Simple Harmonic Oscillator (SHO)

• Any situation where the force exerted on a mass is directly proportional to the negative of the object’s position from an equilibrium point is a simple harmonic oscillator.

• English: A simple harmonic oscillator is an object that gets pushed toward some point, and the farther the object is from that point, the stronger the force pushing it back.

kxF

An Example of a SHO: The spring

X = 0 m

FG

FN

NFFF GN 0

Equilibrium Position

An Example of a SHO: The spring

An Example of a SHO: The springEPEKETotal ...

22

2

1

2

1kxmvETotal

2

2

1.. kAEP Max

2

2

1.. MaxMax mvEK

MaxMax EKEP ....

22

2

1

2

1MaxmvkA

22 Am

kvMax

222

2

1

2

1

2

1kAkxmv

222 kAkxmv 222 kxkAmv

)1()(2

22222

A

xkAxAkmv

)1(2

222

A

x

m

kAv

)1(2

222

A

xvv Max

MaxTotal EPE ..

)1(2

2

A

xvv Max

An Example of a SHO: The spring

An Example of a SHO: The spring

For all SHO:

])2cos[()( itfAtx

])2sin[()( max itfvtv

])2cos[()( max itfata

Am

ka max

m

kAAfv 2max

Where:

Ai i when x0

Sinusoidal Curve Generator

)2cos()( tfAtx )2sin()( max tfvtv

)2cos()( max tfata

0 ,x i isoA Let:

* is measured in radians

)cos()( Atx

)sin()( max vtv

)cos()( max ata

tftT

t 22

Since ,

The Simple Pendulum

FG

FT

NFFF GT 0

Equilibrium Position

0x

m = mass of bob

L = length of cord

FG = weight, mg

FT= tension

The Simple Pendulum

0x FG

Lx

FT

m = mass of bob

L = length of cord

FG = weight, mg

FT= tension

is measured along the arcx

x

The Simple Pendulum

0x FG

Lx

FT

m = mass of bob

L = length of cord

FG = weight, mg

FT= tension

is measured along the arcx

xThe Assumption: For sufficiently small , sin

FGYFGX

FGY=mg cos(θ)

FGX=-mg sin(θ)

mgmgFGX )sin(

xL

mgFGX This is in the format of Hooke’s

law, and thus we have a SHO

T of a Simple Pendulum

kxF (definition of a SHO)

xL

mgFGX (restoring force of a pendulum)

L

mgk

k

mT 2 (eqn. 11-7)

g

L

Lmgm

T 22 (Period of a simple Pendulum)

Pendulum Lab Expected Results:

21

21

2 LgT

21

21

)2( LgT

2/1axy

21

2 ga

g

LT 2 (Period of a simple Pendulum)

s

m

s

m

Tv 8.2

0.3

5.8

Pg. 343, #34: A fisherman notices that wave crests pass the bow of his anchored boat every 3.0 s. He measures the distance between two crests to be 8.5 m. How fast are the waves traveling?

T=3.0 s

λ=8.5 m

v=?

1Tf

11 33.)0.3( ssf

fv Eqn. (11-12)

s

msmv 8.233.5.8 1

Oar. . .

TTv

1

Pg. 343, #35: A sound wave in air has a frequency of 262 Hz and travels with a speed of 330 m/s. How far apart are the wave crests (compressions)?

f=262 Hz

v= 330 m/s

?

fv Eqn. (11-12)

f

v

s

sm

Hzsm

f

v1

26.1262

330

ms

s

m

ss

m

s

sm

1

11

m

s

sm

Hzsm

f

v26.1

126.1

262

330

Pg. 344, # 36: AM radio signals have frequencies between 550 kHz and 1600 kHz and travel with a speed of 3x108 m/s. On FM, the frequencies range from 88.0 MHz to 108 MHz and travel at the same speed. What are the wavelengths of these signals?

fAM high=1600 kHz

fAM low=550 kHz

fFM low=88.0 MHz

fFM high=108 MHz

f

v

v= 3x108 m/s

λAM high f= 188 m

λAM low f= 545 m

λFM low f= 2.78 m

λFM high f= 3.41 m

Pg. 344, #39: A cord of mass 0.55 kg is stretched between two supports 30 m apart. If the tension in the cord is 150 N, how long will it take a pulse to travel from one support to the other?

t

xv

m = 0.55 kg

L = 30 m = Δx

FT = 150 N

Δt=?v

xt

LmF

v T Eqn. 11-13

v = 90.5 m/s

s

smm

v

xt 33.

5.90

30

Pg. 344, #44: Compare the (a.) intensities and (b.) the amplitudes of an earthquake wave as it passes two points 10 km and 20 km from the source.

Area

PowerI

Let the wave have power P.

