HW2 Solutions Notice numbers may change randomly in your assignments and you may have to recalculate solutions for your specific case. Tipler 15.P.041

The wave function for a harmonic wave on a string is

y(x, t) = (0.0030 m) sin((58.8 m-1)x + (312 s-1)t).

(a) In what direction does this wave travel? What is its speed? (b) Find the wavelength of this wave. Find its frequency. Find its period. (c) What is the maximum speed of any string segment? Solution: (a) The wave function is: y(x, t) = (0.0030 m) sin((58.8 m-1)x + (312 s-1)t) = A sin(kx-ωt). So the wave is traveling in the direction of –x with speed

!

v ="

k=312

58.8= 5.31m /s

(b) the wavelength, frequency and period are

!

" =2#

k= 0.107m;f =

$

2#= 49.7Hz;T =

1

f= 0.02s

(c) the speed of any string segment is

v =

!

dy

dt= A" cos(kx #"t) and the maximum is for cos(kx-ωt)=±1,

hence vmax = Aω = 0.0030 x 312 rad/s = 0.936 m/s. Tipler 15.P.043

Waves of frequency 200 Hz and amplitude 1.2 cm move along a

20 m string that has a mass of 0.060 kg and a tension of 50 N.

(a) What is the average total energy of the waves on the string?

(b) Find the power transmitted past a given point on the string. Solution:

(a) The average energy on a segment of length L is the average power Pav multiplied by the time the wave takes to travel the distance L:

!

("E)av

= Pav"t =

1

2µ# 2

vA2"t =

1

2µ# 2

A2"x = 0.5 $

0.06

20$ (2% $ 200)2 $ (1.2 $10&2)2 $ 20 = 6.82J

(b) The wave speed is:

!

v =FT

µ=129.10m /s

Pav

=1

2µ" 2

vA2 = 44.04W

Tipler 16.P.031

A transverse wave of frequency 32 Hz propagates down a string.

Two points 8 cm apart are out of phase by π/6.

(a) What is the wavelength of the wave?

(b) At a given point, what is the phase difference between two

displacements for times 5 ms apart?

(c) What is the wave velocity? Solution: (a) The phase difference is related to the displacement between the 2 points by the relationship:

!

" = 2#$x

%& % = 2#

$x

"= 2#

8 '10(2

# /6= 0.96m

(b)

!

v = "f #$ = 2%v&t

"= 2%f&t = 2% ' 32 ' 5 '10(3 =1.005rad

(c) the wave velocity is:

!

v = "f = 30.7m /s Tipler 16.P.027

Two compact sources of sound oscillate in phase with a

frequency of 120 Hz where the speed of sound is 340 m/s. At a

point that is 3.20 meters from one source and 3.54 meters from

the other, the amplitude of the sound from each source

separately is A. (a) What is the phase difference in the sound waves from the two sources at that point?

(b) What is the amplitude of the resultant wave at that point? Solution:

(a)

!

" = 2#$x

%= 2#f

$x

v= 2# &120 &

(3.54 ' 3.20)

340= 0.75rad

(b) the amplitude of the resultant wave at that point is: E1 = A sin ωt E2 = A sin (ωt + δ) EP = E1+ E2 = A[sin ωt + sin (ωt + δ)]

!

EP

= 2Acos"

2sin #t +

"

2

$

% &

'

( )

where we used

!

sin" + sin# = 2cos1

2(" $#)sin

1

2(" + #)

Hence,

!

EP

= 2Acos"

2sin #t +

"

2

$

% &

'

( ) * A

P= 2Acos

"

2

$

% & '

( ) = 2Acos(0.75 /2) =1.86A

Tipler 16.P.050

The normal range of hearing is about 20 to 20000 Hz. Assume

that your hearing range is 30 to 20000 Hz. (Use 340 m/s as the

speed of sound.)

(a) What is the greatest length of an organ pipe that would have

its fundamental note in your hearing range if it is closed at one

end?

(b) What is the greatest length of an organ pipe that would have

its fundamental note in your hearing range if it is open at both

ends? Solution:

(a)If the pipe is closed at one end there is a pressure node near

the opening at x = 0 and a pressure antinode at the closed end x

= L, the length of the pipe. Hence the amplitude 2Asinkx is

maximum for:

!

kL = n"

2# $

n=4

nL,n =1,3,5...

