This document discusses the displacement of two ropes being moved in alternating waves by the left and right arms. It asks for the speed of the rope in the right arm and the displacement equation for that rope. It is determined that the ropes have the same speed of 5.6 m/s since they are π radians out of phase, meaning their displacements are equal and opposite. The displacement equation for the rope in the right arm is determined to be D(x,t) = -(1.0)sin(0.28x-1.57t) based on the properties of harmonic motion and using information from the graph shown.

1. He wants you to perform alternating waves, such that if
we were to plot the displacement of the air particles
through which the disturbance passed, as a function of
time it would look something like… (next slide)

3. Suppose you are displacing the blue rope with
you’re left arm at a speed of 5.6 m/s:
a) What is the speed of the rope being displaced with
you’re right arm?
b) Determine the equation for the displacement as a
function of time for the rope in you’re right arm.
(Answer may depend on x and t)
QuestionTime!

4. Part A - Answer: 5.6m/s
From the graph, we can see that any given crest of
the red line falls exactly half way between 2 crests
of the blue line (λ/2). This means that the ropes in
your right and left arm are π radians out of phase
(ΔФ = π). When two waves have a phase difference
of π they have equal and opposite displacements.
Since the frequency of these two waves is the same,
the speed of both ropes must be the same.

5. Part B - Answer: D(x,t) = -(1.0)sin(0.28x-1.57t)
The answer for a displacement function for any
wave exhibiting harmonic motion (follows a
sinusoidal curve) should be in the form:
D(x,t) = (A)sin(kt-ωt±ф)
From the graph, we speculate that the amplitude
(A) is 1.0 metres.We can also see that the period (T)
is 4.0 seconds and that x is increasing.

6. Part B - Continued
Using these known variables, we can now
determine the rest as follows:
1) ω = 2π/T 3) k = 2π/λ
= 2π/4.o s = 2π/22.4 m
= 1.57 rad/s = 0.28 rad/m
2) λ = νT 4) ф = arcsin(Dₒ/A)
= (5.6m/s)(4.0s) = arcsin(0/1)
= 22.4 m = 0

7. Part B – Continued
But wait! Since the displacement of the right arm
rope is equal and opposite to that of the left arm
rope, it must be negative.This comes from the fact
that if two sine functions are π radians out of phase
we can use the trigonometric identity:
sin(θ+π) = -sin(θ)
So finally plugging in all our values gives us:
D(x,t) = -(1.0)sin(0.28x-1.57t)
And we studied and worked out all at once!