Suggested+Solution++Test+FyBNVC08+Ch11 12+Waves Sound

8/3/2019 Suggested+Solution++Test+FyBNVC08+Ch11 12+Waves Sound

1/16

Suggested Solutions FyBNVC08 Ch11-12 Mechanical waves Sound NV-College

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In the multi-choice problems below just circle the correct answer (s) in the test paper.

1. If a violin string vibrates at Hz262 as its fundamental frequency, what are thefrequencies of first five harmonics? [1/0]

a. Hz , Hz , Hzf 2621 = f 5242 = f 7863 = , Hzf 04814 = , Hzf 31015 = b. Hz , Hz , Hzf 2621 = f 7863 = f 31015 = , Hzf 83417 = , Hz f 35829 =c. Hz , Hz , Hzf 2621 = f 5243 = f 7865 = , Hzf 04817 = , Hzf 31019 = d. Hz , Hz , Hzf 2621 = f 7862 = f 31013 = , Hzf 83414 = , Hzf 35825 =e. None above, they are:

Answer: Alternative ______

Why? Show a sample of your calculations. [1/0]Suggested solution: Answer: Alternative a

Hzf 2620 = ; ...,5,4,3,2,12620 === nHznfnfn

Hzf 26212621 == First harmonic or Fundamental frequency

Hzf 52422622 == Second harmonic or First overtone frequency

Hzf 78632623 == Third harmonic or second overtone frequency

Hzf 048142624 == Fourth harmonic or third overtone frequency

Hzf 310152625 == Fifth harmonic or fourth overtone frequency

2. An organ pipe is cm7.32 long. What are the fundamental frequency and first fourovertones of the pipe at C20 if it is closed at one end? At C20 , the speed of sound in

air is sm /343 . [1/0]

a. Hz , Hz , Hzf 2621 = f 5242 = f 7863 = , Hzf 04814 = , Hzf 31015 = b.

Hz , Hz , Hzf 2621

=f 7863

=f 31015

=, Hzf 83417

=, Hz f 35829

=

c. Hz , Hz , Hzf 5241 = f 57413 = f 62225 = , Hzf 67137 = , Hzf 72049 =d. Hz , Hz , Hzf 5241 = f 04912 = f 57413 = , Hzf 09824 = , Hz622 f 25 =e. None above, they are:


Why? Show a sample of your calculations. [1/0]

Suggested solution: Answer: Alternative b


2/16



mL 327.0= Hzm

sm

L

vf 262

327.04

/343

40 =

== ;

( ) ( ) ...,5,4,3,2,1,01226212 012 =+=+=+ nHznfnf n

Hzf 26212621 == First harmonic or Fundamental frequency.

Hzf 78632623 == Third harmonic or second overtone frequency.

Hzf 310152625 == Fifth harmonic or third overtone frequency.

Hzf 834172627 == Seventh harmonic or fourth overtone frequency.

Hzf 358292629 == Ninth harmonic or fifth overtone frequency.

3. If the air temperature in a room is increased, the fundamental frequency of the organ pipes

a) will not be effected.b) will be decreased.c) will be increased.d) will be equal to the frequency of the second harmonics.


Why? Show a sample of your calculations. [1/0]

Suggested answer: Answer: Alternative c:According to smTv /60.0331 += , velocity of sound is an increasing

linear function of the temperature, and frequency of the wave is alsoproportional to the velocity of it in the medium. Since the change inthe length of the pipe is minimal in the temperature range of interest,the fundamental wavelength of the sound is independent of thetemperature. Therefore, the frequency of the sound increases (higherpitch) as a function of the temperature:

HzL

Tvf

+==

4

60.0331

0

0

Note that, the incremental increase in the length of the pipe due to itsthermal expansion is negligible in the room temperature range.


3/16



4. What resonant frequency would you expect from blowing an empty soda bottle that iscm0. deep? What would the resonant frequency be if it was two-third filled? [1/0]18

a. Hz476 , and Hz715 respectively.b. Hz476 , and Hz4301 respectively.c. Hz953 , and Hz4301 respectively.d. Hz953 , and Hz8582 respectively.e. None above, they are:


Why? Show your calculations in the space below. [1/1]

Suggested solution: Alternative b: , and 1 Hz476 Hz430

respectively.

mL 180.0= . One end closed pipe: Hzm

sm

L

vf 476

180.04

/343

40 =

== ;

Since two-third of the deep bottle is filled, one-third of it is empty.

