西南交通大学：《大学物理》课程教学资源（讲稿，双语）University Physics AI No. 9 Waves

University Physics Al no9 Waves Class Number ame I. Choose the Correct Answer 1. A disturbance can be written y(x, t)=(e-t/"). This disturbance (A)Not a traveling wave (B)traveling wave with speed v=a (C)a traveling wave with speed v=alb. (D)Traveling wave with speed v=b on:y(x,1)=( Obviously, this disturbanc aveling wave with speed I=b 2. A traveling wave is of the form y(, =Acos(lr-or)+ Bsin(lr-ar), which can also be written as y(x, n=Dsin(kx-@t-o),where (C) (A)D=A+B (B)D=A+B (C)D=A+B( D)D=A-B 3. Which of the following function is not a solution to the wave equation _1 ay2(D) (A)y=sin cost B)y= tan(x+o) (C)y=x3-6x21+12xt2-83 (D)y=sin( +1)cos(x-t) Solution (A)y=sin cost=-[sin(x+1)+sin(x-1]=f(x+vo)+f(x-vo) (B)y=tan(x+1)=f(x+1) 2)3=f(x-1), (D)y=sin(x +t)cos(x-1=sin 2x-sin 2t]*f(x+t), it is not the solution of the wave uation 4. Two waves travel down the same string. The waves have the same velocity, frequency (o), and wavelength but different phase constants(o>%)and amplitudes(A1>A2). According to the principle of superposition, the resultant wave has a frequency f such that (A) (A)f=fo (B)fo/2<f <fo C)0<f<J(D)f=2f0 Solution According to the principle of superposition, the resultant wave has a same frequency as any single 5. A standing wave occurs on a string when two waves of equal amplitude, frequency, and wavelength move in opposite direction on a string. If the wavelength of two waves is decreased to one-half the original length while the wave speed remains unchanged, then the angular frequency of oscillation of the standing wave will

University Physics AI No. 9 Waves Class Number Name I．Choose the Correct Answer 1. A disturbance can be written x b xt b t a y(x,t) (e e e ) 2 2 −( / ) 2 / − = . This disturbance is （ D ） (A) Not a traveling wave. (B) A traveling wave with speed v = a. (C) A traveling wave with speed v = a/b. (D) A traveling wave with speed v = b. Solution: ( , ) ( ) ( ) 2 2 2 2 2 ( ) ( ) ( / ) 2 / y x t e e e e e f x vt b x bt t b x x b xt b t a = = = = − − − − − − − Obviously, this disturbance is a traveling wave with speed v=b. 2. A traveling wave is of the form y(x,t) = Acos(kx −ωt) + Bsin(kx −ωt) , which can also be written as y(x,t) = Dsin(kx −ωt −φ), where （ C ） (A) D = A + B (B) D = A + B (C) 2 2 2 D = A + B (D) D = A − B 3. Which of the following function is not a solution to the wave equation 2 2 2 2 2 1 t y x v y ∂ ∂ = ∂ ∂ ?（ D ） (A) y = sin x cost (B) y = tan(x + t) (C) 3 2 2 3 y = x − 6x t +12xt − 8t (D) y = sin(x + t) cos(x − t) Solution: (A) [sin( ) sin( )] ( ) ( ) 2 1 y = sin xcost = x + t + x − t = f x + vt + f x − vt , (B) y = tan(x + t) = f (x + t), (C) 6 12 8 ( 2 ) ( ) 3 2 2 3 3 y = x − x t + xt − t = x − t = f x − t , (D) [sin 2 sin 2 ] ( ) 2 1 y = sin(x + t)cos(x − t) = x − t ≠ f x ± t , it is not the solution of the wave equation. 4. Two waves travel down the same string. The waves have the same velocity, frequency (f0), and wavelength but different phase constants (φ1>φ2) and amplitudes (A1>A2). According to the principle of superposition, the resultant wave has a frequency f such that （ A ） (A) 0 f = f . (B) 0 2 0 f / < f < f (C) 0 0 < f < f (D) 2 0 f = f Solution: According to the principle of superposition, the resultant wave has a same frequency as any single wave. 5. A standing wave occurs on a string when two waves of equal amplitude, frequency, and wavelength move in opposite direction on a string. If the wavelength of two waves is decreased to one-half the original length while the wave speed remains unchanged, then the angular frequency of oscillation of the standing wave will （ C ）

