1) The transverse speed is defined as the rate of change of displacement with respect to time of an individual string element as a wave passes along it. 2) Wave speed depends on the elasticity and inertia of the medium. Greater tension or lesser inertia results in higher wave speed. 3) For a string, wave speed can be calculated as the square root of tension over linear mass density.

1. TRANSVERSE SPEED AND WAVE SPEED
USING A STRING AS AN EXAMPLE
Notice that the string to the right is composed of
individual string elements
The transverse speed, is defined as the rate of change of
the displacement with time (velocity), of an individual
string element, as a wave passes along it
TRANSVERSE SPEED (pg. 201)
Imagine a pulse along a string, whose displacement is defined as:
𝑫 𝒙, 𝒕 =
𝟐.𝟎
𝒙−𝒗𝒕 𝟐+𝟒
If we take x=10m to be our reference string element, and wave
speed is 0.5m/s:
𝑫 𝒙, 𝒕 =
𝟐.𝟎
𝟏𝟎−𝟎.𝟓𝒕 𝟐+𝟒
 see Displacement graph
After taking the derivative of the displacement with respect to
time (transverse speed = u), at x = 10m:
u 𝒙, 𝒕 =
𝟐.𝟎(𝟏𝟎−𝟎.𝟓𝒕)
𝟏𝟎−𝟎.𝟓𝒕 𝟐+𝟒
𝟐  See Velocity graph
X = 10m
Displacement Graph
Velocity Graph
D (m)
V (m/s)

2. WAVE SPEED (pg. 202-203)
Wave speed is the velocity with which a wave passes through a medium
Wave speed is dependant on the elasticity and inertial property of the rope
• Elasticity: dependent on the tension
 Stronger the tension, the stronger the restorative force, the faster the velocity
• Intertia: dependent on the linear mass density of the rope
 The greater the intertia, the greater resistance to movement, the slower the velocity
• Linear Mass Density (µ): the mass per unit length of an object
µ = mass/length = Kg/m
Through a number of derivations (see page 203), the
following equation for wave speed can be determined
𝒗 =
𝑻 𝒔
µ
Where Ts is the tension in the rope, and µ is the linear
mass density

3. PRACTICE PROBLEM ON WAVE SPEED
It’s a sunny Saturday afternoon and you find yourself with a massive 20m long string with linear mass density 3.0kg/m. You
hang it from your extraordinarily high ceilings and decide that it would be funny to attach a brick (of mass 3.0kg) to the end
of the string.
In your uncontrollable laughter, you trip and jolt the bottom of the string forwards and back forming a perfect pulse.
When you recover, you notice the pulse is 5.0m up the string.
At that instant, you decide to make a number of calculations in your head and begin to contemplate over the following
questions…
a) What is the speed of the pulse at that instance?
b) As the pulse goes higher, what happens to the velocity?
(slower, no change, or faster?) Explain
c) If you had not attached the brick how would the speed of
the pulse compare? (slower, no change, or faster?) Explain

4. PRACTICE PROBLEM SOLUTION
a) What is the speed of the pulse at that instance?
The speed of the pulse is referring to the wave speed.
Therefore, 𝒗 =
𝑻 𝒔
µ
, so in order to determine v, we need Ts and µ
µ (linear mass density), is given to be 3.0 kg/m
Ts can be determined using a simplified free body diagram:
Net force is zero, so tension must equal the weight
Ts = w
Weight (w) is equal to the weight from the brick and the weight
of the rope beneath the 5m point.
Therefore, 𝑤 = 𝑚 𝑏𝑟𝑖𝑐𝑘 𝑔 + 𝑚 𝑟𝑜𝑝𝑒 𝑏𝑒𝑛𝑒𝑎𝑡ℎ 𝑔 = Ts
From the linear mass density we can determine 𝑚 𝑟𝑜𝑝𝑒 𝑏𝑒𝑛𝑒𝑎𝑡ℎ
Ts
w
µ =
𝑚
𝑥
→ 𝑚 𝑟𝑜𝑝𝑒 𝑏𝑒𝑛𝑒𝑎𝑡ℎ = µ𝑥, 𝑤ℎ𝑒𝑟𝑒 𝑥 𝑖𝑠 𝑡ℎ𝑒 𝑙𝑒𝑛𝑔𝑡ℎ 𝑜𝑓 𝑟𝑜𝑝𝑒 𝑏𝑒𝑛𝑒𝑎𝑡ℎ
𝑇ℎ𝑒𝑟𝑒𝑓𝑜𝑟𝑒: 𝑇𝑠 = 𝑚 𝑏𝑟𝑖𝑐𝑘 𝑔 + µ𝑥𝑔
𝒗 =
𝑻 𝒔
µ
=
𝑚 𝑏𝑟𝑖𝑐𝑘 𝑔 + µ𝑥𝑔
µ
=
(3.0𝑘𝑔)(
9.8m
s2 ) + (
3𝑘𝑔
𝑚
)(5.0𝑚)(
9.8𝑚
𝑠2 )
3.0𝑘𝑔
𝑚
=
7.7m
s

5. PRACTICE PROBLEM SOLUTION
b) As the pulse goes higher, what happens to the velocity?
(slower, no change, or faster?) Explain
The pulse would be going faster.
Looking at the equation from part a:
We see that the tenstion ,Ts, increases as the length of rope beneath the pulse (x) increases.
Therefore, as the pulse goes higher, the velocity will increase.
𝒗 =
𝑻 𝒔
µ
=
𝑚 𝑏𝑟𝑖𝑐𝑘 𝑔 + µ𝑥𝑔
µ
c) If you had not attached the brick how would the speed of
the pulse compare? (slower, no change, or faster?) Explain
The pulse would be going slower.
Again, looking at the equation from part a:
We see that the mass of the brick (mbrick ) contributes to the tensions (Ts).
Therefore, removing the brick would reduce the tension, ultimately reducing the velocity.
𝒗 =
𝑻 𝒔
µ
=
𝑚 𝑏𝑟𝑖𝑐𝑘 𝑔 + µ𝑥𝑔
µ