Solutions to Problems: Mechanical Waves‐1 P347: 6, 19, 20, 24, 25, 27, 35, 46, 49, 71 6. A rope of mass 0.65 kg is stretched between two supports, 30 m apart. If the tension in the rope is 120 N, how long will it take a pulse to travel from one support to the other? Solution: The speed of the pulse traveling in the rope is v = T ρ = TL m = 74.4 m/s . So, the time required is t = L v = 0.40 s . 19. Suppose at t = 0, a wave shape is represented by D = DM sin ( 2π x λ + φ ) ; that is, it differs from Eq. 13‐9 by a constant phase factor φ . What then will be the equation for a wave traveling to the left along the x axis as a function of x and t? Solution: Since the phase of particles locating at position x is larger than the phase of the particles at the origin, and the phase difference is Δϕ = kx , the wave function for a wave traveling to the left reads: D ' = DM sin (ωt + 2π x λ + φ ) . 20. A transverse traveling wave on a cord is represented by D = 0.48sin ( 5.6 x + 84t ) where D and x are in meters and t in seconds. For this wave determine (a) the wavelength, (b) frequency, (c) velocity (magnitude and direction), (d) amplitude, and (e) maximum and minimum speeds of particles of the cord. Solution: From the wave function, one getsthe wave number and angular frequency of wave are k = 5.6 m‐1 , and ω = 84 s‐1 . So, (a) the wave length reads λ = 2π k =1.122 m, (b) the frequency reads f = ω 2π = 13.37 Hz , (c) the speed is v = ω k = 15.0 m/s , and the direction is along the ‐x axis, (d) the amplitude is 0.48 m, and (d) the maximum speeds of particles is vmax = Aω = 40.32 m/s , and the minimum speeds of the particles is zero. 24. A 440‐Hz longitudinal wave in air has a speed of 345 m/s. (a) What is the wavelength? (b) How much time is required for the phase to change by 90°at a given point in space? (c) At a particular instant, what is the phase different (in degrees) between two points 4.4 cm apart? Solution: (a) The wave length is λ = v f = 0.784 m .
(b) Δ ϕ = ω Δ t ⇒ Δ t = Δ ϕ ω = Δ ϕ ( 2π f ) = 1 4 f = 5.68 × 10 −4 s .
(c) Δϕ = k Δx = 2πΔx λ = 0.3526 = 20.20° . 25. Write down the equation for the wave in Problem 24 if its amplitude is 0.020 cm and at t = 0, x = 0, and D = −0.020 cm . Solution: From the initial conditions given in the problem, one can figure out ϕ = π . And from the Problem 24, one can obtain that: ω = 2π f = 880π s‐1 , k = ω v = 8.01 m‐1 , and thus the wave equation of the wave is y ( x, t ) = D cos (ωt − kx + ϕ ) = 0.020 cm cos ( 880π t − 8.01x + π ) . 27. Determine if the function D = DM sin kx cos ωt is a solution of the wave equation. Solution: Since ∂2 D ∂t 2 = −ω2 DM sin kx cos ωt = −ω2 D, and ∂2 D ∂x2 = −k 2 sin kx cos ωt = −k 2 D , then we get ∂2 D ∂t 2 = ω2k −2 ∂2 D ∂x2 = v2 ∂2 D ∂x2 , QED. 71. Figure 13‐40 shows the wave shape of a sinusoidal wave traveling to the right at two instants of time. What is the mathematical representation of this wave? Solution: From the wave shape, we can obtain directly: A = 3.5 cm, λ = 6 cm , and we can read from the figure the distance the wave traveling from t = 0, t = 3.0 s is 4 cm. The wave speed is thus v = Δx Δt = 1.33 cm/s, and k = 2π λ = 1.047 cm ‐1 , and ω = kv = 1.39 s‐1 . From the wave form at t = 0, we get ϕ = 0 . The wave function of the wave is y ( x, t ) = 3.5 cm cos (1.39t s‐1 − 1.047 x cm‐1 ) .