(A)decrease to one-half (B)remain the same (C)double The wave length is=vT, if the v keeps unchanged, the wavelength decrease to one-half, then the frequency of oscillation of the standing wave will double 6. In a standing wave on a string, the spacing between nodes is Ar. If the tension in the string is doubled but the frequency of the standing waves is fixed, then the spacing between the nodes will change to (A)2Ax (B)√2Ax (C)△x/2 (D)△x/√2 Solution If the tension in the string is doubled, the velocity of the wave is v=-= Wu VH So the spacing between the nodes will change to Ard v_√2yf I. Filling the blanks 1. a transverse wave on a string is described by the wavefunction V(x, t)=(0.300 m)cos((2.57 rad/m)x-(251.3 rad/s)r] The amplitude of the wave is 0.3m; the angular wavenumber of the wave is 12.57rad/m; the wavelength of the wave is_ 0.5m; the angular frequency of the wave is _251. 3rad/s; the frequency of the wave is 40.0Hz, the period of the wave is 0.025s_; the speed of the wave is 20.0m/s the wave disturbance at x=3.00 m when (= 1.50 x 10s is 0. 24m Solution 3(m) (b)k=12.57(rad/m) (c)k 22 2丌2丌 k12.570.5(m) (d)o=251.3rad/s _251.3 =40.0(Hz) 0025(s) 251.3 251.3 (g)v k12.57

(A) decrease to one-half . (B) remain the same (C) double Solution: The wave length isλ = vT , if the v keeps unchanged, the wavelength decrease to one-half, then the frequency of oscillation of the standing wave will double. 6. In a standing wave on a string, the spacing between nodes is∆x . If the tension in the string is doubled but the frequency of the standing waves is fixed, then the spacing between the nodes will change to （ B ） (A) 2∆x (B) 2∆x (C) ∆x / 2 (D) ∆x / 2 Solution: If the tension in the string is doubled, the velocity of the wave is v T T v 2 2 = = ′ ′ = µ µ So the spacing between the nodes will change to x v f vf x = = = ∆ ′ = ′ ∆ ′ = 2 2 2 2 2 2 2 λ λ II. Filling the Blanks 1. A transverse wave on a string is described by the wavefunction ψ (x,t) = (0.300 m) cos[(12.57 rad/m)x − (251.3 rad/s)t] The amplitude of the wave is 0.3m ; the angular wavenumber of the wave is 12.57rad/m ; the wavelength of the wave is 0.5m ; the angular frequency of the wave is 251.3rad/s ; the frequency of the wave is 40.0Hz ; the period of the wave is 0.025s ; the speed of the wave is 20.0m/s ; the wave disturbance at x = 3.00 m when t = 1.50 × 10-3 s is 0.24m . Solution: (a) A=0.3(m) (b) k=12.57(rad/m) (c) 0.5(m) 12.57 2 2 2 = ⇒ = = = π π λ λ π k k (d) ω = 251.3rad/s (e) 40.0(Hz) 2 251.3 2 = = = π π ω ν (f) 0.025(s) 251.3 2 2 = = = π ω π T (g) 20.0(m/s) 12.57 251.3 = = = k v ω

(h)y=0.3cos(12.57×3-251.3×1.5×10-)=0.24m) 2. A simple harmonic transverse wave is propagating shows a plot of the displacement as a function of y(m) along a string toward the left (or -x)direction. Fig osition at time F0. The string tension is 3.6N and its 4 linear density is 25g/m. The wave amplitude is 5m,2 the wavelength Om the wave speed is 0 144m/s,the wave period is_0.14s, and the .2 maximum speed of a particle in the string -4 write an equation describing the traveling wave H(x,D)=5c0s[2m(+121)--丌] Solution See the figure 1, we know the wave amplitude is 5m, and the wavelength is 40m 3.6 The wave speed is v= 12 V4-V25×10 The wave period is T =3.33s The phase constant of point O is-0.2T according to the reference circle According to the equation y(x, t)=Acos[2T(+vt)+o, the wave function is Y(x, t)=5cos[2T( x+12) 3. A traverse wave pulse on a string has the wavefunction 0.250m3 (xD)=200m2+[x-(500mSn)2 The speed of the pulse is 5 m/s Solution a-y 1 a-y Since the classical wave equation is ay 0.5(x-51) ay2(x-51)20.5 ax[2+(x-5) ax2[2+(x-51)2][2+(x-5)]

