Lecture 24 1

PHY121 – Physics for the Life Sciences I

Lecture 241. Review: Kinematics and Dynamics2. Traveling Waves3. Intensity Level4. Doppler Effect

Note: set your Clicker to Channel 215/1/2014

A Quick Review of Kinematics & Dynamics1. Generalized Velocity & Acceleration are basic to

physics and other sciences …– they describe how ‘things’ CHANGE with time or position –

life would be extremely dull otherwise … – derivatives are everywhere: velocity, acceleration, …, cancer

growth (dN/dt), radioactivity (dN/dt), …

2. Not too complicated mathematically see BB Course Documents/Lectures/1st document, and: http://sbhepnt.physics.sunysb.edu/~rijssenbeek/Differentiation.pdf

3. Can use ‘work-arounds’ such as Δx/Δt , etc. or simply ‘memorize’ results …

5/1/2014 Lecture 24 2

KinematicsMotion: position as function of time x(t), s(t)(x(t),y(t),z(t))• Kinematics = describing the motion• e.g. acceleration: a(t)dv/dt

• and velocity: v(t)ds/dt

5/1/2014 Lecture 24 3

ddt

v a d dt v a

0

( )

0

t t

d dt v

v

v a 00

( )t

t dt v v a 00

( )t

t dt v v a

0

For on ly: t

dt tconstan at aa 0 constant acceleration( )t t v v a

Cross check: ddtv 0

d tdt

v a

ddt

s v d dt s v

0

( )

0

t t

d dt s

s

s v 00

( )t

t dt s s v 00

( )t

t dt s s v

0

For :t

dtcons vtant a 00

( )t

t dt v a 20

12

t t v a

20 0

constant acceleration

12

( )t t t s s v a

0 a

KinematicsNotes:• Nothing special about the first, second time-derivatives,

we can go to higher and higher derivatives of position w.r.t. time if needed …

• In mechanics, the second time derivative a is ‘special’ because it correlates (Newton’s Law) directly with the net force Fnet !

• constants s0=(x0,y0,z0) and v0=(v0x,v0y,v0z) are the ‘initial conditions’ of the (linear) motion!

5/1/2014 Lecture 24 4

Angular Kinematicsmany other variables to describe the motion …• e.g. ANGULAR variables useful for curved motion:

• note: for any ‘curved’ motion, there must be a ‘centripetal’ acceleration (and therefore centripetal force):

5/1/2014 Lecture 24 5

( )( ) s ttR

( ) dtdt

( ) dtdt

θ(t)O

R

s(t)1 dsR dt

/ /

Rv

/ /1 dvR dt

/ /

Ra

2 2/ /ca v R R

‘//’ means parallel to the curved path …

Transformations between Angular and …One can always TRANSFORM between variables …• e.g. for ANGULAR variables to/from RECTANGULAR

coordinates:

5/1/2014 Lecture 24 6

0

0

cos( ) cos( ...)sin( ) sin( ...)

xy

tA AtA A

2 2

arctan

A x yyx

θ(t)O

R

s(t)

x

y

DynamicsDynamics: Motion as result of Forces• Newton:

• The forces: Fnet=F1 + F2 + … Fi can be many types ! – e.g. constant force: e.g. Fnet= W = mg

simple:therefore motion:

– position (s) dependent: e.g. Fnet= –kxmore complicated:

constants A and follow from the ‘initial conditions’ (x0 and v0):

5/1/2014 Lecture 24 7

net mF a2

2

dmdt

s dm

dt

v

mg m a 0,x ya a g

kx ma 2

2

d xmdt

diff. equation

( ) cos( )x At t , with km

i.e. constant a 0 0xx t x v t

0

0

atan vx

2

2200

v xA

20 0

12

, –yy t y v t gt

ExampleExample: A mass m = 0.20 kg on spring with k = 20 N/m is set into motion at x0 = 0.30 m and v0 = 4.0 m/s; calculate the position and velocity at t=2.0 s:

– Initial conditions:

– Result: x(t) = (0.50 m) cos(10t + 0.93)– using t=2.0 s: x(t=2.0 s) = (0.50 m) cos(20.93) = 0.24 m– velocity: v(t) = (0×0.50 m/s) sin(10t + 0.93) v(t=2.0 s) = 4.3 m/s

5/1/2014 Lecture 24 8

stre

tche

d

m x0

mx

v0

k m –220 kg s 0.20 kg –2100 s 10 rad/s

0 cosx A 0, sinv A 0

0

tan vx

4.0 m/s10 0.30 m/s

0.93

2200

vA x

2 24.0 3 5 0.50 m10 10 10

2 =0.62 sT

Dynamics (continued)Dynamics: Motion as result of Forces• Newton:

• The forces: Fnet=F1 + F2 + … Fi can be many types ! – velocity (v) dependent: Fnet = – bv

– velocity-dependent + constant (mg) force … (free fall):

• note, from filling in solution we find: mg=bv∞ and v(t→∞)=v∞=mg/b the terminal velocityby differentiating: initial acceleration a(t=0) = g

– time (t) dependent force: e.g. Fd (t) =Fmaxcost(ωdt) • see resonance in SHM …

5/1/2014 Lecture 24 9

net mF a2

2

dmdt

s dm

dt

v

bv ma dvmdt

diff. equatio 0n

( ) tt evv , with bm

bv mg ma dvmdt

diff. equation

( ) (1 )tv t ev

, with bm

HW ProblemA point on a string undergoes simple harmonic motion as a sinusoidal wave passes. • When a sinusoidal wave with speed v, wavelength λ cm

and amplitude A passes, what is the maximum speed of a point on the string?

• Solution:

5/1/2014 Lecture 24 10

cos( )y A kx t sin( )yv A kx t

,maxyv A 2 Af 2 ...v A

Lecture 24 11

Energy Transport in Traveling Wave: P ω2A2

Consider, as before, a piece of the string on which a wave is traveling in +ve x-direction:– The RATE of energy transfer (Power)

transmitted into the direction of propagation of the wave

– Don’t worry about the derivation, but look at the behavior:• Power oscillates with x and t ! • Maximum: Pmax = √(μTS) ω2A2 ; Minimum: Pmin = 0, varies as ω2 and A2

• Average power (average of sin2=½): Pavg = ½ Pmax

– Intensity is defined as Power-per-unit-Area at a particular point…

Ts

F(x) Fy(x)

x

vy, ay

y

P dWdt

sin( ) sin( )ST kA kx t A kx t 2 22sin ( )ST kx tA

y yF v S ydyT vdx

Sdy dyTdx dt

d

dt

F y

2 2sin ( )ST k A kx t

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cos( )y A kx t

Lecture 24 12

Intensity = Power / AreaIntensity I is Power-per-unit-Area (“” to the wave): – for energy emission from a point-source of power Pavg, radiating in all

direction equally (3-d), the intensity at distance r away from the source is:

I(at distance r) = Pavg/Area = Pavg/4πr2

– Even if the source does not radiate equally in all directions, still intensity often drops as the inverse square of the distance ! • true for many types of radiation: Sun light, flash light beams, loudspeaker

sound, ….• e.g. the intensity of sunlight received by Earth varies over its orbit,

because distance Earth-Sun varies slightly over the course of the year…

– Notes:• we ignored absorption or reflection of radiation along the way…• by focusing a beam, we can make the intensity inside the beam INCREASE

with distance up to the focal point; beyond that, the intensity drops off again with the square of the distance…

• laser beams keep a constant cross section and intensity is thus constant …5/1/2014

avg2(at distance )

PI r

r

HW ProblemAt a rock concert, the sound intensity R1=1.0 m in front of the bank of loudspeakers is I1=0.10 W/m2. A fan is R2=30 m from the loudspeakers. Her eardrum has an area A2=50 mm2.• How much energy is transferred to each eardrum in

one second?

5/1/2014 Lecture 24 13

1 givenI

2 2 2P I A

2 22 2 1

2 21 1 2

1,1

I R RI R R

21

2 122

RI IR

A point source delivers an intensity of 36 W/m2 at 2 m distance; what is the received

intensity at 6 m distance (in the same direction)?

