Unformatted text preview: θ = λ
D = v
343 m/s
=
= 0.404
f D ( 607 Hz )( 2 × 0.700 m ) θ = sin −1 ( 0.404 ) = 23.8° 16. REASONING The diffraction angle θ depends upon the wavelength λ of the sound wave
and the width D of the doorway according to sin θ = λ (Equation 17.1). We will determine
D
the wavelength from the speed v and frequency f of the sound wave via v = f λ
(Equation 16.1). We note that, in part (a), the frequency of the sound wave is given in
kilohertz (kHz), so we will use the equivalence 1 kHz = 103 Hz. SOLUTION a. Solving sin θ = λ
D (Equation 17.1) for θ, we obtain 914 THE PRINCIPLE OF LINEAR SUPERPOSITION AND INTERFERENCE PHENOMENA λ D θ = sin −1 Solving v = f λ (Equation 16.1) for λ yields λ =
(1), we find that (1) v
. Substituting this result into Equation
f λ
−1 v = sin D Df θ = sin −1 (2) Therefore, when the frequency of the sound wave is 5.0 kHz = 5.0×103 Hz, the diffraction
angle is
343 m/s
v −1 o = sin ( = 5.1
3
Df ) ( 5.0 × 10 Hz ) 0.77 m θ = sin −1 b. When the frequency of the sound wave is 5.0×102 Hz, the diffraction angle is
343 m/s
v −1 o = sin ( = 63
2
Df 0.77 m ) ( 5.0 × 10 Hz ) θ = sin −1 17. REASONING AND SOLUTION At 0 °C the speed of sound is air is given as 331 m/s in
Table 16.1 in the text. This corresponds to a wavelength of λ1 = v/f = (331 m/s)/(3.00 × 103 Hz) = 0.1103 m
The diffraction angle is given by Equation 17.2 as 1.22 λ –1 1.22 ( 0.1103 m ) = 50.3° = sin 0.175 m
D θ1 = sin –1 For an ideal gas, the speed of sound is proportional to the square root of the Kelvin
temperature, according to Equation 16.5. Therefore, the speed of sound at 29 °C is b v = 331 m / s K
g 302 K = 348 m / s
273 The wavelength at this temperature is λ2 = (348 m/s)/(3.00 × 103 Hz) = 0.116 m. This gives
a diffraction angle of θ2 = 54.0°. The change in the diffraction angle is thus
∆θ = 54.0° − 50.3° = 3.7° Chapter 17 Problems 18. REASONING AND SOLUTION
The figure at the right shows the
geometry of the situation. Diffraction horn The tone will not be heard at seats located 8.7 m
at the first diffraction minimum. This
occurs when sin θ = λ
D = 915 v
fD Stage θ θ
First row
of seats C x That is, the angle θ is given by v
343 m/s
−1 = 27.2° = sin 4 f D (1.0 ×10 Hz)(0.075 m) θ = sin −1 From the figure at the right, we see that tan 27.2° = x
8.7 m ⇒ x = (8.7 m)(tan 27.2°) = 4.47 m 8.7 m θ x
Thus, seats at which the tone cannot be heard are a distance x on either side of the center
seat C. Thus, the distance between the two seats is 2x = 2(4.47 m) = 8.9 m 19. SSM REASONING The beat frequency of two sound waves is the difference between
the two sound frequencies. From the graphs, we see that the period of the wave in the upper
text figure is 0.020 s, so its frequency is f1 = 1/ T1 = 1/(0.020 s) = 5.0 × 101 Hz . The
frequency of the wave in the lower figure is f 2 = 1/(0.024 s)=4.2 ×101 Hz .
SOLUTION The beat frequency of the two sound waves is f beat = f1 − f 2 = 5.0 × 101 Hz – 4.2 × 101 Hz = 8 Hz 916 THE PRINCIPLE OF LINEAR SUPERPOSITION AND INTERFERENCE PHENOMENA 20. REASONING The time between successive beats is the period of the beat frequency. The
period that corresponds to any frequency is the reciprocal of that frequency. The beat
frequency itself is the magnitude of the difference between the two sound frequencies
produced by the pianos. Each of the piano frequencies can be determined as the speed of
sound divided by the wavelength.
SOLUTION According to Equation 10.5, the period T (the time between successive beats)
that corresponds to the beat frequency fbeat is T = 1/ f beat . The beat frequency is the
magnitude of the difference between the two sound frequencies fA and fB, so that f beat = f A − f B . With this substitution, the expression for the period becomes 1 T= f beat = 1
fA − fB (1) The frequencies fA and fB are given by Equation 16.1 as
fA = v λA and fB = v λB Substituting these expressions into Equation (1) gives T= 1
1
1
=
=
= 0.25 s
343 m/s 343 m/s
fA − fB
v
v
−
−
0.769 m 0.776 m
λ
λ
A B 21. REASONING When two frequencies are sounded simultaneously, the beat frequency
produced is the difference between the two. Thus, knowing the beat frequency between the
tuning fork and one flute tone tells us only the difference between the known frequency and
the tuningfork frequency. It does not tell us whether the tuningfork frequency is greater or
smaller than the known frequency. However, two different beat frequencies and two flute
frequencies are given. Consideration of both beat frequencies will enable us to find the
tuningfork frequency.
SOLUTION The fact that a 1Hz beat frequency is heard when the tuning fork is sounded
along with the 262Hz tone implies that the tuningfork frequency is either 263 Hz or
261 Hz. We can eliminate one of these values by considering the fact that a 3Hz beat
frequency is heard when the tuning fork is sounded along with the 266Hz tone. This
implies that the tuningfork frequency is either 269 Hz...
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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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