(h) 0.3cos(12.57 3 251.3 1.5 10 ) 0.24(m) 3 = × − × × = − Ψ 2. A simple harmonic transverse wave is propagating along a string toward the left (or -x) direction. Fig. 1 shows a plot of the displacement as a function of position at time t=0. The string tension is 3.6N and its linear density is 25g/m. The wave amplitude is 5m , the wavelength is 20m , the wave speed is 144m/s , the wave period is 0.14s , and the maximum speed of a particle in the string is , write an equation describing the traveling wave ] 5 1 12 ) 40 Ψ( , ) = 5cos[2π ( + t − π x x t . Solution: See the figure 1, we know the wave amplitude is 5m, and the wavelength is 40m. The wave speed is 12 m/s 25 10 3.6 3 = × = = − µ T v . The wave period is 3.33 s 12 40 = = = v T λ The phase constant of point O is -0.2π according to the reference circle. According to the equation ( , ) cos[2 ( ) φ] λ = π + vt + x Ψ x t A , the wave function is ] 5 1 12 ) 40 ( , ) = 5cos[2π ( + t − π x Ψ x t 3. A traverse wave pulse on a string has the wavefunction 2 2 3 2.00m [ (5.00 m/s) ] 0.250m ( , ) x t x t + − ψ = . The speed of the pulse is 5 m/s . Solution: Since the classical wave equation is 0 1 2 2 2 2 2 = ∂ ∂ − ∂ ∂ t Ψ x v Ψ 2 3 2 2 2 2 2 2 2 [2 ( 5 ) ] 0.5 [2 ( 5 ) ] 2( 5 ) [2 ( 5 ) ] 0.5( 5 ) x t x t x t x Ψ x t x t x Ψ + − − + − − = ∂ ∂ ⇒ + − − = − ∂ ∂ Fig.1 x(m) 10 30 50 70 90 0 -2 -6 2 6 -4 4 y(m) φ

ay2.5(x-5) ay50(x-5) a2Y The speed of the pulse is v= a-y 5(m/s) 4. Figure 2 depicts the waveform of a traveling sinusoidal wave at two instants, when t=0 s and 0. 25 s later. The wave is traveling toward increasing values of x. ux, n) 0.5n(-x(8n+6t+ ux)m) en fOs when f0250s Solution From the figure we can know /=2m, A=0.5m The distance the wave traveling from FOs to 0.25s is 0 x(m n+1.5mn=0,1,2 therefore m 0.2 0.25A 丌(8n+ 2丌 For k d/m) Then we have Y= Acos(lr-or)= Asin(lr-ot+)=0.5sin(-(8n+6)r+) IIL. Give the solutions of the following problems se wave moving at 4.00 m/s toward decreasing values of x has the waveform indicated in Figure 3 A ux,Os)ym) 0.250 (a)What is the wavelength? Indicate the wavelength on the

2 3 2 2 2 2 2 2 2 [2 ( 5 ) ] 12.5 [2 ( 5 ) ] 50( 5 ) [2 ( 5 ) ] 2.5( 5 ) x t x t x t t Ψ x t x t t Ψ + − − + − − = ∂ ∂ ⇒ + − − = ∂ ∂ The speed of the pulse is 5(m/s) 2 2 2 2 = ∂ ∂ ∂ ∂ = x Ψ t Ψ v 4. Figure 2 depicts the waveform of a traveling sinusoidal wave at two instants, when t = 0 s and 0.25 s later. The wave is traveling toward increasing values of x. ψ(x, t) = ) 2 0.5sin( (8 6) π πx − π n + t + . Solution: From the figure, we can know λ = 2m, A = 0.5m The distance the wave traveling from t=0s to t=0.25s is nλ +1.5m n = 0,1,2L therefore (8 6) 2 4 3 1 1.5m 0.25 0.25 1.5m ⇒ = = + + = + ⇒ = + = n T n n T n T π π ω λ λ λ λ For (rad/m) 2 π λ π k = = Then we have ) 2 ) 0.5sin( (8 6) 2 cos( ) sin( π π π π Ψ = A kx −ωt = A kx −ωt + = x − n + t + III. Give the Solutions of the Following Problems 1. A periodic transverse wave moving at 4.00 m/s toward decreasing values of x has the waveform indicated in Figure 3 when t = 0 s. (a) What is the wavelength? Indicate the wavelength on the Fig.3 0 3 6 x(m) ψ(x, 0s)(m) 0.250 -3 λ 1.0 2.0 3.0 x(m ψ(x)(m) 0.2 0.4 -0.2 -0.2 0 when t=0s when t=0.250s Fig.2

waveform (b) Draw a graph of the disturbance at x=0 m as a function of time. Indicate the period of the wave on this graph. What is the period in seconds? (c)Use the results of part(b)to determine the frequency and confirm that the speed is the product of the frequency and wavelength as Equation 12 10 states (d)What is the maximum transverse speed (a) The wavelength is 3m b)from the figure, the period in seconds is T=0.75(s) 51(s) (c) The frequency is V=T 0.75-3Hzy A=-3=4(m/s) Thus the speed is the product of the frequency and wavelength: v=v.1 (d) The maximum transverse speed is dy(o, t max= maximum slop of the 025=1(m/s) d 0.25 2. When t=0 S, a pulse on a string is described by the equation y(x,0s)= 2.50m2+x (a)Sketch u(x, Os)as a function of x over the domain.00 m <x< 10.00 m (b) The pulse moves parallel to -i with a speed of 8.00 m/s. Find u(x, t) Solution (a) The graph is shown in figure (b) The wave function is