Lecture 24 14A. B. C. D.

15%

3%

66%

16%

A. 12 W/m2

B. 9 W/m2

C. 4 W/m2

D. 1 W/m2

I(at distance r) = Pavg/Area Pavg/r2

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Lecture 24 15

Speed of Sound Waves v ≈ 343 m/s (T=293K)The medium in which the sound wave (= pressure/compression wave) propagates is characterized by B [N/m2] and density ρ= m/V [kg/m3].The only way to make a velocity out of this is: v √(B/ρ)

– It can be shown that :

For sound in air, treating air as an ideal gas and the pressure waves propagating adiabatically (no flow of heat; Q=0):

– where M is the molar mass (mass per mole) of air, – and T is the absolute temperature in Kelvin–Then:– varies like √T ! – (T is temperature here; in K !)

v B

adiabatic:pV C

dp dp dCV nRT mRTV V CV pdV V dV d

RTBMV V MV

1.40 8.31 292 0. 343028 m8 /sB R Mv T

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Lecture 24 16

Sound Intensity I P/Area (Δpmax)2

Intensity at the position of the receiver-of-sound is defined as the average power (P) emitted by the (point-like) source that is received per unit of Area (A) at the receiver:

2 2

2

22 2 max

2 2

2

( , )( , ) ( , ) ( , )Area Area

sin( ) sin( )

sin ( )

1sin ( )Area 21 1 12 2 2

yP dy x tp x t v x t p x t

dxBkA kx t A kx t

Bk A kx t

PI Bk A kx t Bk A

BBk Av

pI AB

I

F v

•overbar indicates

overbarsindicate Average

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Lecture 24 17

Sound Intensity Level β(10 dB)log10(I/I0)Our hearing is very sensitive: – young ears are sensitive to sound intensities of only I0 = 10–12 W/m2,

the ‘threshold of hearing’, at the ‘best’ frequency around 1000 Hz. – The “pain level” is about 1 W/m2

– thus, our ears have a very large dynamic range of 1012 !

Sound Intensity Level is defined as a logarithmic base-10 scale, with unit dB, the deci-Bel– tenth of a Bel, after A. Bell, defined as:

– with I0≡10–12 W/m2

– hearing goes from 0 dB (I=10–12 W/m2) to 120 dBAlexander Graham Bell

(1847 –1922)

100

( ) 10 dB log III

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Lecture 24 18

Review Logarithms …Logarithms (base 10):– if y=10x

– then: log10 y = x– and vice versa:

if log10 y = x, then y=10x

Thus: – log1010=1, log10100=2, log101000=3– log101=0, log100.1= –1, log100.01= –2– Properties: alogx=logxa ; logx + logy=log(xy); and logx – logy=log(x/y)

A 3 dB Increase in sound level corresponds to:

– often: log10 is written as log– and loge is written as: ln (natural log; e ≡ 2.71828 …)

new

o1

ld0lo3 dB 10 dB g I

I 3 1 0.3new

old

01 20 10 .0II

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x →

log10 x

x10

Lecture 24 19

Examplea 20% efficient, 200 W (electrical) loudspeaker is blasting at nominal power: – What is the distance at which I reach pain level?

(Assume the speaker radiates sound uniformly in the frontal hemisphere only)

– What distance do I need for only 0.1 W/m2?

– What is the Sound Level difference between the two intensities above?

– NOTE: This is simplified: it ignores the non-point source character of the loudspeaker, and effects from absorption and reflection off nearby surfaces

pain 1I I P Area 2 2pain200 0.20W (2 ) 1 W/m r

2pain 40 W / (2 )r pain 2.5 mr

2 AreaI P 2 22200 0.20W (2 ) 0.1 W/mr

22 210 40W (2 ) 8 mr r

1 2( ) – ( )I I 1 2

0 0

10log – 10logI II I

1 0

2 0

10log I II I

1

2

10log II

1 0log 10 10 dB

100

12 20

2pain

( ) 10 dB log

10 W/m1.0 W/m

III

II

5/1/2014

HW ProblemYour ears are sensitive to differences in pitch, but they are not very sensitive to differences in intensity. You are not capable of detecting a difference in sound intensity level of less than 1 dB.• By what factor does the sound intensity increase if the

sound intensity level β increases from 60 dB to 61 dB ?

