = 2 p
L g f = 1 = 1 g
T 2 p
L
The Period and Frequency of a pendulum depends only on its length

Swings and the Pendulum
• To go fast, you need a high frequency
• Short length (tucking and extending your legs) f = 1 g
2 p
L decrease the denominator

Example 1: Pendulum
What would be the period of a grandfather clock with a 1.0 m long pendulum?
T = 2 p
L g
Ans: 2.0 s

Example 2: Pendulum
Estimate the length of the pendulum of a grandfather clock that ticks once per second (T =
1.0 s).
T = 2 p
L g
Ans: 0.25 m

Damped Harmonic Motion
•Most SHM systems slowly stop
•For car shocks, a fluid “dampens” the motion

Resonance: Forced Vibrations
• Can manually move a spring (sitting on a car and bouncing it)
• Natural or Resonant frequency (f o
)
• When the driving frequency f = f o
, maximum amplitude results
– Tacoma Narrows Bridge
– 1989 freeway collapse
– Shattering a glass by singing

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 = l f

Velocity of Waves in a String
• Depends on:
• Tension (F
T
) [tighter string, faster wave]
• Mass per unit length (m/L)
[heavier string, more inertia] v = F
T m/L

Example 1: Strings
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 = 1000 N
15 kg/300 m
= 140 m/s v = l f f = v/ l
= 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 v long
= E r
Elastic modulus
• Wave moving through a liquid or gas v long
= B r
Bulk modulus

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?
v long
= E = (2.0 X 10 11 /7800 kg/m 3 ) 1/2 r v long
= 5100 m/s (much fast than in air) v = x/t t = x/v
=
1000m/5100m/s = 0.20 s

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

Energy Transported by Waves
Intensity = Power transported across a unit area perpendicular to the wave’s direction
I = Power = P
Area 4 p r 2
Comparing two distances:
I
1 r
1
2 = I
2 r
2
2

Intensity: Example 1
The intensity of an earthquake wave is 1.0 X 10 6
W/m 2 at a distance of 100 km from the source.
What is the intensity 400 km from the source?
I
1 r
1
2 = I
2 r
2
2
I
2
= I
1 r
1
2 /r
2
2
I
2
= (1.0 X 10 6 W/m 2 )(100 km) 2 /(400 km) 2
I
2
= 6.2 X 10 4 W/m 2

Reflection of a Wave
•Hard boundary inverts the wave
•Exerts an equal and opposite force
•Loose rope returns in same direction

•Continue in same direction if using another rope boundary

Constructive and Destructive
Interference
Destructive
Interference
Constructive
Interference

Constructive and Destructive
Interference : Phases
Waves
“in phase” “out of phase” in between

Resonance
Standing Wave
– a wave that doesn’t appear to move
Node – Point of destructive interference
Antinode – Point of constructive interference (think “Antinode,Amplitude)
“Standing waves are produced only at the natural (resonant) frequencies.”

Resonance: Harmonics
Fundamental
•Lowest possible frequency
•“first harmonic”
•L = ½ l
First overtone (Second Harmonic)
Second overtone (Third Harmonic)

L = n l n
2
Resonance: Equations n = 1, 2, 3…..
f = nv = nf
1
2L v = l f v = F
T m/L

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 l n
2 l
1 =
2L = 2.20 m
1 v = l f = (2.20 m)(131 Hz) = 288 m/s

v = F
T m/L v 2 = F
T m/L
F
T
= v 2 m = (288 m/s) 2 (0.009 kg) = 676 N
L (1.10 m)

What are the frequencies of the first four harmonics of this string?
f
1
= 131 Hz 1 st Harmonic f
2
= 262 Hz 2 nd Harmonic f
3
= 393 Hz 3 rd Harmonic f
4
= 524 Hz 4 th Harmonic
1
2
3 st nd rd
Overtone
Overtone
Overtone

Hitting a Boundary
• Both reflection and refraction occur
• Angle of incidence = angle of reflection q
1 q
2 q
1
= q
2
Reflected wave air water
Refracted wave

Refraction
•Velocity of a wave changes when crossing between substances
•Soldiers slow down marching into mud sin q
1
= v
1 sin q
2
= v
2

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 30 o , what is the angle of refraction?
sin q
1 sin q
2
= v
1
= v
2 q
2
= 38 o sin 30 o = 6.5 km/s sin q
2
8.0 km/s

Example 2: Refraction
A sound wave travels through air at 343 m/s and strikes water at an angle of 5 0 . If the refracted angle is 21.4
o , what is the speed of sound in water?
(Ans: 1440 m/s)

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