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.
To find the values of A, T, ω, ϕ0, λ, and k, we can use the following equations:
Amplitude (A): A = maximum displacement = 6 cm
Period (T): T = time for one complete cycle = 2 * 0.2 s = 0.4 s
Angular frequency (ω): ω = 2π / T = 2π / 0.4 s = 5π rad/s
Phase constant (ϕ0): The particle starts with a displacement of 3 cm above equilibrium, so it has a phase of 0 at t = 0. This means ϕ0 = 0.
Wavelength (λ): The wave moves left at 2 m/s, so the distance between two consecutive points with the same displacement is 2 m. Since the particle has reached a maximum of 6 cm below equilibrium, the wavelength is λ = 2 m + (6 cm / 100 cm/m) = 2.06 m.
Wave number (k): k = 2π / λ = 2π / 2.06 m ≈ 3.06 m⁻¹.
The equation for D(x, t) can be written as:
D(x, t) = A * sin(ωt - kx + ϕ0)
D(x, t) = 6 * sin(5πt - 3.06x)
To draw the snapshot of the wave at t = 0, we can set t = 0 in the equation:
D(x, 0) = 6 * sin(-3.06x)
The "history graph" at x = 0 can be obtained by setting x = 0 in the equation:
D(0, t) = 6 * sin(5πt)