Unformatted text preview: bstituting in the numbers for A, f, and λ, we have 2π x 2π x −2
y = A sin 2π f t + = 3.6 × 10 m sin 2π (12 Hz ) t + λ
0.35 m ( ) ( ) = 3.6 × 10−2 m sin ( 75 rad/s ) t + (18 m −1 ) x ______________________________________________________________________________
30. REASONING Using either Equation 16.3 or 16.4 with x = 0 m, we obtain y = A sin (2 π f t )
Applying this equation with f = 175 Hz, we can find the times at which y = 0.10 m.
SOLUTION Using the given values for y and A, we obtain
0.10 m = ( 0.20 m ) sin (2π f t ) or 0.10 m = sin –1 ( 0.50 )
2 π f t = sin –1 0.20 m The smallest two angles for which the sine function is 0.50 are 30° and 150°. However, the
values for 2π ft must be expressed in radians. Thus, we find that the smallest two angles for
which the sine function is 0.50 are expressed in radians as follows: 2π 360° (30°) and 2π 360° (150° ) The times corresponding to these angles can be obtained as follows: 2π 2 π f t1 = (30° ) 360° t1 = 30°
f ( 360°) 2π 2 π f t 2 = (150° ) 360° t2 = 150°
f (360° ) Subtracting and using f = 175 Hz gives (150° – 30° )
150°
30°
–
=
= 1.9 × 10 –3 s
f (360°) f ( 360° ) (175 Hz )(360° )
______________________________________________________________________________
t2 – t1 = Chapter 16 Problems 31. 851 SSM REASONING AND SOLUTION The speed of sound in an ideal gas is given by
text Equation 16.5
γ kT
v=
m
where m is the mass of a single gas particle (atom or molecule). Solving for T gives T= mv 2
γk (1) The mass of a single helium atom is
4.003 g/mol 1 kg −27
kg = 6.650×10
6.022×1023 mol−1 1000 g The speed of sound in oxygen at 0 °C is 316 m/s (see Table 16.1). Since helium is a
monatomic gas, γ = 1.67. Then, substituting into Equation (1) gives ( 6.65 ×10−27 kg ) ( 316 m/s )2 = 28.8 K
T=
1.67 (1.38 × 10−23 J/K )
______________________________________________________________________________
32. REASONING The wavelength λ of a sound wave is equal to its speed v divided by its
frequency f (Equation 16.1):
v
(1)
λ=
f
The speed of sound in a gas is given by Equation 16.5 as v = γ kT / m , where T is the
Kelvin temperature and m is the mass of a single air molecule. The Kelvin temperature is
related to the Celsius temperature Tc by T = Tc + 273.15 (Equation 12.1), so the speed of
sound can be expressed as v= γ k (Tc + 273.15 )
m (2) We know that the speed of sound is 343 m/s at 20.0 °C, so that 343 m/s =
Dividing Equation (2) by (3) gives γ k ( 20.0 °C + 273.15)
m (3) 852 WAVES AND SOUND γ k (Tc + 273.15 ) v
=
343 m/s Tc + 273.15
m
=
20.0 °C + 273.15
γ k ( 20.0 °C + 273.15 )
m (4) SOLUTION Solving Equation (4) for v and substituting the result into Equation (1) gives
Tc + 273.15
( 343 m/s ) 35 °C + 273.15
v
20.0 °C + 273.15 =
20.0 °C + 273.15 = 3.9 ×10−3 m
λ= =
f
f
91×103 Hz
______________________________________________________________________________ ( 343 m/s ) 33. REASONING The speed v, frequency f, and wavelength λ of the sound are related
according to v = f λ (Equation 16.1). This expression can be solved for the wavelength in
terms of the speed and the frequency. The speed of sound in seawater is 1522 m/s, as given
in Table 16.1. While the frequency is not given directly, the period T is known and is
related to the frequency according to f = 1/T (Equation 10.5). SOLUTION Substituting the frequency from Equation 10.5 into Equation 16.1 gives 1
v = f λ = λ
T Solving this result for the wavelength yields ( ) λ = vT = (1522 m/s ) 71×10−3 s = 110 m
______________________________________________________________________________
34. REASONING A rail can be approximated as a long slender bar, so the speed of sound in
the rail is given by Equation 16.7. With this equation and the given data for Young’s
modulus and the density of steel, we can determine the speed of sound in the rail. Then, we
will be able to compare this speed to the speed of sound in air at 20 °C, which is 343 m/s.
SOLUTION The speed of sound in the rail is
v= Y ρ = 2.0 × 1011 N/m 2
7860 kg/m 3 = 5.0 × 10 3 m/s This speed is greater than the speed of sound in air at 20 °C by a factor of
v rail
v air 5.0 × 10 3 m/s
=
= 15
343 m/s Chapter 16 Problems 853 ______________________________________________________________________________
35. SSM REASONING If we treat the sample of argon atoms like an ideal monatomic gas
( γ = 1.67 ) at 298 K, Equation 14.6 ( 1
2
mvrms
2 3
2 ) = kT can be solved for the rootmeansquare speed v rms of the argon atoms. The speed of sound in argon can be found from
Equation 16.5: v = γ kT / m .
SOLUTION We first find the mass of an argon atom. Since the molecular mass of argon is
39.9 u, argon has a mass per mole of 39.9 × 10 –3 kg /mol. Thus, the mass of a single argon
atom is
39.9×10−3 kg/mol
m=
= 6.63×10−26 kg
23
−1
6.022×10 mol a. Solving Equation 14.6 for v rms and substituting the data given in the problem statement,
we find
vrms ( ) 3 1.38 × 10−23 J/K ( 298 K )...
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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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