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∑ Fx = F − m1 g (sin 30.0°) = 0
(1) ∑ Fy = F − m2 g = 0 (2) (2) 896 WAVES AND SOUND where F is the tension in the wire. In Equation (1) we have taken the direction of the +x
axis for block 1 to be parallel to and up the incline. In Equation (2) we have taken the
direction of the +y axis to be upward for block 2. This set of equations consists of two
equations in three unknowns, m1, m2, and F. Thus, a third equation is needed in order to
solve for any of the unknowns. A useful third equation can be obtained by solving Equation
16.2 for F: F = ( m / L ) v2 (3) Combining Equation (3) with Equations (1) and (2) leads to (m / L)v 2 − m1g sin 30.0° = 0 (4) (m / L)v 2 − m2 g = 0 (5) Equations (4) and (5) can be solved directly for the masses m1 and m2.
SOLUTION Substituting values into Equation (4), we obtain
m1 = ( m / L )v 2
(0.0250 kg/m)(75.0 m/s) 2
=
= 28.7 kg
g sin 30.0°
(9.80 m/s 2 ) sin 30.0° Similarly, substituting values into Equation (5), we obtain
(m / L )v 2 (0.0250 kg/m)(75.0 m/s)2
= 14.3 kg
=
g
(9.80 m/s 2 )
______________________________________________________________________________
m2 = 106. REASONING A particle of the string is moving in simple harmonic motion. The maximum
speed of the particle is given by Equation 10.8 as vmax = Aω, where A is the amplitude of the
wave and ω is the angular frequency. The angular frequency is related to the frequency f by
Equation 10.6, ω = 2π f, so the maximum speed can be written as vmax = 2π f A. The speed v
of a wave on a string is related to the frequency f and wavelength λ by Equation 16.1,
v = f λ. The ratio of the maximum particle speed to the speed of the wave is vmax 2π f A 2π A
=
=
v
fλ
λ
The equation can be used to find the wavelength of the wave. Chapter 16 Problems 897 SOLUTION Solving the equation above for the wavelength, we have 2 π ( 4.5 cm )
2π A
=
= 9.1 cm
3.1 vmax v ______________________________________________________________________________ λ= 107. REASONING AND SOLUTION
a. According to Equation 16.2, the speed of the wave is
v= F
315 N
=
= 2.20 × 10 2 m/s
−3
m/ L
6.50 × 10 kg/m b. According to Equations 10.6 and 10.8, the maximum speed of the point on the wire is ( ) v max = (2 π f )A = 2 π (585 Hz ) 2.50 × 10 –3 m = 9.19 m/s
______________________________________________________________________________
108. REASONING We must first find the intensities that correspond to the given sound
intensity levels (in decibels). The total intensity is the sum of the two intensities. Once the
total intensity is known, Equation 16.10 can be used to find the total sound intensity level in
decibels.
SOLUTION Since, according to Equation 16.10, β = (10 dB ) log (I / I 0 ) , where I0 is the ( ) reference intensity corresponding to the threshold of hearing I 0 = 1.00 ×10−12 W/m 2 , it
follows that I = I 0 10 β /(10 dB ) . Therefore, if β1 = 75.0 dB and β 2 = 72.0 dB at the point in
question, the corresponding intensities are
I1 = I 0 10 β1 / (10 dB ) I 2 = I 0 10 β 2 / (10 dB ) = (1.00 × 10 –12 W/m 2 ) 10 (75.0 dB) / (10 dB ) = 3.16 × 10−5 W/m 2 (72.0 dB) / (10 dB ) = 1.58 × 10 −5 W/m 2 = (1.00 × 10 –12 W/m 2 ) 10 Therefore, the total intensity Itotal at the point in question is ( )( ) I total = I1 + I 2 = 3.16 ×10−5 W/m 2 + 1.58 ×10−5 W/m 2 = 4.74 × 10−5 W/m 2
and the corresponding intensity level βtotal is 898 WAVES AND SOUND I total 4.74 ×10−5 W/m2 = (10 dB) log = 76.8 dB
−12 W/m 2 1.00 ×10 I0 ______________________________________________________________________________ β total = (10 dB ) log 109. REASONING AND SOLUTION The sound emitted by the plane at A reaches the person
after a time t. This time, t, required for the sound wave at A to reach the person is the same
as the time required for the plane to fly from A to B.
The figure at the right shows the relevant
geometry. During the time t, the plane travels the
distance x, while the sound wave travels the
distance d. The sound wave travels with constant
speed. The plane has a speed of v0 at A and a
speed v at B and travels with constant
acceleration. Thus, x
A B d d = vsound t
and
x= 1
2 θ
Person ( v + v0 ) t From the drawing, we have x
sin θ = =
d 1
2 ( v + v0 ) t =
vsound t v + v0
2vsound Solving for v gives
v = 2vsound sinθ – v0 = 2 ( 343 m/s ) sin 36.0° –164 m/s = 239 m/s ______________________________________________________________________________
110. REASONING AND SOLUTION The speed of sound in fresh water is
vfresh = Bad ρ = 2.20 ×109 Pa
= 1.48 × 103 m/s
3
3
1.00 × 10 kg/m The speed of sound in salt water is vsalt = Bad ρ The ratio of these two speeds is = 2.37×109 Pa
= 1.52×103 m/s
3
1025 kg/m Chapter 16 Problems vsalt
vfresh = 899 1.52×103 m/s
= 1.03
1.48×103 m/s Thus, the sonar unit erroneously calculates that the objects are 1.03 times the distance
measured in salt water: d actual = (1.03)(10.0 m) = 10.3 m
______________________________________________________________________________ CHAPTER 17 THE PRINCIPLE OF LINEA...
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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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