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Unformatted text preview: nic must be n + 2.
Taking the ratio of these two relations gives Chapter 17 Problems 750 Hz
=
450 Hz 937 ( n + 2) v 4L = n + 2
n
v
n 4L Solving this equation for n gives n = 3 .
v
b. Solving the equation 450 Hz = n for L and using n = 3, we find that the length of 4L the tube is 343 m/s v
L = n = 3 = 0.57 m
4f 4 ( 450 Hz ) n 58. REASONING AND SOLUTION According to the principle of linear superposition, when
two or more waves are present simultaneously at the same place, the resultant wave is the
sum of the individual waves. Therefore, the shape of the string at the indicated times looks
like the following:
t=1s t=2s t=3s t=4s 0 2 4 6 8 10 12 cm 938 THE PRINCIPLE OF LINEAR SUPERPOSITION AND INTERFERENCE PHENOMENA 59. SSM REASONING According to Equation 17.3, the fundamental (n = 1) frequency of a
string fixed at both ends is related to the wave speed v by f1 = v / 2 L , where L is the length
of the string. Thus, the speed of the wave is v = 2Lf1. Combining this with Equation 16.2,
v = F /(m / L) , we have, after some rearranging,
F
= 4 f12 (m / L)
2
L Since the strings have the same tension and the same lengths between their fixed ends, we
have
2
2
f1E ( m / L) E = f1G (m / L )G
where the symbols “E” and “G” represent the E and G strings on the violin. This equation
can be solved for the linear density of the G string. SOLUTION The linear density of the string is
( m / L )G = 2
f1E
2
f1G f
(m / L) E = 1E
f 1G 659.3 Hz = 196.0 Hz 2 2 ( m / L) E (3.47 ×10–4 kg/m ) = 3.93 × 10 –3 kg/m 60. REASONING The frequencies fn of the standing waves allowed on a string fixed at both v
ends are given by Equation 17.3 as f n = n , where n is an integer that specifies the 2L harmonic number, v is the speed of the traveling waves that make up the standing waves,
and L is the length of the string. The speed v is related to the tension F in the string and the
F
(Equation 16.2). Therefore, the frequencies of the
linear density m/L via v =
m/ L
standing waves can be written as F n
F
v
fn = n = n m/ L = 2L 2L m / L 2L The tension F in each string is provided by the weight W (either WA or WB) that hangs from
the right end, so F = W. Thus, the expression for fn becomes f n =
this relation to find the weight WB. n
W
. We will use
2L m / L Chapter 17 Problems A B WA 939 WB SOLUTION String A has one loop so n = 1, and the frequency f1A of this standing wave is 1 WA
. String B has two loops so n = 2, and the frequency f 2B of this standing
2L m / L
2 WB
wave is f 2B =
. We are given that the two frequencies are equal, so
2L m / L
f1A = WB
1 WA
2
=
2 L 24L 2 L 24L
14 m /3 14 m /3
f 2B f1A Solving this expression for WB gives
WB = 1 WA =
4 1
4 ( 44 N ) = 11 N 61. REASONING The fact that a loud sound is heard implies constructive interference, which
occurs when the difference in path lengths is an integer number (1, 2, 3, K) of
wavelengths. This difference is 50.5 m − 26.0 m = 24.5 m. Therefore, constructive
interference occurs when 24.5 m = nλ, where n = 1, 2, 3, …. The wavelength is equal to the
speed v of sound divided by the frequency f; λ = v/f (Equation 16.1). Substituting this
relation for λ into 24.5 m = nλ, and solving for the frequency gives
f= nv
24.5 m This relation will allow us to find the two lowest frequencies that the listener perceives as
being loud due to constructive interference. SOLUTION The lowest frequency occurs when n = 1: f= (1) ( 343 m/s ) = 14 Hz
nv
=
24.5 m
24.5 m 940 THE PRINCIPLE OF LINEAR SUPERPOSITION AND INTERFERENCE PHENOMENA This frequency lies below 20 Hz, so it cannot be heard by the listener. For n = 2 and n = 3,
the frequencies are 28 and 42 Hz , which are the two lowest frequencies that the listener
perceives as being loud. 62. REASONING AND SOLUTION The speed of the speakers is
vs = 2π r/t = 2π (9.01 m)/(20.0 s) = 2.83 m/s
The sound that an observer hears coming from the right speaker is Doppler shifted to a new
frequency given by Equation 16.11 as
f OR = fs
1–vs /v = 100.0 Hz
= 100.83 Hz
1– ( 2.83 m/s ) / ( 343.00 m/s ) The sound that an observer hears coming from the left speaker is shifted to a new frequency
given by Equation 16.12 as
f OL = fs 1 + vs /v = 100.0 Hz
= 99.18 Hz
1 + ( 2.83 m/s ) / ( 343.00 m/s ) The beat frequency heard by the observer is then
100.83 Hz − 99.18 Hz = 1.7 Hz 63. SSM WWW REASONING The beat frequency produced when the piano and the other instrument sound the note (three octaves higher than middle C) is f beat = f − f 0 ,
where f is the frequency of the piano and f 0 is the frequency of the other instrument
( f 0 = 2093 Hz ). We can find f by considering the temperature effects and the mechanical
effects that occur when the temperature drops from 25.0 °C to 20.0 °C.
SOLUTION The fundamental frequency f0 of the wire at 25.0 °C is related to the tension
F0 in the wire by
f0 = v
=
2 L0 F0 /(m / L )
2 L0 where Equations 17.3 and 16.2 have been combined. (1) Chapter 17 Problems 941 The amount ∆L by which the piano wire attempts to contract is (see Equation 12.2)
∆L = α L0 ∆T , where α is the coefficient of linear expan...
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This note was uploaded on 04/30/2013 for the course PHYS 1100 and 2 taught by Professor Chastain during the Spring '13 term at LSU.
 Spring '13
 CHASTAIN
 Physics, The Lottery

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