A particle on a wave starts with a displacement of 3 cm above equilibrium. As the wave moves
left at 2 m/s, the particle reaches a maximum displacement of 6 cm below the equilibrium in
0.2 s.
(a) Find A, T, ω, ϕ0, λ, k. From this write the equation for D(x, t).
(b) Draw a snapshot of the wave at t = 0 and a ”history graph” (i.e. graph of D(x = 0, t) vs.
t) at x = 0. Show 2 cycles and label them appropriately.
(a) To find the amplitude A, we take the maximum displacement, which is 6 cm below equilibrium.
A = 6 cm
To find the period T, we can use the information that the wave moves left at 2 m/s. The period is the time it takes for one complete cycle of the wave to pass a given point. In 0.2 seconds, the wave moves 2 m/s * 0.2 s = 0.4 m to the left. This means that one complete cycle of the wave has a length of 0.4 m.
T = 0.4 m
To find ω, we use the equation ω = 2π/T.
ω = 2π/0.4 rad/s
To find ϕ0, we use the fact that the wave starts with a displacement of 3 cm above equilibrium. This means that at t = 0, D(x, t) = 3 cm. The equation for D(x, t) is given by D(x, t) = A sin(kx - ωt + ϕ0). Substituting the given values, we have 3 cm = 6 cm * sin(0 + ϕ0), which gives us:
sin(ϕ0) = 0.5
Since sin(30°) = 0.5, we can conclude that ϕ0 = 30°.
To find the wavelength λ, we can use the equation λ = vT = 2 m/s * 0.4 s = 0.8 m.
λ = 0.8 m
To find k, we can use the equation k = 2π/λ.
k = 2π/0.8 rad/m
From this information, the equation for D(x, t) is given by:
D(x, t) = 6 cm * sin((2π/0.8) * x - (2π/0.4) * t + 30°)
(b) To draw the snapshot of the wave at t = 0, we can use the equation D(x, t) = 6 cm * sin((2π/0.8) * x + 30°), where t = 0. This equation gives us the displacement as a function of x when t = 0. We can choose different values of x and plug them into the equation to find the corresponding displacement.
For example, if we choose x = 0, the equation becomes D(0, 0) = 6 cm * sin(30°) = 3 cm. This means that at x = 0, the displacement is 3 cm below equilibrium. Similarly, if we choose x = 0.2 m, the equation becomes D(0.2, 0) = 6 cm * sin((π/0.4) + 30°) ≈ 3.87 cm. This means that at x = 0.2 m, the displacement is 3.87 cm below equilibrium. We can continue this process for different values of x to get a snapshot of the wave.
To draw the history graph at x = 0, we can use the equation D(x = 0, t) = 6 cm * sin((-2π/0.4) * t + 30°), where x = 0. This equation gives us the displacement as a function of t when x = 0. We can choose different values of t and plug them into the equation to find the corresponding displacement.
For example, if we choose t = 0, the equation becomes D(0, 0) = 6 cm * sin(30°) = 3 cm. This means that at t = 0, the displacement is 3 cm below equilibrium. Similarly, if we choose t = 0.1 s, the equation becomes D(0, 0.1) = 6 cm * sin(-2π * 0.1 + 30°) ≈ -1.84 cm. This means that at t = 0.1 s, the displacement is 1.84 cm above equilibrium. We can continue this process for different values of t to get the history graph.