Physics 231 Lecture 28

• Concepts for today’s lecture”• Spherical waves

PI =

• Dopper shift

24 rI

π=

+

• Interference of sound waves

−+=

s

o

vvvvƒƒ'

• Standing waves on string: edestructiv )2/1(0,1,2)(n veconstructi

21

21

λλ

+=−==−

nLLnLL

g g

,,1,2,3, 2 2 n === n

nL

Lnvfn λ

Intensity of spherical waves.

If d di i l• If one suspends non-directional sound source with power P in air, the sound will radiate in all directions All points on a sphere ofdirections. All points on a sphere of radius r will see the same sound intensity I=P/A:

r24 r

PIπ

=Fig 14.4, p. 433

Slide 8

Example

Wh h li t i h i 1100 di tl h d b• When a helicopter is hovering 1100 m directly overhead, an observer on the ground measures a sound intensity I1. Assume that sound is radiated uniformly as a spherical wave from the helicopter and that ground reflections are negligible How far must the helicopter fly in aground reflections are negligible. How far must the helicopter fly in a straight line parallel to the ground before the observer measures a sound intensity of I2 =I1 /5?

2

1

I 1/ 5I

=

2 2

PI 4 rπ

= 1 2

P I4 rπ

=dxz

y

hr2

24 rπ 14 rπ

2 2 22r h d = + 2

15r= 25h=

z

2 2d 4h = d 2h 2200m = =

Doppler effect for moving observer

Fi 14 8 435

• An observer is moving toward a stationary source

• An observer is moving away from a stationary source

Fig 14.8, p. 435

Slide 12Fig 14.9, p. 436

Slide 13

y• Due to his movement, the

observer detects an additional number of wave fronts

y• The observer detects fewer

wave fronts per second• The frequency heard is

• The frequency heard is increased, the pitch is higher

The frequency heard is decrease, the pitch is lower

v v+ ov vƒ ' ƒ

v+ =

Doppler Effect, Source in Motion – general formula

+= ovvƒƒ'

• Both the source and the

− svv

ƒƒ

Both the source and the observer could be moving

• Use positive values of voand v when the motion is

• As the source moves toward the observer (A), the wavelength appears shorter and the

and vs when the motion is toward – Frequency appears

higherfrequency increases

• As the source moves away from the observer (B), the

higher• Use negative values of vo

and vs when the motion is away

wavelength appears longer and the frequency appears to be lower

away– Frequency appears

lower

Example

A t i t t it d t f f 1 000 H A• A train at rest emits a sound at a frequency of 1 000 Hz. An observer in a car travels away from the sound source at a speed of 30.0 m/s. What is the frequency heard by the observer? (assume the sound velocity v=343 m/s)(assume the sound velocity v=343 m/s)– a) 513 Hz– b) 713 Hz– c) 913 Hz– d) 1013 Hz– e) 1113 Hz)

sv 0, =ov 30m / s= −

+ vv

( )o' v vf f

v+

=3131000Hz 913Hz343

= =

−+=

s

o

vvvvƒƒ'

Example

A h f t i it d ith f f 300 H h• A horn of train emits sound with a frequency of 300 Hz when the train is at rest. The train is moving at a speed of 100 km/h towards an observer that is at rests. What is the frequency heard by the observer? (assume the sound velocity v=343 m/s)heard by the observer? (assume the sound velocity v=343 m/s)– a) 275Hz– b) 293 Hz

Ov 0, =– c) 300 Hz– d) 313 Hz– e) 327 Hz

O

S1000m 1hv 100km / h 27.8m / s

km 3600s= =

343)

'

s

vf fv v

=−

343 300Hz 327Hz343 28

= =−

+ vv

−+=

s

o

vvvvƒƒ'

Conceptual quiz• Three observers,A,B, and C are listening to a moving source ofThree observers,A,B, and C are listening to a moving source of

sound. The diagram below shows the location of the wavecrestsof the moving source with respect to the three observers.Whichof the following is true?g– a. The wavefronts move faster at A than at B and C.– b. The wavefronts move faster at C than at A and B.

c The frequency of the sound is highest at A

Wavefronts moveat sound speed 343 m/s

– c. The frequency of the sound is highest at A.– d. The frequency of the sound is highest at B.– e. The frequency of the sound is highest at C.

soundvf

λ=

λ is smallest at Cf is largest at C

Standing waves on a string

T t li i id l• Two traveling sinusoidal waves of the same frequency moving in opposite directions can superimposed so as to make asuperimposed so as to make a standing wave:

• Because the string is held at each end those points areeach end, those points are “nodes” of the oscillator motion, where the amplitude of motion vanishes.motion vanishes.

• Anti-nodes are the points where the motion is maximal. “fundamental” “harmonics”Fig 14.18, p. 443

Slide 25

• Motion with one antinode has λ/2 = L, two antinodes has λ=L...

• The general relationship is:

n=1 fund. Cn=2 2nd Har. Cn=3 3d Har. G

nnL2

λ=n=4 4th Har. Cn=5 5th Har. E n

2L ; n

λ = nn

v fλ

= nv n F n 1,2,3,,,2L 2L μ

= = =

Example

A t t h d t i fi d t h d h f 40 0 d• A stretched string fixed at each end has a mass of 40.0 g and a length of 8.00 m. The tension in the string is 49.0 N. (a) Determine the positions of the nodes and antinodes for the 2nd

harmonic (b) What is the vibration frequency for this harmonic?harmonic. (b) What is the vibration frequency for this harmonic?

Second harmonic (n 2) has nodes at0 4 and 8 m and anti nodes

=0,4, and 8 m and anti-nodes at 2 and 6 m.

2vf F2f

2L= v

m / L=

49N

Fig 14.18, p. 443

Slide 25

49Nv 98m / s0.04kg / 8m

= =

2v 98m / sans. f 12.4Hz= = =2ans. f 12.4HzL 8m

Standing waves in an air columns: both ends open

If d f th i l i l d d t i t t thi d• If one end of the air column is closed, a node must exist at this end since the movement of the air is restricted

• If the end is open, the elements of the air have complete freedom of t d ti d i tmovement and an antinode exists

,,,,,nnL

Lnvfn 321 2 2 n === λ

Conceptual question• A string is clamped at both ends and plucked so it vibrates in a

standing mode between two extreme positions a and b Let upwardstanding mode between two extreme positions a and b. Let upward motion correspond to positive velocities.When the string is in position c, the instantaneous velocity of points along the string:– a is zero everywherea. is zero everywhere.– b. is positive everywhere.– c. is negative everywhere.

d d d l ti– d. depends on location.

dd

f

ff

d

Standing waves in an air columns: one end closed

If d f th i l i l d d t i t t thi d• If one end of the air column is closed, a node must exist at this end since the movement of the air is restricted

• If the end is open, the elements of the air have complete freedom of t d ti d i tmovement and an antinode exists

,,,,,nnL

Lnvfn 531 4 4 n === λ

Beats

B lt ti i l d d t i t f• Beats are alternations in loudness, due to interference• Waves have slightly different frequencies and the time between

constructive and destructive interference alternates

21 fffbeat −=