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Unformatted text preview: . The
2
2
wavelength is, according to Equation 16.1, is equal to the speed v of sound divided by the
frequency f ; λ = v/f .
SOLUTION Substituting λ = v/f into d = 1 λ gives
2 v 343 m/s d = 1λ = 1 = 1
= 0.700 m
2
2f
2 245 Hz 3. SSM REASONING AND SOLUTION The shape of the string looks like t=1s t=2s t=3s t=4s 0 2 4 6 8 10 904 THE PRINCIPLE OF LINEAR SUPERPOSITION AND INTERFERENCE PHENOMENA 4. REASONING According to the principle of linear superposition, the resultant displacement
due to two waves is the sum of the displacements due to each wave. In order to find the net
displacement at the stated time and positions, then, we will calculate the individual
displacements y1 and y2 and then find their sum. We note that the phase angles are measured
in radians rather than degrees, so calculators must be set to the radian mode in order to yield
valid results.
SOLUTION
a. At t = 4.00 s and x = 2.16 m, the net displacement y of the string is
y = y1 + y2
= ( 24.0 mm ) sin ( 9.00π rad/s )( 4.00 s ) − (1.25π rad/m )( 2.16 m ) + ( 35.0 mm ) sin ( 2.88π rad/s )( 4.00 s ) + ( 0.400π rad/m )( 2.16 m ) = +13.3 mm b. The time is still t = 4.00 s, but the position is now x = 2.56 m. Therefore, the net
displacement y is
y = y1 + y2
= ( 24.0 mm ) sin ( 9.00π rad/s )( 4.00 s ) − (1.25π rad/m )( 2.56 m ) + ( 35.0 mm ) sin ( 2.88π rad/s )( 4.00 s ) + ( 0.400π rad/m )( 2.56 m ) = +48.8 mm 5. SSM REASONING The tones from the two speakers will produce destructive
interference with the smallest frequency when the path length difference at C is onehalf of a
wavelength. From Figure 17.7, we see that the path length difference is ∆s = sAC – sBC .
From Example 1, we know that sAC = 4.00 m , and from Figure 17.7, sBC = 2.40 m . Therefore, the path length difference is ∆s = 4.00 m – 2.40 m = 1.60 m . SOLUTION Thus, destructive interference will occur when λ
2 = 1.60 m or λ = 3.20 m This corresponds to a frequency of
f= v λ = 343 m / s
= 107 Hz
3.20 m Chapter 17 Problems 6. 905 REASONING In Drawing 1 the two speakers are equidistant from the observer O. Since
each wave travels the same distance in reaching the observer, the difference in
traveldistances is zero, and constructive interference will occur for any frequency. Different
frequencies will correspond to different wavelengths, but the path difference will always be
zero. Condensations will always meet condensations and rarefactions will always meet
rarefactions at the observation point. Since any frequency is acceptable in Drawing 1, our
solution will focus on Drawing 2.
In Drawing 2, destructive interference occurs only when the difference in travel distances
for the two waves is an odd integer number n of halfwavelengths. Only certain frequencies,
therefore, will be consistent with this requirement.
SOLUTION The frequency f and wavelength λ are related by λ = v/f (Equation 16.1),
where v is the speed of the sound. Using this equation together with the requirement for
destructive interference in drawing 2, we have λ
v
2
L2 244
L
14 + L − 3 = n 2 = n 2 f
4
Difference in travel where n = 1, 3, 5, ... distances Here we have used the Pythagorean theorem to determine the length of the diagonal of the
square. Solving for the frequency f gives f= 2 ( nv ) 2 −1 L The problem asks for the minimum frequency, so we choose n = 1 and obtain f= 2 ( v ) 2 −1 L = 2 ( 343 m/s ) 2 − 1 ( 0.75 m ) = 550 Hz 906 THE PRINCIPLE OF LINEAR SUPERPOSITION AND INTERFERENCE PHENOMENA 7. SSM WWW REASONING The geometry of the positions of the loudspeakers and the
listener is shown in the following drawing.
C d 1 = 1.00 m d2
y
60.0° A B x1 x2 The listener at C will hear either a loud sound or no sound, depending upon whether the
interference occurring at C is constructive or destructive. If the listener hears no sound,
destructive interference occurs, so
nλ
(1)
d 2 − d1 =
n = 1, 3, 5, K
2
SOLUTION Since v = λ f , according to Equation 16.1, the wavelength of the tone is λ= v 343 m/s
=
= 5.00 m
f 68.6 Hz Speaker B will be closest to Speaker A when n = 1 in Equation (1) above, so d2 = nλ
5.00 m
+ d1 =
+ 1.00 m = 3.50 m
2
2 From the figure above we have that,
x1 = (1.00 m) cos 60.0° = 0.500 m
y = (1.00 m) sin 60.0° = 0.866 m
Then
2
2
x2 + y 2 = d 2 = (3.50 m)2 or x2 = (3.50 m)2 − (0.866 m)2 = 3.39 m Therefore, the closest that speaker A can be to speaker B so that the listener hears no sound
is x1 + x2 = 0.500 m + 3.39 m = 3.89 m . Chapter 17 Problems 8. 907 REASONING The two speakers are vibrating exactly out of phase. This means that the
conditions for constructive and destructive interference are opposite of those that apply
when the speakers vibrate in phase, as they do in Example 1 in the text. Thus, for two wave
sources vibrating exactly out of phase, a difference in path lengths that is zero or an integer
number (1, 2, 3, …) of wavelengths leads to destructive interference; a difference in path
lengths that is a halfinteger number ( 1
1
,1 ,
2
2 1
2 ) 2 , ... of wavelengths leads to con...
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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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