Physics Solution Manual for 1100 and 2101

# 46 103 s it follows that l2 l1 vt 1 second the

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Unformatted text preview: e terms 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.

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