(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