16. Sound and Light Waves

16.18 Since sound wave amplitude increases with pis-ton displacement, low-frequency speakers needbigger displacements for comparable power out-put. Mechanical design requires a bigger systemto produce the greater displacements.

16.19 The wave corresponding to the fundamentalfrequency is sketched hi each tube.

open organ pipe

close one end

In the open pipe case:

Ai =2L

closed pipe case:

! = 41

So, since the wavelength is inversely propor-tional to frequency, it follows that the frequencyis cut in half when the pipe is closed.

0 p* = 6.3 mg/m3; v0 = 1.7 mm/s; s0 = 53 nm\

16.21/At 0° C the speed of sound is 330 m / s. Speedof the jet craft,

f i km '1000 m

1«jet - (0.82) (330 m/s)

= 970^h

16.22 L = 1.3 m; 196 Hz; 327 Hz; 458 Hz

16.23 a) Wavelength is AI = 2L.

open organ pipe

Frequency:

Thus L is:

VIA

1L

2/i 2(32.7 Hz)— 5,2 in

b) Wavelength:

closed at one end

AI =4L

Frequency:

/i = -2- = ~ .\ L = —~AI 4L 4/1

This is half the answer of part a): L — 2.6 m

16.24 /i = 3.1 kHz

16.25 a) Pressure amplitude P* is given by:

p* _ _ o f

^SQ = P*[povs2irfrl

10.2 Pa_(1.3 kg / m3) (330 m / s) 27r (103 Hz)

= 3.8 x 10~6 m

b)

so = (3.8 x 10-6 m) (1""OHz) = 76 x 10~6

50 Hzm

16.27 The chamber must have a displacement nodeat each end.

Vacuum chamber cylinder

F mdamental

" L = 3.5 m "

In general the wavelength of a standing wave is:

An = 2L \

X _ 2LAS - -3-

\ — LA4 — - -

3r

fn

Frequency is related to wavelength and thespeed of sound vs by:

\ f e vs _nvsAn/n - V. -> /„ - — - —

if vs = 340 m/s, then the fundamental fre-quency is

340 m / s2(3.5 m) -

The next harmonic has A = f £ and

/3 = 5/0 = 5 (65.4 Hz) = 327 Hz

The third harmonic has frequency:

. /4 = 7/0 = 7(65.4 Hz) =45 8 Hz

16.24 The ear canal acts like a closed pipe. Thus the fundamental frequency is relatedto the length by

v' u» 340 m/s

damental frequency is about 3 kHz in this model.

*f Q_16Jj6^The pipe is "open at one end", so it is presumably closed at the other end. Thusin the fundamental mode, A = 4£ and /„ = vs/X, so

v, = \f = 4Lf = 4 (25 x 10~2 m) (340 Hz) = 340 m/s

Using the given formula:

So:

ThusT = (546°C) (1.0303 -1) = 16.5°C

But we only have 2 sig fig, so T = 17°C.

16.28 The refractive index is n — c/v, so

c 3.0 x 108 m/s2.4

The speed of light in diamond is 1.3 x 10s m/s.

= 1.25x 10s m/s

16.30 In 1 y light travels a distance:

d = ct = (3.00 x 10a m/s) (365.25 d) * 3 = 9.47 x 1015 m

So 1 light year equals 9.47 xlO15 m.

Tf- Cjl6.32/rhe radio communications between astronauts and mission control travelled at thespeecTof light In each communcation, signals had to travel twice the Earth/Moon distance.Thus

d 4 x l O « m _ 8 ^ _ 2 -4-2

C-23TIo5^-3S~2- 'SThe time delay experienced by the astronauts was 3 s.

1634 Let's assume that gaps and teeth have equal width. If the wheel rotates so that the

16. Sound and Light Waves 315

L6.40 /visible = 0.066 W/m2

16.41 Spherical loudspeaker

radius.

frequency,

R= 10.0 cm = .100 m

/ = 3.0x 102Hz

amplitude,

s0 = .050 mm = 5.0 x 10~5 m

SIL is given by:

I.

Where the reference intensity is

/ref = 10-12 W/m2

The difference in SIL at the speakerand at distance D(SILo) is:

SILD - SILR = I

-10dBlog10

10dBlog10

According to the inverse square law, intensity isinversely proportional to distance squared.

/ oc -s

Thus,

20dBlog10(-\

At the speaker, intensity is related to frequencyand displacement amplitude, so by Eqn. 16.16:

J = -p0vs(us0)2

where air density po and the speed of sound vs

are given typical values:

po =1.3kg

mvs = 330 m/s

SILD = 10 dB Iog10

+20dBlog10(^

SILD =

Now

us0R = (27r)(3.0 x 102 Hz)x(5.0 x 10~5 m)(.10 m)

= 9.43 xlO3 —s

So

SILD = 10dBlog10

2(iO-ia -2

= 53 dB

Yes, this is audible.