Therefore:

cm0.18

mm

L 060.03

180.0== Hz

m

sm

L

vf 4301

060.04

/343

40 =

==

5. The velocity of waves on a string is sm /0. . If the frequency of the standing wavesis Hz475 , how far apart are two adjacent nodes? [2/0]

92

a. m968. 0b. m68.9 c. cm68. ,9d. m194. 0e. None above its ___________


Suggested solution: Answer: Alternative c cm68.9

cmmf

vvf 4.19194.0

475

0.92=====

The nodes of the standing wave are cmmf

v68.90968.0

4752

0.92

22==

==

apart.


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6. The mass of a harmonic oscillator of period s0.1 is tripled, i.e. 12 3mm = . The periodof the new harmonic oscillator is [2/0]

a) s3

1

b) s3 c) s3 .d) s

3

1

e) s9 Answer: Alternative : _______________

Suggested solution: Answer: Alternative c:

k

mT = 2 ,

1

2

1

2

m

m

T

T= , s

T

T3

1

2 = , sTnew 3=

7. In an earthquake, it is noted that a footbridge oscillates at fundamental frequency ofperiod s5.2 . What other possible resonant periods of motion are there for this bridge?

[1/0]

a) ...,3,2 ,15.2 = nsnb) ...,3,2,15.2

1

= nsn

c) ...,3,2,12

5=ns

n

d) ...,3,2,15

2=ns

n.

e) None above. It is ..Answer: Alternative: _______________

Why? Show the detail of your calculations in the space provided below: [0/1]

Suggested solution: Answer: Alternative c i.e.: ...,3,2,12

5=

= ns

nnT

Tf

1= ; ...,3,2,10 == nfnfn ; Hzf

5.2

10 = ; ...,3,2,1

5.2

1== nHznfn

...,3,2,12

5

10

255.2

5.2

1

11=

=

==

== nHzn

Hzn

Hzn

Hz

nf

Tn

n


5/16



8. Of the following procedures, which one (s) will higher the pitch of a guitar string? [1/0]a) Decrease string tension.

b) Increase string tension.c) Shorten the string.d) Lengthen the string.e) Use a heavier string.f) Use a lighter string.

Answer: Alternative: _______________


Suggested solution: Answer: Alternative b, c, and f: usingLm

Fv T

/= , L21 = ,

andL

vvf

vf

200 ===

.

If the velocity of the propagation of the wave in the stringLm

Fv T

/= is increased the

frequency

vf = will be increased. This may be achieved, either by increasing the

tension TF (alternative b) or by using a lighter string (alternative f). The higher pitch

(frequency) also may be achieved by shortening the wire which in turn increases the

wavelength of the sound: Lm

F

LL

vv

fT

/2

1

200

=== .

9. Calculate the beat frequency if E above the middle C frequency at Hz330 and F abovethe middle C frequency at Hz349 are played simultaneously. Will the beat be

audible? What if each played two octaves lower? Will the beat be audible? Note that a

frequency will be lowered by a factor of two if it is played at one octave lower.

a. The beat frequencies are Hz679 , audible, and Hz170 audible, respectively.b. The beat frequencies are Hz679 , not audible, and Hz340 not audible,

respectively.

c. The beat frequencies are Hz19 , audible, and Hz75. audible, respectively.4d. The beat frequencies are Hz19 , not audible, and Hz75. audible, respectively.4

Answer: Alternative: _________


Suggested solution: Alternative d: Respectively, beat frequencies

are and .HzfB 19= HzfB 75.4= HzfB 19= is not audible (may not be

heard) while is audible (may be heard.)HzfB 75.4=

HzfB 19330349 == No the beat will not be heard it is more than .

At two octaves below, the beat frequency:

Hz10

HzfB 75.44

19

4

330

4

349

===


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10. The siren of a police car at rest emits a sound wave at Hz6001 . What frequency willyou hear, if you are at rest and the police car

is approaching you at hkm /108 .

a. I will hear the siren at Hz4711 .b.

I will hear the siren at Hz7531 .

c. I will hear the siren at Hz7401 .d. I will hear the siren at Hz4601 .e. I will hear the siren at Hz460 .2f. I will hear the siren at Hz1851 .g. None above. I will hear the siren at .

Answer: Alternative: _________


Suggested solution: Alternative b

smv /3036001000108 ==

Hz

v

v

ff

s

7531

343

301

6001

1

=

=

=

11. Each figure below illustrates a transmitted wave, T , and a reflected wave, R .Determine which medium is heavy (ttare) section and which medium is light (tunnare)

section. Note that the vertical dashed line just indicates the change of medium. Your

answer must include the direction of motion of the pulse, and the reason for your

conclusion in sufficient detail.i. [1/1]Answer:The pulse moves from right toleft. The left section is heavierthan the right section. Thereason is the fact that thereflected pulse is inverted whilethe transmitted pulse is effected only in its amplitude. The heavier thesecond section, the less is transmitted pulse. This is according to

Newtons third law. The pulse reaching the boundary exerts a forceupward on the heavier section, in respond the heavier section exertsan equal but opposite force on the lighter cord which generates theinverted reflected pulse which is going to move to the right.

