Answer, Key Homework 4 David McIntyre 1

This print-out should have 10 questions, checkthat it is complete. Multiple-choice questionsmay continue on the next column or page:find all choices before making your selection.The due time is Central time.Chapter 18 problems.

001 (part 1 of 2) 5 pointsA standing wave is formed on a string thatis 27 m long, has a mass per unit length0.00641 kg/m, and is stretched to a tension of25 N.Find the fundamental frequency.

Correct answer: 1.1565 s1.Explanation:

Basic Concepts: standing waveSolution:

f1 =v

2L

=(62.4512 m/s)

(2) (27 m)

= 1.1565 s1 ,

where

v =

F

=

(25 N)

(0.00641 kg/m)

= 62.4512 m/s .

002 (part 2 of 2) 5 pointsFind the next frequency that could cause astanding wave pattern.Correct answer: 2.31301 s1.Explanation:

f2 = 2 f1

= 2.31301 s1 .

003 (part 1 of 1) 0 pointsIf the air in an organ pipe with one end closedis in resonance with a tuning fork so that theair column (including the two ends) containstwo nodes and two antinodes, the length ofthe pipe is approximately:

1. 2

2.3

4correct

3.

4.

4

5.

2

Explanation:

The neighboring node and antinode are

4apart. Since there are two nodes and twoantinodes, the length of the pipe is given by

4 3 = 3

4.

004 (part 1 of 2) 0 pointsGiven: The speed of sound in air is 344 m/s.If an organ pipe is to resonate at 28.8 Hz,

what is its required length if it is open at bothends?Correct answer: 5.97222 m.Explanation:

The wavelength of the standing sound waveis

=vsf

=344 m/s

28.8 Hz= 11.9444 m .

If the pipe is open at both ends, then itslength Lo must be half of the wavelength ofthe standing sound wave

Lo =

2

=(11.9444 m)

2= 5.97222 m .

005 (part 2 of 2) 0 pointsWhat is its required length if it is closed atone end?Correct answer: 2.98611 m.

Answer, Key Homework 4 David McIntyre 2

Explanation:

When the pipe is closed at one end, itslength Lc has to be one fourth of the wave-length

Lc =

4

=(11.9444 m)

4= 2.98611 m .

006 (part 1 of 1) 10 pointsA tuning fork vibrating at frequency f =868 Hz is held above a long cylindrical pipewith an open top end. The pipe is partiallyfilled with water up to a slowly rising heighth.

R

f

For some values of the water level h =h1, h2, h3, . . . the air column in the pipe (abovethe water) resonates to the tuning fork. Astudent performing this experiment noticesequal height differences between successiveresonances,

h2 h1 = h3 h2 = h4 h3 = = h.Given the speed of sound in air vs = 344 m/s,calculate the height difference h.Correct answer: 19.8157 cm.Explanation:

The resonance happens when the standingsound wave of wavelength

=vsf

has a pressure node at the open top of thepipe and a displacement node at the waterlevel which acts as a closed bottom of

the air column. The pressure nodes of astanding wave are displacement antinodes andvice verse, hence the length of the resonant aircolumn should be a half-integral multiple ofthe internode distance /2. Thus,

L =1

2

2,3

2

2,5

2

2, . . . .

Suppose the first resonance happens when

L = htop hwater1 = (n+ 12)2 ,

then the next resonance happens when thewater rises by /2 and the air column shortensby the same amount,

htop hwater1 = (n+ 12)2 ,htop hwater2 = (n 12)2 ,htop hwater3 = (n 32)2 ,. . . . . . . . . . . . . . . . . . . . . . . .

htop hwatern = (12)2 ,

and therefore

h2 h1 = h3 h2 = = hn hn1 = 2.

In other words,

h =

2=

vs2f

= 19.8157 cm.

007 (part 1 of 1) 0 pointsAn open pipe 0.55 m in length is placed ver-tically in a cylindrical bucket with a cross-sectional area of 0.1 m2. Water is poured intothe bucket until a tuning fork of frequency774 Hz, placed over the pipe, produces reso-nance.

