2. Transverse Speed and Wave speed
This learning object will focus on the displacement function of a
pulse and how we can use this to determine the velocity of the
medium in which the pulse is travelling at a specific time and
position.
A continuously travelling wave can be thought of as a series of
wave pulses.
3. Displacement of a pulse in One
Dimension
A waveform can be described by giving the displacement of each
element of a medium at a fixed time.
𝐷 𝑥, 𝑡 =
2
(𝑥−𝑣𝑡)4+3
4. Warm up question!
Which of the following equations describes a pulse moving towards
the direction of decreasing x?
A) 𝐷 𝑥, 𝑡 =
2
(𝑥−4𝑡)4+3
B) 𝐷 𝑥, 𝑡 =
2
(𝑥+9𝑡)4+3
C) 𝐷 𝑥, 𝑡 =
7
(𝑥−.6𝑡)4+3
5. Correct Answer is B!
In the case of choice B, we see that in the denominator velocity
multiplied by time is being added instead of subtracted from the
position x of the pulse.
𝐷 𝑥, 𝑡 =
2
(𝑥−𝑣𝑡)4+3
vs. 𝐷 𝑥, 𝑡 =
2
(𝑥+𝑣𝑡)4+3
The presence of an addition symbol says that the pulse is moving
towards the negative values of x, a good way to rationale this is by
thinking about shifts of a graph.
6. Below are 3 graphs of the same function differing only by addition or
subtraction symbols
As you can see, the graph that has the addition on the denominator is
shifted left, whereas that with the subtraction is shifted right (ie
increasing x)
7. Lets try this question
A pulse travelling through a string is described by :
𝐷 𝑥, 𝑡 =
5
(𝑥 − 0.7𝑡)2+3
A) What is the velocity of the pulse
B) At t = 4.0s what are the speeds of the string elements located at
x = 0.3m and x = 1.4m?
(Hint: think about speed as the rate of change of displacement)
8. Solution!
A) The equation for the displacement of a pulse is given by:
𝐷 𝑥, 𝑡 =
5
(𝑥−𝑣𝑡)2+3
The given equation was: 𝐷 𝑥, 𝑡 =
5
(𝑥−0.7𝑡)2+3
Therefore the velocity of this pulse is 0.7m/s
9. Solution!
B) To determine speed (rate of change) we take the derivative of the
displacement function with respect to time holding x as a constant
𝜕𝐷(𝑥,𝑡)
𝜕𝑡
=
7(𝑥−0.7𝑡)
[(𝑥−0.7𝑡)2+3]2
Now we can evaluate the speeds at (0.3m, 4.0s) and (1.4m, 4.0s)
7(0.3−0.7(4))
[(0.3−0.7(4))2+3]2 = -0.205m/s
7(1.4−0.7(4))
[(1.4−0.7(4))2+3]2 = -0.398m/s
Therefore a waves velocity can differ at any point depending on the
position and time.
