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in the given mathematical description of the wave and Equation 16.4 will permit us to
determine the wavelength λ and frequency f.
SOLUTION The dimensionless term 2π f t in Equation 16.4 corresponds to the term 8.2π t
in the given wave equation. The time t is measured in seconds (s), so in order for the
quantity 8.2π t to be dimensionless, the units of the numerical factor 8.2 must be s−1 = Hz,
and we have that
2 π f t = ( 8.2 Hz ) π t or 2 f = 8.2 Hz or f= 8.2 Hz
= 4.1 Hz
2 Similarly, the dimensionless term 2π x/λ in Equation 16.4 corresponds to the term 0.54π x in
the mathematical description of this wave. Because x is measured in meters (m), the term
0.54π x is dimensionless if the numerical factor 0.54 has units of m−1. Thus, Chapter 16 Problems 2π x λ ( ) = 0.54 m −1 π x or 2 λ = 0.54 m −1 λ= or 847 2
= 3.7 m
0.54 m −1 Applying v = f λ (Equation 16.1), we obtain the speed of the wave:
v = λ f = ( 3.7 m ) ( 4.1 Hz ) = 15 m/s 25. SSM REASONING Since the wave is traveling in the +x direction, its form is given by
Equation 16.3 as
2π x y = A sin 2π ft − λ 2π
We are given that the amplitude is A = 0.35 m. However, we need to evaluate 2πf and
. λ Although the wavelength λ is not stated directly, it can be obtained from the values for the
speed v and the frequency f, since we know that v = f λ (Equation 16.1). SOLUTION Since the frequency is f = 14 Hz, we have 2πf = 2π(14 Hz) = 88 rad/s
It follows from Equation 16.1 that
2π λ = Using these values for 2πf and 2π f 2π (14 Hz )
=
= 17 m −1
v
5.2 m/s 2π λ in Equation 16.3, we have 2π x y = A sin 2π ft − λ ( ) y = ( 0.35 m ) sin ( 88 rad/s ) t − 17 m −1 x ______________________________________________________________________________
26. REASONING AND SOLUTION Referring to the drawing that accompanies the problem
statement, we find from the graph on the left that λ = 0.060 m – 0.020 m = 0.040 m and
Α = 0.010 m; from the graph on the right we find that Τ = 0.30 s – 0.10 s = 0.20 s. Then,
f = 1/(0.20 s) = 5.0 Hz. Substituting these into Equation 16.3 we get
2π x y = A sin 2π f t – λ and y = ( 0.010 m ) sin (10π t – 50π x ) 848 WAVES AND SOUND ______________________________________________________________________________
27. REASONING The mathematical form for the displacement of a wave traveling in the –x
2π x direction is given by Equation 16.4: y = A sin 2π f t +
. λ
SOLUTION Using Equation 16.1 and the fact that f = 1 / T , we obtain the following
numerical values for f and λ :
f= 1
1
=
= 1.3 Hz
T 0.77 s and λ= v 12 m/s
=
= 9.2 m
f 1.3 Hz Using the given value for the amplitude A and substituting the above values for f and λ into
Equation 16.4 gives 2π x 2π y = A sin 2π f t + = ( 0.37 m ) sin 2π (1.3 Hz ) t + x
λ 9.2 m ( ) −1
y = ( 0.37 m ) sin ( 8.2 rad/s ) t + 0.68 m x ______________________________________________________________________________ F
(Equation 16.2),
mL
where v is the speed of a wave traveling along the string and m/L is the string’s linear
density. The speed v of the wave depends upon the frequency f and wavelength λ of the
wave, according to v = f λ (Equation 16.1). We will determine the frequency and
wavelength of the wave by comparing the given equation for the displacement y of a particle
2π x from its equilibrium position to the general equation y = A sin 2π ft − λ (Equation 16.3). Once we have identified the frequency and wavelength, we will be able to
determine the wave speed v via Equation 16.1. We will then use Equation 16.2 to obtain the
tension F in the string. 28. REASONING The tension F in a string can be found from v = Comparing y = ( 0.021 m ) sin ( 25t − 2.0 x ) with the general equation
2π x y = A sin 2π ft − (Equation 16.3), we see that 2π ft = 25t , or 2π f = 25 , and that
λ 2π x
2π
= 2.0 x , or
= 2.0 . Therefore, we have that
λ
λ SOLUTION f= 25
Hz
2π and λ= 2π
m =π m
2.0 (1) Chapter 16 Problems Squaring both sides of v = 849 F
(Equation 16.2) and solving for F, we obtain
mL v2 = F
mL or F = v2 ( m L ) (2) Substituting v = f λ (Equation 16.1) into Equation (2) yields
F = v2 ( m L ) = ( f λ ) 2 (m L) (3) Substituting Equations (1) and the linear density into Equation (3), we find that
F = ( f λ) 29. 2 2 25 −2
( m L ) = 2 π Hz ( π m ) (1.6 × 10 kg/m ) = 2.5 N SSM REASONING The speed of a wave on the string is given by Equation 16.2 as F
, where F is the tension in the string and m/L is the mass per unit length (or
m/ L
linear density) of the string. The wavelength λ is the speed of the wave divided by its
frequency f (Equation 16.1).
v= SOLUTION
a. The speed of the wave on the string is v= F
15 N
=
= 4.2 m/s
( m / L ) 0.85 kg/m b. The wavelength is λ= v 4.2 m/s
=
= 0.35 m
f
12 Hz c. The amplitude of the wave is A = 3.6 cm = 3.6 × 10−2 m. Since the wave is moving along
the −x direction, the mathematical expression for the wave is given by Equation 16.4 as
2π x y = A sin 2π f t + λ 850 WAVES AND SOUND Su...
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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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