5/1/2014 Lecture 24 20

2 1

2 110

0 0

(10 dB) log I II I

0.12

1

10 ...II

2 110 10

0 0

(10 dB) log (10 dB) logI II I

1 dB

210

1

(10 dB) log II

1 dB

Real Loudspeakers• Bosch 36W column loud-

speaker polar pattern• Monsoon Flat Panel

speaker:(5 dB grid)– 400 Hz:

– 1 kHz:

– 3 kHz:

– 7 kHz:

Lecture 24 215/1/2014

A 10 dB increase in the sound level corresponds to …

5/1/2014 Lecture 24 22A. B. C. D.

25% 25%25%25%

A. an increase in intensity of factor 2B. an increase in intensity of factor 3C. an increase in intensity of factor 5D. an increase in intensity of factor 10

β(I) ≡ (10 dB) log10(I /I0) with I0≡10–12 W/m2

Δβ(I) ≡ β(If) – β(Ii) = (10 dB) log10(If /Ii)

Lecture 24 23

Doppler Effect: fL=fS(v+vL)/(v+vS)Doppler effect: frequency of received sound depends on the relative motion of the source or receiver with respect to the medium (e.g. air):

λ’<λf ’>f

λ’

λ’’>λf ’’<f

λ’’λ

5/1/2014

+ve for Lapproaching S

+ve for Smoving away

from L

LL S

S

v vf fvv

Doppler Formula:

Lecture 24 24

Doppler Effect - DerivationDoppler effect: frequency of received sound depends on the relative motion of the source or receiver with respect to the medium (e.g. air):– E.g. moving towards the source of the sound wave, my ears will “catch” more

pressure variations per second than if I stay still or move away. • Thus, the frequency fL I perceive depends on my velocity vL with respect to the air.• Similarly, the motion of the sound source with respect to the air vS affects the

wavelength λ of the air waves.For the simple case where all motions are along the x-direction: – Assume a sound wave in –ve x-direction; the speed of sound is v = 343 m/s– LISTENER who has velocity vL in +ve x-direction:

– SOURCE with velocity vS in +ve x-direction: the source travels a distance vSTper period, so that the effective wavelength is increased by that amount:

Combining: Doppler Formula:

relative speedwavelengthLf

S

LLf v v

+ve for Lapproaching S

+ve for Smoving away

from L

( )Lv v

Lvv

Sv T S

S

v vf

S

S S

vvf f

LS

S

vfv

vv

vvL

x

vx

λ

λ vS

5/1/2014

Lecture 24 25

Example

– Note, that if vL = vS then fL = fS (e.g. when the wind blows from source to receiver, nothing changes…

a bat emits a high-pitched “chirp” at 80 kHz (ultra-sound) when it approached a fixed wall with velocity vBat=10 m/s– calculate the frequency of the reflected chirps the bat receives…

– Note that the SIGN of the velocities is CRUCIAL!– the problem is even more complex when the wall is a MOVING INSECT!

This is an example of SONAR

incident on wall BatBat

vf fv v

Batreceived by Bat reflected from wall

v vf fv

BatBat

Bat

35480 kHz 84.8 kHz334

v vfv v

+ve for L approaching S

+ve for S moving away from LLL S

S

v vf fvv

reflected from wallf

BatBat

Bat

v vvfv v v

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A strong wind is blowing from a stationary source towards a stationary listener …

5/1/2014 Lecture 24 26A. B. C.

0% 0%0%

A. The received tone is lower than the emitted tone … B. The received tone is equal to the emitted tone …C. The received tone is higher than the emitted tone …

+ve for Lapproaching S

+ve for Smoving away

from L

LL S

S

v vf fvv

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