So the maximum length can be obtained for the maximum

wavelength

!

"max

=v

fmin

=340

30=11.3m for the fundamental note

corresponding to n=1:

!

Lmax

="max

4=11.3

4= 2.83m

(b) If the organ pipe is open at both ends, there are 2 pressure

nodes at the 2 extreme ends (min amplitude):

!

kL = n" # $n

=2

nL,n =1,2,...

Lmax

=$max

2=11.3

2= 5.65m

for n=1.

Tipler 16.P.069

Working for a small gold mining company, you stumble across an

abandoned mine shaft that, because of decaying wood shoring,

looks too dangerous to explore in person. To measure its depth,

you employ an audio oscillator of variable frequency. You

determine that successive resonances are produced at

frequencies of 63.58 and 89.25 Hz. Estimate the depth of the

shaft. Solution:

We’ll model the shaft as a pipe of length L with one end open.

We can relate the frequencies of the harmonics to their

wavelengths and the speed of sound by v = fnλn and the depth of

the mine shaft to the resonance wavelengths using the standing

wave condition for a pipe with one end open:

!

kL = n"

2# $n =

4

nL,n =1,3,5,...# fn =

v

$n=n

4Lv

Knowing that they are successive resonances we can make the

ratio of the two corresponding frequencies (take care that since n

= 1,3,5,…two successive resonances correspond to n and n+2):

!

fn

fn+2

=n

n + 2" n =

2 fn

fn+2 # fn= 5.07 $ 5

And so knowing that the sped of sound in air is 343 m/s:

!

fn = f5 = 63.58Hz =5v

4L" L =

5(343)

4 f5= 6.74m

Tipler 33.P.040

Two small loudspeakers are separated by a distance of d = 5.50

cm, as shown in the figure below. The speakers are driven in

phase with a sine wave signal of frequency 16 kHz. A small

microphone is placed a distance L = 2.00 m away from the

speakers on the axis running through the middle of the two

speakers, and the microphone is then moved perpendicular to

the axis. Where does the microphone record the first minimum

and the first maximum of the interference pattern from the

speakers? The speed of sound in air is 343 m/s. Solution:

Constructive interference condition:

!

d sin" = m#,m = 0,1,2,...

Destructive interference condition:

!

d sin" = (m +1

2)#,m = 0,1,2,..

We cannot use the small angles approximation:

!

"max

= sin#1 $

d% y

max= L tan"

max= 84.6cm

"mim = sin#1 $

2d% y

min= L tan"

min= 39.7cm

Tipler 33.P.022

Light of wavelength 418 nm is incident normally on a film of

water 1.0 µm thick. The index of refraction of water is 1.33.

(a) What is the wavelength of the light in the water?

(b) How many wavelengths are contained in the distance 2t,

where t is the thickness of the film? (Do not round your answer

to a whole number.)

(c) What is the phase difference between the wave reflected from

the top of the air-water interface and the one reflected from the

bottom of the water-air interface in the region where the two

reflected waves superpose? Solution:

(a)

!

"water

="

n= 314.3nm

(b)

!

N =2t

"water

= 6.4

(c) Light is going from a medium of smaller index of refraction

(air n = 1) to a medium with higher one (water n = 1.33) so the

reflected wave is out of phase by π respect to the transmitted

wave:

!

" = #$ + 2$2t

%water

= #$ + 2$ & 6.4 =11.8$

and subtracting 10π, we get 1.8π = 5.65 rad

Tipler 33.P.028

A drop of oil (n = 1.22) floats on water (n = 1.33). When

reflected light is observed from above as shown in the figure

below, what is the thickness of the drop at the point where the

second red fringe, counting from the edge of the drop, is

observed? Assume red light has a wavelength of 660 nm. Solution:

There is a change of phase of π for the reflected wave going from

air to oil and another phase change of π for the refracted wave

that cancel and so the condition for constructive interference is:

!

" = 2#2t

$water

= 2#m% 2t = m$

noil

% t =2 & 660

2 &1.22= 536nm

Tipler 33.P.041

Light of wavelength 510 nm is incident on a long, narrow slit.

Find the angle of the first diffraction minimum for each of the

following widths of the slit.

(a) 1 mm

(b) 0.1 mm

(c) 0.01 mm

Solution: Diffraction minima condition:

!

asin" = m#,m =1,2,...

