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16 - Wave Motion

When we use this procedure to analyze the total

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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 small-angle 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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