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 iswaveform. (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