Note: If more water is poured into thebucket after reaching this resonance, no far-ther resonance occur.Find the mass of water in the bucket at

this moment. The speed of sound in air is344 m/s.Correct answer: 43.8889 kg.Explanation:

Answer, Key Homework 4 David McIntyre 3

The wavelength of the standing sound waveis

=vsf

=344 m/s

774 Hz= 0.444444 m .

The length of the resonant air column is

d =

4

=0.444444 m

4= 0.111111 m ,

so the height of the water column is

h = L d= (0.55 m) (0.111111 m)= 0.438889 m .

The mass of the water in the bucket is equalto the density water times the volume V ofthe water

m = water V

= water Ah

= (1000 kg/m3) (0.1 m2) (0.438889 m)

= 43.8889 kg .

008 (part 1 of 2) 5 pointsAn organ pipe of length L is open at one endand closed at the other. A tuning fork isbrought near the pipe, exciting the air in thepipe with three nodes as represented in thefigure.

LWith the speed of the sound denoted by v,

what is the frequency of the tuning fork?

1.2

Lv

2.5

4Lv correct

3.1

4Lv

4.3

Lv

5.7

4Lv

6.1

2Lv

7.1

Lv

8.3

4Lv

9.5

2Lv

10.3

2Lv

Explanation:

We must have a node (stationary point)at the closed end of our air column, and ananti-node (maximum oscillation point) at theopen end of our air column. The distancebetween nodes is a half-wavelength, and thedistance between a node and an anti-node is aquarter-wavelength. Therefore the length ofthe air column must correspond to an integernumber of half-wavelengths plus one quarter-wavelength. This is equivalent to an odd inte-ger number of quarter wavelengths. Therefore

L = (2n 1) 4

(n = 1, 2, 3, . . .).

This is not a harmonic series since the higherfrequencies are not integral multiples of thefundamental frequency. We can express thisas the odd members of a harmonic series as

L = m

4(m = 1, 3, 5, . . .).

The fifth harmonic is the third lowest fre-quency of the organ pipe, which correspondsto n = 3 or m = 5. The frequency is justf = v/. Combining these equations andsolving for the frequency gives:

f =5

4Lv .

009 (part 2 of 2) 5 points

Answer, Key Homework 4 David McIntyre 4

A string is stretched to a length l and bothends are fixed. The density of the string is and its tension is F . A standing wave mode ofthe string with six nodes, including both endpoints (as pictured), is generated.

lWhat is the frequency of this oscillation?

1.5

2 l

F

correct

2.3

2 l

F

3.1

2 l

F

4.5

4 l

F

5.7

2 l

F

6.2

l

F

7.9

4 l

F

8.1

l

F

9.3

l

F

Explanation:

Both ends of the string are fixed (immo-bile), so any standing waves on the stringmust have nodes (stationary points) at bothends. This requires the wavelength be suchthat an integer number of half-wavelengths(the distance between nodes) completely fillsthe string, i.e.

n

2= l (n = 1, 2, 3, ...).

The resulting possible frequencies are:

fn =v

=n

2 lv

=n

2 l

F

(n = 1, 2, 3, ...).

In this case, we can clearly see that five half-wavelengths fill the string, therefore n = 5.

010 (part 1 of 1) 10 pointsIn certain ranges of a piano keyboard, morethan one string is tuned to the same note toprovide greater intensity. For example, thenote at 134 Hz has two strings at this pitch.If one string slips from its normal tension of

628 N to 546 N, what beat frequency will beheard when the two strings are struck simul-taneously?Correct answer: 9.0543 beats/s.Explanation:

The beat frequency f is equal to the dif-ference of the frequencies f1 and f2

f = f1 f2 .

To calculate the frequency f2, we use the factthat the fundamental frequency of vibrationsin a string is proportional to the velocity ofthe waves along the string, which in turn isproportional to the square root of the tensionT in the string

f v T ,

therefore

f2 =

T2T1

f1

=

546 N

628 N134 Hz

= 124.946 Hz .

So for the beat frequency we obtain

f = |f1 f2|= |(134 Hz) (124.946 Hz)|= 9.0543 beats/s .