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: If we take x=10m to be our reference string element, and wave speed is 0.5m/s: see Displacement graph

After taking the derivative of the displacement with respect to time (transverse speed = u), at x = 10m:

u See Velocity graph

X = 10m

Displacement Graph

Velocity Graph

D (m)

V (m/s)

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 objectµ = mass/length = Kg/m

Through a number of derivations (see page 203), the following equation for wave speed can be determined

Where Ts is the tension in the rope, and µ is the linear mass density

PRACTICE PROBLEM ON WAVE SPEEDIt’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 ofthe pulse compare? (slower, no change, or faster?) Explain

PRACTICE PROBLEM SOLUTIONa) What is the speed of the pulse at that instance?The speed of the pulse is referring to the wave speed.Therefore, , 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 = wWeight (w) is equal to the weight from the brick and the weight of the rope beneath the 5m point. Therefore, From the linear mass density we can determine

Ts

w

µ=𝑚𝑥→𝑚𝑟𝑜𝑝𝑒 h𝑏𝑒𝑛𝑒𝑎𝑡 =µ 𝑥 , h𝑤 𝑒𝑟𝑒 𝑥𝑖𝑠 h𝑡 𝑒 h𝑙𝑒𝑛𝑔𝑡 𝑜𝑓 𝑟𝑜𝑝𝑒 h𝑏𝑒𝑛𝑒𝑎𝑡

h𝑇 𝑒𝑟𝑒𝑓𝑜𝑟𝑒 :𝑇𝑠=𝑚𝑏𝑟𝑖𝑐𝑘𝑔+µ 𝑥𝑔𝒗=√𝑻 𝒔

µ=√𝑚𝑏𝑟𝑖𝑐𝑘𝑔+µ 𝑥𝑔

µ=√(3.0𝑘𝑔)( 9.8m

s2)+( 3𝑘𝑔

𝑚)(5.0𝑚)( 9.8𝑚

𝑠2)

3.0𝑘𝑔𝑚

=7.7ms

PRACTICE PROBLEM SOLUTIONb) As the pulse goes higher, what happens to the velocity? (slower, no change, or faster?) ExplainThe 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=√𝑚𝑏𝑟𝑖𝑐𝑘𝑔+µ 𝑥𝑔

µ

c) If you had not attached the brick how would the speed ofthe pulse compare? (slower, no change, or faster?) ExplainThe 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=√𝑚𝑏𝑟𝑖𝑐𝑘𝑔+µ 𝑥𝑔

µ