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SOLUTION Since the smallest value for L is a quarter of a wavelength, we have L = λ = 4L. According to Equation 16.1, the speed of sound is 1λ
4 or v = f λ = f ( 4 L ) = ( 485 Hz )( 4 × 0.264 m ) = 512 m/s 45. REASONING The fundamental frequency f1A of air column A, which is open at both
v
ends, is given by Equation 17.4 with n = 1: f1A = (1) , where v is the speed of sound 2L A
in air and LA is the length of the air column. Similarly, the fundamental frequency f1B of air
column B, which is open at only one end, can be expressed using Equation 17.5 with n = 1:
v
. These two relations will allow us to determine the length of air column B.
f1B = (1) 4L B SOLUTION Since the fundamental frequencies of the two air columns are the same
f1A = f1B , so that v
v = (1) 4L 2 LA 24 1 24 1
43
4B
3 (1) f1A or LB = 1 LA =
2 1
2 ( 0.70 m ) = 0.35 m f1B 46. REASONING For a tube open at only one end, the series of natural frequencies is given by
f n = nv / ( 4 L ) (Equation 17.5), where n has the values 1, 3, 5, etc., v is the speed of sound,
and L is the tube length. We will apply this expression to both the airfilled and the heliumfilled tube in order to determine the desired ratio.
SOLUTION According to Equation 17.5, we have f n, air = nvair
4L and f n, helium = nvhelium
4L Chapter 17 Problems 931 Dividing the expression for helium by the expression for air, we find that
nvhelium
f n, helium
v
1.00 ×103 m/s
= 4 L = helium =
= 2.92
nvair
f n, air
vair
343 m/s
4L 47. SSM REASONING AND SOLUTION
a. For a string fixed at both ends the fundamental frequency is f1 = v/(2L) so fn = nf1. f 2 = 800 Hz, f3 = 1200 Hz, f 4 = 1600 Hz b. For a pipe with both ends open the fundamental frequency is f1 = v/(2L) so fn = nf1.
f 2 = 800 Hz, f3 = 1200 Hz, f 4 = 1600 Hz c. For a pipe open at one end only the fundamental frequency is f1 = v/(4L) so fn = nf1 with
n odd.
f3 = 1200 Hz, f5 = 2000 Hz, f7 = 2800 Hz 48. REASONING According to Equation 16.1, the frequency f of the vibrations in the rod is
related to the wavelength λ and speed v of the sound waves by
f= v λ (16.1) The distance between adjacent antinodes of a longitudinal standing wave is equal to half a
wavelength 1 λ . This rod has antinodes at the ends only, so the length L of the rod is
2 () equal to half a wavelength. Put another way, the wavelength is twice the length of the rod: λ = 2L (1) The speed v of sound in the rod depends upon the values of Young’s modulus Y and the
density ρ of aluminum via Equation 16.7:
v= Y ρ SOLUTION Substituting Equation (1) into Equation 16.1 yields (16.7) 932 THE PRINCIPLE OF LINEAR SUPERPOSITION AND INTERFERENCE PHENOMENA f= v λ = v
2L (2) Substituting Equation 16.7 into Equation (2), we obtain f= v
1Y
1
6.9 ×1010 N/m 2
=
=
= 2100 Hz
2 L 2 L ρ 2 (1.2 m )
2700 kg/m3 49. REASONING We will make use of the series of natural frequencies (including the first
overtone frequency) given by f n = nv / ( 2 L ) (Equation 17.4), for a tube open at both ends.
In this expression, n takes on the integer values 1, 2, 3, etc. and has the value of n = 2 for the
first overtone frequency that is given. We can solve Equation 17.4 for L, but must deal with
the fact that no value is given for the speed v of sound in gas B. To obtain the necessary
value, we will use the fact that both gases are ideal gases and utilize the speed given for gas
A and the masses of the two types of molecules.
SOLUTION According to Equation 17.4 as applied to gas B, we have f n = nvB / ( 2 L ) ,
which can be solved for L to show that
nv
L= B
(1)
2 fn Since both gases are ideal gases, the speed is given by v = γ kT / m (Equation 16.5), where
γ is the ratio of the specific heat capacities at constant pressure and constant volume, k is
Boltzmann’s constant, T is the Kelvin temperature, and m is the mass of a molecule of the
gas. Noting that γ and T are the same for each gas, we can apply this expression to both
gases:
γ kT
γ kT
vA =
and vB =
mA
mB
Dividing the expression for gas B by that for gas A gives γ kT
vB
vA = mB γ kT = mA
mB or vB = vA mA
Substituting this result for vB into Equation (1), we find that mA
mB Chapter 17 Problems L= nvB nvA
=
2 fn 2 fn 933 mA 2 ( 259 m/s ) 7.31×10−26 kg
=
= 0.557 m
mB
2 ( 386 Hz ) 1.06 × 10−25 kg 50. SSM WWW REASONING The pressure P2 at a depth h in a static fluid such as the mercury column is given by P2 = Patm + ρ gh (Equation 11.4), where Patm = 1.01×105 Pa is the air pressure at the surface of the fluid, ρ is the density of the fluid, and g is the
magnitude of the acceleration due to gravity. Because the airfilled portion of the tube is
v
open at one end, it can have standing waves with natural frequencies given by f n = n 4L (Equation 17.5), where n can take on only odd integral values (n = 1, 3, 5, …), and v is the
speed of sound in air. The mercury decreases the effective length of the airfilled portion of
the tube from its initial length L0 = 0.75 m to its final length L. The third harmonic f3,0 of the
original tube is found by choosing n = 3 in Equation...
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 Spring '13
 CHASTAIN
 Physics, The Lottery

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