waveform. (b) Draw a graph of the disturbance at x = 0 m as a function of time. Indicate the period of the wave on this graph. What is the period in seconds? (c) Use the results of part (b) to determine the frequency and confirm that the speed is the product of the frequency and wavelength as Equation 12.10 states. (d) What is the maximum transverse speed? Solution: (a) The wavelength is 3m. (b) from the figure, the period in seconds is T = 0.75(s) . (c) The frequency is (Hz) 3 4 0.75 1 1 = = = T ν As 3 4(m/s) 3 4 ν ⋅λ = ⋅ = Thus the speed is the product of the frequency and wavelength: v =ν ⋅λ . (d) The maximum transverse speed is 1(m/s) 0.25 0.25 ) maximum slop of the graph d d (0, ) ( = max = = = t Ψ t vtrans 2. When t = 0 s, a pulse on a string is described by the equation 2 2 3 2.50m 1.00m ( ,0s) x x + ψ = (a) Sketch ψ(x, 0s) as a function of x over the domain -10.00 m ≤ x < 10.00 m. (b) The pulse moves parallel to i − ˆ with a speed of 8.00 m/s. Find ψ(x, t). Solution: (a) The graph is shown in figure. (b) The wave function is t(s) Ψ(m) -0.5 1.5 0 0.25 T v← -10 -5 5 10 0.1 0.2 0.3 0.4

y(x25+(x+8)2 3. A policeman is chasing a speeder along a straight interstate highway. Both the speeder and trooper are moving at 150.0 km/h. The sire on the police cruiser has a frequency of 2.0 x10'Hz What frequency is detected: (a) by the policeman?(b) by the speeder?(c) by a doomed, dumb possum(Didelphis marsupialis) watching the cars approach from a vantage point in the middle of the road? Solution (a) The frequency is detected by the policeman is (b) The frequency is detected by the speeder is J-0 =V y=2×103(Hz) (c)v=vl 343 )=以( 343-15.、l03)=2.28×103(Hz) 3600 4. A wave is described by the wavefunction v(x,)=(200×102m)cos180rad/m)x-(24.0rad (a)What wave, when added to the given wave, will form a standing wave pattern (b) For such a standing wave, what is the distance between the nodes of the wave? (c)For such a standing wave pattern, what is the distance between the antinodes? (d)What is the distance between a node and the nearest antinodes? Solution (a)Thewavefuntonisv(x:1)=(200×102m)cos(80radm)x+(240rads] (b)The distance between the nodes 12 z 3.14 =0.017 k18 (c) The distance between the antinodes is the same as the nodes (d) The distance between a node and the nearest antinodes is (n+1)-n==00087m

2 2.5 ( 8 ) 1 ( , ) x t Ψ x t + + = 3. A policeman is chasing a speeder along a straight interstate highway. Both the speeder and trooper are moving at 150.0 km/h. The sire on the police cruiser has a frequency of 3 2.0×10 Hz. What frequency is detected: (a) by the policeman? (b) by the speeder? (c) by a doomed, dumb possum (Didelphis maarsupialis) watching the cars approach from a vantage point in the middle of the road? Solution: (a) The frequency is detected by the policeman is ( ) 2 10 (Hz) 3 ν ′ =ν =ν = × V V (b) The frequency is detected by the speeder is ) 2 10 (Hz) 0 0 ( ) ( 3 = = × − − = − − ν ′ =ν ν ν V V V V V V souce speeder (c) ) 2.28 10 (Hz) 3600 10 343 15 343 ( ) ( 3 3 = × − ⋅ = − ν ′ =ν ν V Vsouce V 4. A wave is described by the wavefunction (x,t) (2.00 10 m)cos[(180rad/m)x (24.0rad/s)t] 2 = × − − ψ (a) What wave, when added to the given wave, will form a standing wave pattern? (b) For such a standing wave, what is the distance between the nodes of the wave? (c) For such a standing wave pattern, what is the distance between the antinodes? (d) What is the distance between a node and the nearest antinodes? Solution: (a) The wavefunction is (x,t) (2.00 10 m)cos[(180rad/m)x (24.0rad/s)t] 2 = × + − ψ . (b) The distance between the nodes is 0.017 m 180 2 3.14 2 1 2 = × = = = k k λ π π (c) The distance between the antinodes is the same as the nodes. (d) The distance between a node and the nearest antinodes is 0.0087 m 4 4 4 ( +1) − = = λ λ λ n n