210 2 r

PI km

2220 8)2(2 r

P

r

PI km

r10 km = 10,000 m

r20 km=20,000 m = 2r10 km

4

812

8

2

2

2

20

10

rPrP

I

I

km

km

I10km is 4 times greater than I20km

?20

10 km

km

I

I

Pg. 344, #44: Compare the (a.) intensities and (b.) the amplitudes of an earthquake wave as it passes two points 10 km and 20 km from the source.

Let the wave have power P.r10 km = 10,000 m

r20 km= 20,000 m = 2r10 km

I10km = 4 I20km

rA

1

2

21

1

10

10

20

10

km

km

km

km

r

r

A

A?

20

10 km

km

A

A

Pg. 344, #45: The intensity of a particular earthquake wave is measured to be 2.0 x 106 J/m2s at a distance of 50 km from the source.

a.) What was the intensity when it passed a point only 1 km from the source?

I50 km= 2.0 x 106 J/m2s

r50 km = 50,000 mArea

timeEnergy

Area

PowerI

250505050 2 kmkmkmkm rIAreaIP

25050 kmkm rIP

sm

J

r

rI

r

P

Area

PI

km

kmkm

kmkmkm

2

92

1

25050

211

1 100.52

2

2

r1 km= 1000 m

?1 kmI

Pg. 344, #45: The intensity of a particular earthquake wave is measured to be 2.0 x 106 J/m2s at a distance of 50 km from the source.

b.) What was the rate energy passed through an area of 10.0 m2 at 1.0 km?

I50 km= 2.0 x 106 J/m2s

r50 km = 50 kmArea

timeEnergy

Area

PowerI

25050 2 kmkm rIP

sm

JI km

2

91 100.5

r1 km= 1 km )(1 AreaItimeEnergy

km

Area = 10.0 m2

Wattsmsm

Jtime

Energy 1022

9 105)10(100.5

Pg. 344, #46: Show that the amplitude A of circular water waves decreases as the square root of the distance r from the source. Ignore damping.

rA

1In other words, show that:

1. The same energy that passes through the small circle each second must pass through the big circle each second, so P for each whole circle is constant.

2. Since and ,r

PI

2 2AI

r

PA

22

The only variable is r, so r

A1

Pg. 344, #50:

Standing Waves; Resonance• Standing waves occur on a string of length L

when the waves have a wavelength,λ, in which L is a multiple of .5λ

2nn

L

Standing Waves; Resonance

Only certain wavelengths can create standing waves for a given length of cord, but there are an infinite number of wavelengths that can create a standing wave on any given cord.

n

Ln

2

2nn

L

Standing Waves; ResonanceA little algebra to find resonant frequencies:

v

f

11 )2

()( n

Ln

L

n

n 2

1

fv Eqn. (11-12)

L

nvfn 2

n

Ln

2 Eqn. (11-19)

For the first harmonic, n=1:

L

vf

21 so: 1nffn

Pg. 344, #51: If a violin string vibrates at 440 Hz as its fundamental frequency, what are the frequencies of the first four harmonics?

fn=1 = 440 Hz

fn=2 = ?

fn=3 = ?

fn=4 = ?

1nffn

fn=2 =880 Hz

fn=3 = 1320 Hz

fn=4 = 1760 Hz

(derived on pg. 336)

Sound Intensity• The more powerful a wave is when it reaches the

ear, the more energy per unit time it delivers to the ear, and as a consequence, a more powerful sound is perceived as being louder.

• The human ear can perceive a wide range of sound intensities:– 10-12 W/m2 – threshold of hearing– 1 W/m2 – threshold of pain

• Notice that the quietest sound a human can hear is almost a trillion times less intense than a sound that is so intense that it causes physical pain.

The (deci) bel

• The human ear can hear a wide range of intensities, but can not distinguish between small changes in intensity.

• We connect the measurable quantity, Intensity, I, to the perceived quantity of loudness, called intensity level, β, using the equation:

0

log10I

I Where I0 is the threshold

of hearing, 1x10-12 W/m2

Logarithms Math Crash-course:

Logarithms Math Crash-course:

Arithmetic scale Logarithmic scale

Logarithms Math Crash-course:xb y yxb )(logIf then

Logarithms are discussed in your text, Appendix A, pg. 1046

Logarithm identities:

)log()log()log( yxxy

)log()log( xyx y

)log()log()log( yxy

x

)log()log( xy yx y

xxy )log(

)log(

The (deci) bel continued

• A change in the intensity level is associated with a change in a sound’s loudness.

• A human ear can perceive a change in intensity level of about 1 dB. – How great of an increase in intensity is an

increase in intensity level of 1 dB?– How great of an increase in intensity is an

increase in intensity level of 10 dB?