T R

sectionleft sectionright



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ii. [1/1]Answer:The pulse moves from left toright. The left section is heavierthan the right one. The reason is

the fact both reflected andtransmitted pulses are uprightindicating the fact that thereflected wave is not inverted. Therefore the left section where thepulse is generated initially must be heavier than the right section.

iii. [1/1]Answer:The pulse moves from right toleft. The right section is heavier

than the left one. The reason isthe fact both reflected andtransmitted pulses are similarindicating the fact that the reflected wave is not inverted. Thereforethe left section where the pulse is generated initially must be heavierthan the right one.

iv. [1/1]Answer:The pulse moves from left to

right. The right section is heavierthan the left one. The reason isthe fact that the reflected pulseis inverted while the transmittedpulse is altered only in itsamplitude. The heavier the second section, the less is the amplitudeof the transmitted pulse. This is according to Newtons third law. Thepulse reaching the boundary exerts a force upward on the heaviersection, in respond the heavier section exerts an equal but oppositeforce on the lighter cord which generates the inverted reflected pulsewhich is going to move back to the left.

TR


TR


T R



8/16


12. The highest key on a grand piano corresponds to a frequency about 150 times that of thelowest key.

a.If the string for the highest frequency is cm0.4 , how long would the length of thestring corresponding to the lowest frequency have to be if it had identical linear

density and was under the same tension? Even for grand piano strings are not longer

than m3 . [1/1]

b.What should be done in order to reduce the length of the string corresponding to thelowest frequency to m0.3 . Explain clearly your solution. [0/2]

Suggested solution: Data: mcmLhigh 040.00.4 == ;L

vvf

2==

mLLf

fL

L

L

L

v

L

v

f

f

L

vf

L

vf

lowhigh

low

high

low

high

low

low

high

low

high

lowest

lowest

highest

highest

0.6040.0150

2

2

2

2====

/

/

/

/

=

=

=

The problem is usually solved using a heavier string (higher Lm ) which

will reduce theLm

Fv T

/= and therefore, reduce the length requirement. If

the tension on the strings are identical, using a wire whose linear massdensity is four times larger than that of the wire used for the highfrequency note, the required length for the lowest frequency note will dropto . Problem may be solved also by using different tensions. The

velocity of the propagation of wave in a wire of lower tension is lower and

therefore the fundament frequency of the wire would be lower:

m0.3

L

vf

2= .

Second method:If the wires and the tension in them are identical, velocity of propagationof the sound wave in the wires are therefore identical, i.e.: 21 vv = .

UsingL

vvf

2==

Lffv 2==

1

221221121 22f

fLLfLfLvv ===

mLf

fLL low 0.615004.0

1

221 ===

To reduce the unreasonably long length of the wire from m0.6 to the

required m0.3 , we must reduce the velocity of propagation of the wave in

the wire by a factor of two. Considering the fact thatLm

Fv T

/= there are

few methods:

Use a wire whose linear density,l

m, is four times higher. i.e. use a

wire four times heavier, but use identical tension. Use identical wire, but reduce the tension in the wire by a factor of

four. Use a wire whose linear density is twice times higher.

Simultaneously reduce the tension in the wire by a factor of two.



9/16



13. Show that the amplitude A of circular water waves decreases as the square root of thedistance r from the source

rA

1 . Ignore damping. [0/2/]

Suggested solution:

The intensity of a circular wave is r

P

r

timeEnergy

Area

PowerI 22

/

=== .

The energy of a simple harmonic oscillation as well as any (sinusoidal)mechanical wave is proportional to the square of the amplitude of the

oscillation- wave: 2

2

1kAE= . Therefore, the intensity must be proportional

to the square of the amplitude of the oscillation- wave: 2AI

Therefore:

=

2

2

AI

r

PI

r

PA

2

2 r

A12

rA

1

Answer: The amplitude of the water wave is inversely proportional to the

square root of the distance:r

A1


10/16


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14. A balsa wood block of mass gmBlock

0.90= floats on a lake, bobbing up and down at a

frequency of HzfBlock

0.3=

a What is the value of the effective spring constant of water? [1/1]b A partially filled water bottle of mass gmbottle 160= and almost the same size and

shape of the balsa block is tossed into the water. Calculate the period at which the bottle may bob up and down? Assume simple harmonic oscillation. [1/1]