" = asin#

a

$

% & '

( )

(a) 5.1e-4 rad, (b) 5.1e-3 rad, (c)=5.1e-2

Tipler 33.P.060

The ceiling of your lecture hall is probably covered with acoustic

tile, which has small holes separated by about 4 mm.

(a) Using light with a wavelength of 540 nm, how far could you

be from this tile and still resolve these holes? The diameter of the

pupil of your eye is about 5 mm.

(b) Could you resolve these holes better with violet light or with

green light? Solution:

(a) For Rayleigh criterium and supposing holes are circular

apertures:

!

"c

=1.22#

D

$

% &

'

( ) * tan"c

=x

L+ L =

xD

1.22#=4 ,10-3 , 5 ,10-3

1.22 , 540 ,10-9= 30.4m

(b) Since αc is directly proportional to the wavelength the

resolving power can be increased by decreasing the wavelength

so it is better for violet light than for green because smaller

angles can be resolved.

Tipler 33.P.069

Sodium light of wavelength 589 nm falls normally on a 4 cm

square diffraction grating ruled with 3700 lines per centimeter.

The Fraunhofer diffraction pattern is projected onto a screen at

1.8 m by a lens of focal length 1.8 m placed immediately in front

of the grating.

(a) Find the positions of the first two intensity maxima on one

side of the central maximum.

(b) Find the width of the central maximum.

(c) Find the resolution in the first order. Solution:

(a) Interference maxima:

!

d sin" = m#,m = 0,1,2,...

!

dym

L= m" # y

1="L

d=589 $10

%9$1.8

2.7 $10%6

= 0.393m,y2

= 2"L

d= 0.785m

where the slit separation is

!

d =1/3700 = 2.7 "10#4cm

(b) The distance from the first slit to the last is (N-1)d≈Nd. The

diffraction minimum is located at Ndsinθmin=mλ and so

!

"min

= sin"min

=#

Nd$%y =

2L#

Nd= 53µm

(c) The resolving power is R = mN hence the resolution of the

first order with m=1 is R= 14800 since the total number of lines

is N=3700*4=14800.

Tipler 33.P.080

A radio telescope is situated at the edge of a lake. The telescope

is looking at light from a radio galaxy that is just rising over the

horizon. If the height of the antenna is 16 m above the surface of

the lake, at what angle above the horizon will the radio galaxy be

when the telescope is centered in the first intensity interference

maximum of the radio waves? Assume the wavelength of the

radio waves is 34 cm. Hint: The interference is caused by the

light reflecting off the lake, and remember that this reflection will

result in a 180° phase shift. Solution:

The radio waves from the galaxy reach the telescope coming

directly from the galaxy and from the surface of the lake

(reflected and shifted in phase by 180 deg). We use the condition

for constructive interference of 2 waves to find the angle above

the horizon at which the radio waves from the galaxy interfere

constructively.

!

"r = m +1

2

#

$ %

&

' ( ),m = 0,1,2,...

!

sin" #$r

d%" = sin

&1 $r

d

'

( )

*

+ , = sin

&1

m +1

2

'

( )

*

+ , -

d

.

/

0 0 0 0

1

2

3 3 3 3

For m = 0: θ=0.609degrees

Problems for practice: Tipler 15.P.039 A harmonic wave on a string with a mass per unit length of 0.050 kg/m and a tension of 85 N has an amplitude of 5.0 cm. Each section of the string moves with simple harmonic motion at a frequency of 14 Hz. Find the power propagated along the string.

Solution: The sinusoidal wave has the form:

!

y(x, t) = Asin(kx "#t) = 5.0 $10"2sin(2.2x " 88.0t) in m and with x in m

and t in s, where

!

" = 2#f = 87.96rad/s

v =FT

µ= 41.23m /s = f$ % $ =

v

f= 2.94m% k =

2#

$= 2.14m&1

Hence, the average power propagated along the string by a harmonic wave is

!

Pav

=1

2µ" 2

vA2 = 0.5 # 0.05 # 88.02 # 41.2 # (5 #10$2)2 =19.91W

Tipler 15.P.033 (a) Compute the derivative of the speed of a wave on a string with respect to the tension dv/dFT and show that the differential dv and dFT obey dv/v = ½dFT/FT. (b) A wave moves with a speed of 260 m/s on a string that is under a tension of 495 N. Using the differential approximation, estimate how much the tension must be changed to increase the speed to 314 m/s. (c) Calculate ΔFT exactly and compare it to the differential approximation result in Part (b). Assume that the string does not stretch with the increase in tension. Solution:

a)

!

v =FT

µ" dv =

1

2 µFT

#1/ 2dF

T"

dv

v=

1

2v µFT

dFT

=1

2FT

dFT"

dv

dFT

=v

2FT

b)

!