16.42 5.6 m"^^x16.43/The explosion releases 1.0 x 107 J in 1 second,

so the power of the explosion is:

of the energy is energy is converted tolund waves (the other half presumably is con-

verted to light and mechanical deformation).The sound waves have power:

Psound = ( .5)(1.0X10 7W)

' = 5.0 x 106 W

Assuming spherical wave fronts, the intensity is

,- Jsound4-Trr2

where r is the distance at which we wish to fineme intensity, r = no iu

I = 5.0 x 106 W47r(110 m)2

= 33 W/m2

116 Lea & Burke Physics: The Nature of Things

Sound intensity in decibels is the SIL:

SIL - 10dBl0glo(10_127

w/m2)

32.88

= 140 dB

L6.44 130 dB; 130 Pa; 2.5 x 10~4 m; 500 W

16.45 Signal with frequency / = 250 Hz.

area ATube of diameter

10 cm. = dfilled withair

Amplifier5.0 W of power

Intensity of the sound wave generated: Thepower incident on any cross sectional area ofthe tube, A, is P = 5.0 W. Cross-sectional areais

Thus intensity is-(I)'

5.0 WA f (.1 m)2

W= 636.6

WI = 640-^

Displacement amplitude and pressure amplitudeof wave: Intensity is related to pressure ampli-tude P, by:

P2,1 = •P.= ^2Ip0vs

16.46 = 1.5 x

16.47 SLL is given by:

= 10dBloglo — ,-'ref

where reference intensity

Intensity is proportional to inverse square of dis-tance:

I oc -=•r2

The difference in two sound intensity levels

A5IL -

- 10dBlog10 (A") _ 10dBloglo (£-V-'ref/ Vrrf

In terms of distance:

If r2 = 2r1; then

AS7L = 20 dBlog10 ( i ) = -6 dB

Using air density po = 1.2 kg / m3 and speed ofsound va = 340 m/s

P. = N/2(637W/m2)(1.2 kg/m3)(340 m/s)

= 720 Pa

Intensity is related to displacement amplitudeso by:

1 '^Wvr2/2^

16.48 20%

16.49 The speed of light is the same in all frames

c = 3 x 108 m / s

16.50 If light is blue-shifted, then the object is ap-proaching. If light is redshifted, then the objectis receding. Jupiter is spinning with respect tcEarth. The blueshifted side is spinning towardsEarth.

16.51

T = period

' vwT

637 W/m2

2(1.2 kg/m3)(340 m/s)= 1.1 mm

Above the threshold of pain.

If the man with whistle bicycles toward the friend at 10.0 m/s, then we have a sourcewith velocity 3 m/s (in the air frame) toward the observer, and the observer has a velocity-7 m/s in the air frame. Thus the Doppler shifted frequency is (equation 16.22a):

1 + 7/330/ = (2.00 kHz)

1 - 3/330

If we use the approximate formula 16.22b, we get

= (2.00 kHz) 1.0306 = 2.06 kHz

/ = (2.00 kHz) l -

which is the same result to 3 sig fig.

= (2.00 kHz) = 2. 06 kHz

f 16.50 Following the method of Example 16.9:

1 +

Thus:

(3.8T7)2 =

3.8T72 - 1= 0.87524

Thus:c 3.87T2 +1

v, = 0.87524 (2.998 x 10s m/s) = 2.624 x 10s m/s

16.58 U&ve leaving A at time t = 0 reaches 0 after a time Ai = r/uw = tt

leaving B at time t = T travels a shorter distance rj and arrives at time ta where:

Since s = veT -C r, we may expand the square root:

= ri/l - 2- cos 5 + f-)2 ~ r f l - -V r XT-/ \ r

cos d

Thus:

A S BThe time interval between the arrival times of waves from A and B is:

r f a \ r s veTtg—t^ = TH 11 cost)} = T cos9 = T cos9 ••vw \ r I vw vw vw

The observed frequency is (time between passage of wavefronts)"1, so

/ = - : - /0

v. • vu

equation

16.6p^Mfe divide the problem into two parts: outgoing sound and returning sound.

/.- "/ _^ JZ

&-?/

Outgoing: ^fe have bom a moving source (the torpedo) and a moving observer (thesubmarine), so we use equation 16.22. The speed of sound in water is 1531 m/s (Table16.1), much greater than the speed of either source or observer, so let's use 16.22b. Thus:

(16-32) m/sN / 16 \1531 m/s )~ \ ' 1531 /

Returning: The reflected pulse is Doppier shifted again. Both source and observer aremoving opposite the direction that the wave is moving. Now the submarine is the sourceand the torpedo is the observer. Thus:

Thus the frequency received by me torpedo is:

2 - 12-3kHz

16.62, 64, 66 See solutions manual

16.68 When light is incident at the critical angle, the angle of refraction is 90°. ThusSnelFs law becomes:

m this case, HI — 1.31 (ice) and n2 = 1.00 (air), so:

am*.. IS .ig.. 0.78338HI l.ol

Thus0e = sin-1 0.76336 = 0. 8685 rad = 49. 8°

From Example 16.13, 9a = 41° for glass. So glass is a better material for light pipes sincemore of the light is totally internally reflected.

16.70 At the first side, ni = 1 (air) and n-i = n (glass), so the angle of refraction <p isgiven by:

sin# = nsin<£ (1)At the far side of the slab, n\ = n and n2 = 1 . The angle of incidence is <b (see diagram)

son3in(p = sin0/ (2)

Comparing equations 1 and 2, we see that 9f = 9. Thus the light emerges at the sameangle that it enters. However it has been displaced, and emerges from the slab at point 3.The undisplaced ray intersects the side of the slab at point C. We want to find the distance

TcL

Jtft