Suggested solution:Data: gmBlock 0.90= , HzfBlock 0.3= , kggmbottle 16.0160 ==

Problem: ?=k , ?=Bottlef

a Answer: The effective spring constant of water is mN/2 .k 24.3 =Using

k

mT 2=

m

k

Tf

2

11== 22

2

2

44

fTm

k==

224 fmk =

( ) ( ) mNk /24.30.3109042232

==

mNk /24.32

= b Answer: The partially filled water of mass kgmbottle 16.0= may oscillate

at the rate of HzfBottle 25.2= . The period of its oscillation is

s44.0 .sTBottle5.4

2=

k

mT 2= sssTBottle 44.0

5.4

2

5.4

2

24.3

16.02

2==

=

HzHzT

fBottle

Bottle 25.22

5.41===

Second method: HzHzfBottle 25.22

5.4

16.0

24.3

2

1 2=

/

/=

=

15. Niklas and Oscar designed an experiment to measure the velocity of sound in air. Theystayed at a considerable distance from a high building. The building was well isolated

and no other building is closed by. Niklas clapped and Oscar measured the time. Niklas

arranged the clapping time such that he clapped as he heard the echo of his previous clap

from the building. Oscar counted the number of Niklas clap. In average Niklas clapped

9 claps in s0.5 . Oscar measured the distance from the point Niklas clapped to the

building. It was m97 . Determine the value that Niklas and Oscar obtained for the

velocity of the sound in air. [0/2]

Suggested solution:

Data:9

0.5 st= , mx 972=

smsms

m

s

m

t

xv /350/349

5

9729

9

0.5

972=

=

==

Answer: The velocity of sound in air according to Niklas-Oscar is.smv /350


11/16



16. At a rock concert, a decibel-meter registered dB135 when placed m0.2 in front of aloudspeaker on the stage.

a) Show that the power output of the speaker, assuming uniform spherical spreading ofthe sound and neglecting absorption in the air is about kw6.1 . [0/2]

b) How far away would the intensity level be a somewhat reasonable dB90 ? [0/2]Suggested solution:

Data: Sound wave dB135= , mr 0.2= ,

Problem: ?=P , ?=r to the intensity dB90=

a) Answer: The power output of the speaker, assuming uniform spherical spreading of thesound and neglecting absorption in the air was: wP 2109.7 .

dB135= dBI

I135log10

0

=

= 5.13log

0

=

I

I 5.13

0

10=I

I

25.1125.13

0

5.13 /10101010 mwII === [0/1]

2/mwA

PI= wIrwIAP == 24

( ) wwwP 35.12 106.15.5891100.24 == kwP 6.1 [0/1]

using:

=

0

log10I

I , ax =log ax 10= , 270 /10 mwI

= , 2/mwA

PI= and

the surface area of a sphere of radius r:24 rA =

b)

Answer: mr 360 away from the speaker the intensity level is a somewhat reasonabledB90 .

dBI

I90log10

0

=

= 9log

0

=

I

I 9

0

10=I

I

23129

0

9 /10101010 mwII === [0/1]

2/mw

A

PI= 2m

I

PA = 22

4m

I

Pr

=

m

I

Pr

=

4

mw

r

=3

104

5.5891

mmr 3606.355 = mr 360 [0/1]


12/16



17. An ambulance carrying a seriously injured patient moves down a highway at a constantvelocity of hkm /126 . Its siren emits a sound at Hz.400 . Determine the frequency

heard by a car driver moving in the opposite direction at constant velocity of

hkm /0 .90

a. as the car approaches the ambulance. [0/2]b. as the car moves away from the ambulance. [0/2]

Suggested solution:

Data: Hzf 400= , smhkmvv sambulance /0.356003

1000126/126 ==== ,

smhkmvv sambulance /0.356003

1000126/126 ==== , and

velocity of sound in air usingsmv /343=

smTv /60.0331+=

and=

20T :

a. As the car approaches the ambulance the observed frequency of thesiren is: HzHzf

vv

vvf

s

o 47840035343

25343'

+=

+= [0/2]

b. As the car moves away from the ambulance the observed frequencyof the siren is: HzHzf

vv

vvf

s

o 337 [0/2]40035343

25343'

+

=

+

=


13/16


18. Two pulses shown in the figure below are moving toward each other. Their speed areidentical

a Sketch the shape of the string at the moment directly overlap. [0/2]b Sketch the shape of the string a few moments later. [2/0]

-6

-4

-2

0

2

4

6

8

0 5 10 15 20 25 30 35 40 45 50 55 60

t [s]

y[cm]

Suggested solution:a. Superposition of the pulses at the moment they overlap.