"FT

=2"vFT

v=(311# 280) $ 2 $ 540

280=119.6N

c)

!

FT ,1 = v1

2µ

FT ,2 = v2

2µ

µ =FT ,1

v12

"FT

= FT ,2 # FT ,1 = µ(v2

2# v1

2) =

FT ,1

v12(v2

2# v1

2) =126.2N

5.2% higher than the result in a) Tipler 15.P.053

In the oceans, whales communicate by sound transmission

through the water. A whale emits a sound of 43 Hz to tell a

wayward calf to catch up to the pod. The speed of sound in water

is about 1500 m/s. (a) How long does it take the sound to reach the calf if he is 1.27 km away?

(b) What is the wavelength of this sound in the water? (c) If the whales are close to the surface, some of the sound energy might refract out into the air. What would be the frequency and wavelength of the sound in the air? Solution:

(a) The time the sound waves take to reach the calf over the distance from the whale to the calf is:

!

"t =d

v=1.27 #10

3

1500= 0.85s

(b)

!

"water =vwater

f=1500

43= 3.49m /s

(c) The frequency does not change from one medium (water) to another and the sound velocity in air is

!

"air =vair

f=343

43= 7m /s

Tipler 16.P.051

The wave function y(x, t) for a certain standing wave on a string

fixed at both ends is given by

y(x, t) = 3.6 sin (0.21x) cos (290t),

where y and x are in centimeters and t is in seconds.

(a) What is the wavelength of this wave?

What is the frequency of this wave?

(b) What is the speed of transverse waves on this string?

(c) If the string is vibrating in its fourth harmonic, how long is it? Solution:

(a) The general expression for a standing wave is

y(x,t)=Asinkxcosωt,

Hence:

!

" =2#

k= 29.9cm

!

f ="

2#= 46.2Hz

(b) the speed of transverse waves on this string is:

!

v ="

k=

290

0.21#102

=13.8m /s

(c) the string is fixed at both ends, hence

!

L = n"n

2# L = 4

29.9

2= 59.8cm

Tipler 16.P.020

It is said that a powerful opera singer can hit a high note with

sufficient intensity to shatter a wine glass by causing the air in an

empty wine glass to resonate at the frequency of her voice.

Estimate the frequency necessary to obtain a standing wave in

an 8.0 cm high glass. (The 8.0 cm does not include the height of

the stem.)

Approximately how many octaves above middle C(262 Hz) is

this? Hint: To go up one octave means to double the frequency. Solution:

We picture the wine glass as a cylinder open at one end: there is a node at the bottom of the glass and an anti-node at the other

end. The fundamental frequency is

!

f0

=v

"0

and since there is an

antinode (amplitude is maximum) in x = L:

!

kL = (2n +1)"

2# $

0= 4L

for n=0 and f0=343/4/8.0x10-2=1071.88 Hz Since 262 x 2 = 524 Hz, and 262 x 4 = 1024 Hz, then 1071.88 Hz is approximately 2 octaves above 262 Hz.

Tipler 33.P.046

A two-slit Fraunhofer interference-diffraction pattern is observed

with light of wavelength 640 nm. The slits have widths of 0.01

mm and are separated by 0.18 mm. How many bright fringes will

be seen in the central diffraction maximum? Solution: Diffraction minima:

!

asin" = m#,m =1,2,...

Interference maxima:

!

d sin" = m#,m = 0,1,2,...

Let’s equate the sin of this angle to the sine of the angle of the

mth interference maximum:

!

"

a=m"

d# m = d /a

If m = d/a (d distance between sources and a slit size) the mth

interference maximum will fall at the first diffraction minimum

and because the mth fringe cannot be seen there will be m-1

fringes on each side of the central maximum, hence N = 2(m-

1)+1 (one is in the center) is the number of interference maxima

in the diffraction central fringe. The number of bright interference

maxima will be N = 2m-1:

!

N = 2d

"#1= 35