Unformatted text preview: ergy.
When we use this procedure to analyze the total kinetic energy in one wavelength
of the string, we obtain the same result:
K dK 1
4 2A2 The total energy in one wavelength of the wave is the sum of the potential and kinetic energies:
E U K 1
2 2A2 (16.20) As the wave moves along the string, this amount of energy passes by a given point
on the string during one period of the oscillation. Thus, the power, or rate of energy transfer, associated with the wave is
1
2 E
t Power of a wave
1
2 2A2 T
2 A2v 1
2 2A2 T
(16.21) This shows that the rate of energy transfer by a sinusoidal wave on a string is proportional to (a) the wave speed, (b) the square of the frequency, and (c) the
square of the amplitude. In fact: the rate of energy transfer in any sinusoidal
wave is proportional to the square of the angular frequency and to the
square of the amplitude. 509 16.9 The Linear Wave Equation EXAMPLE 16.5 Power Supplied to a Vibrating String A taut string for which
5.00 10 2 kg/m is under a tension of 80.0 N. How much power must be supplied to the
string to generate sinusoidal waves at a frequency of 60.0 Hz
and an amplitude of 6.00 cm? Solution oidal waves on the string has the value
2f The wave speed on the string is, from Equation √√ 80.0 N
10 2 kg/m T Because f 5.00 1
2A2v
2
1
2 (5.00 (6.00 40.0 m/s 60.0 Hz, the angular frequency 377 s 1 Using these values in Equation 16.21 for the power, with
A 6.00 10 2 m, we obtain 16.4,
v 2 (60.0 Hz) 2 10
10 kg/m)(377 s 1)2
2 m)2(40.0 m/s) 512 W of the sinus Optional Section 16.9 THE LINEAR WAVE EQUATION In Section 16.3 we introduced the concept of the wave function to represent waves
traveling on a string. All wave functions y (x, t ) represent solutions of an equation
called the linear wave equation. This equation gives a complete description of the
wave motion, and from it one can derive an expression for the wave speed. Furthermore, the linear wave equation is basic to many forms of wave motion. In this
section, we derive this equation as applied to waves on strings.
Suppose a traveling wave is propagating along a string that is under a tension
T. Let us consider one small string segment of length x (Fig. 16.22). The ends of
the segment make small angles A and B with the x axis. The net force acting on
the segment in the vertical direction is
Fy T sin T sin B T(sin A sin B T(tan B tan A) However, the tangents of the angles at A and B are deﬁned as the slopes of the string
segment at these points. Because the slope of a curve is given by y/ x, we have
Fy y
x T y
x B (16.22)
A We now apply Newton’s second law to the segment, with the mass of the segment given by m
x:
2y
F y ma y
x
(16.23)
t2
Combining Equation 16.22 with Equation 16.23, we obtain
x 2y t2
2y T t2 T y
x y
x B ( y/ x)B A ( y/ x)A
x ∆x
θA A) Because the angles are small, we can use the smallangle approximation sin
tan to express the net force as
Fy T (16.24) θB
B A T Figure 16.22 A segment of a
string under tension T. The slopes
at points A and B are given by
tan A and tan B , respectively. 510 CHAPTER 16 Wave Motion The right side of this equation can be expressed in a different form if we note that
the partial derivative of any function is deﬁned as
f
x f (x lim x)
x x:0 f (x) If we associate f (x
x) with ( y/ x)B and f (x) with ( y/ x)A , we see that, in the
limit x : 0, Equation 16.24 becomes
2y T 2y t2 Linear wave equation x2 (16.25) This is the linear wave equation as it applies to waves on a string.
We now show that the sinusoidal wave function (Eq. 16.11) represents a solution of the linear wave equation. If we take the sinusoidal wave function to be of
the form y(x, t ) A sin(kx
t ), then the appropriate derivatives are
2y 2A sin(kx t) k 2A sin(kx t) t2
2y x2 Substituting these expressions into Equation 16.25, we obtain
2 T sin(kx k 2 sin(kx t) t) This equation must be true for all values of the variables x and t in order for the
sinusoidal wave function to be a solution of the wave equation. Both sides of the
equation depend on x and t through the same function sin(kx
t ). Because this
function divides out, we do indeed have an identity, provided that
2 k2
Using the relationship v T /k (Eq. 16.13) in this expression, we see that
2 v2
v T k2 √ T which is Equation 16.4. This derivation represents another proof of the expression
for the wave speed on a taut string.
The linear wave equation (Eq. 16.25) is often written in the form
Linear wave equation in general 2y x2 1
v2 2y t2 (16.26) This expression applies in general to various types of traveling waves. For waves on
strings, y represents the vertical displacement of the string. For sound waves, y corresponds to displacement of air molecules from equilibrium or variations in either
the pressure or the density of the gas through which the sound waves are propagating. In the case of electromagnetic waves, y corresponds to electric or magnetic
ﬁeld components.
We have shown that the sinusoidal wave function (Eq. 16.11) is one solution of
the linear wave equation (Eq....
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 Spring '09
 MIHALISIN
 Physics, pulse, wave function

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