-6

-4

-2

0

2

4

6

8

0 5 10 15 20 25 30 35 40 45 50 55 60

t [s]

y[cm]

[0/2]

b. The configuration of the pulses after passing each other.

-6

-4

-2

0

2

4

6

8

0 5 10 15 20 25 30 35 40 45 50 55 60

t [s]

y[cm]

[2/0]



14/16


19. One end of a horizontal light string is attached to a small-amplitude vibrator. The stringpasses over a frictionless pulley a distance mL away, and a g100 is hung from the

other end. The figure below is printed in the scale 10:1 . The vibrator is connected to a

tone-generator as illustrated in the figure. You may assume the string at the generator is

a node, which is nearly true. According to the manufacture of the string the mass ofm0.50 of the string is g75.1 .

a. Determine the speed of the wave in sing. [0/2]b. Calculate the frequency of the tone generator. [0/2/]c. The length of the string may be adjusted by moving the pulley. If the hanging mass

is fixed at g100 , and the frequency of the vibrator is fixed at Hz500 , how many

different standing wave pattern may be achieved by varying L between cm10 and

cm150 ? [0/2/]

g100vibrator

oscillatortonefrequency

Suggested solution:

Data: mkgmgm

g

L

m/105.3/035.0

0.50

75.1 5=== , 4=L , kggM 1.0100 == ,

measurement from the figure: mcmL 3.11301013 === .

a. smmkg

N

Lm

Fv T /167

/105.3

8.91.0

/ 5

==

[0/2]

b. HzLL

vf 607

1.1

16741674

4

167=

=

===

[0/2]

c. mf

v334.0

500

167=== ; mmL 167.0

2

334.0

2===

half-standing-wave

mmnmnnL 50.1167.050.12

334.0

2==

, 98.8

167.0

50.1 n

m

mn ;

Therefore, eight standing wave pattern may be achieved by varyingbetween and : They are

L

cm10 cm150

wavesstanding4,

2

73,,

2

52,,

2

31,,

2

1.



15/16


20. A particular organ pipe can resonate at Hz735 , Hz0291 , and Hz3231 , but not anyother frequencies between Hz735 and Hz3231 . [0/6/]

a) Is this open or closed organ. Why? Show details of your calculation and explainyour reasoning.

b) What is the fundament frequency of this pipe?Suggested solution: Data: (n-1)th overtone Hzfn 7351 = , nth overtone Hzfn 0291= , and

(n+1)th overtone Hzfn 32311 =+ ,

Both-ends-open pipe mth overtone1mffm = ( ) 1111 1 ffmfmff mm =+=+

Therefore if the pipe is an open tube it is expected that mth overtone is aninteger factor of the fundamental frequency: 1mffm =

Hzfn 7351 = , nth overtone Hzfn 0291= , and (n+1)

th overtone

Hzfn 32311 =+ ,

Hzff nn 294029132311 ==+

Hzff nn 29473502911 ==

As demonstrated above if the pipe is both-end-open it is required that:

...,3,2,1n294

2942941

1111=

==

====+ n

HzfnfHzfHzfffff

n

nnnn



16/16


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Therefore, if Hzfnfn n2941 == the pipe is both-end-open

But integer5.2294

735

294

1 ==Hzfn ; integer5.3

294

0291

294==

Hzfn and

integer5.4

294

3231

294

1 ==+Hzfn

Therefore, the pipe can not be a both-ends-open pipe. [0/2]

One-end-closed pipe (2m+1)th overtone ( ) 12121

12112 +=+=

++ m

f

ffmf mm

( ) ( ) 1111212 21212 ffmfmff mm =+= +

Therefore, if the pipe is a one-end-closedtube, as demonstrated above, the differencebetween two consequent overtones is

11212 2fff mm = + . Therefore, it is expected that

the fundamental frequency of the pipe to be

11212 229402913231 fHzff mm === + , which implies that

HzHzf 1472

2941 == [0/2]

Checking the requirement ( ) integer12121

12112 =+=+=

++ m

f

ffmf mm

5147

735

147

12 ==nf

; 7147

0291

147

12 ==+nf

and 9147

3231

147

32 ==+nf

.

Therefore, the pipe is a one end closed tube with the fundamental

frequency Hzf 1471 = , and three consecutive overtones Hzf 7355 = ,Hzf 02917 = , and Hzf 32319 = . [0/2/]

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