Video Transcript
In this video, we’re going to learn
about the propagation of waves along a string. We’ll see what kind of waves move
on a string. And we’ll also understand how wave
speed relates to string parameters.
To get started, imagine that you
and some friends have devised a communication system that uses ropes with pulses
sent along those ropes to send signals. To make the communication as
efficient as possible, you’d like to be able to send the pulses down the ropes as
fast as you can. To figure out what to do to the
ropes to make this possible, it will be helpful to know about the propagation of
waves along a string.
When we talk about waves on a
string, we’re speaking of a transverse wave, one whose oscillations are
perpendicular to its direction of motion. What we’ve drawn here is a snapshot
of a wave on a string at a particular moment in time. So, if this string, say, was
fastened to two walls on each end, then we can imagine this pulse moving down the
string as time evolves.
But going back to our original wave
pulse, one thing interesting about what we’re seeing is that this is not just an
abstract representation of a wave. This is actually a physical string
being displaced from its equilibrium position by energy that’s passing along it. And because the string, which has
mass, is in motion and is accelerating, that means there’s forces at play acting on
the string.
To get a sense for what those
forces are at a microscopic scale, let’s zoom way in on a little piece of this
string. So, imagine that this segment of
string we’re seeing here is a tiny segment of this whole overall rope. And we ask ourselves, what are the
forces that are acting on this little segment of string?
We know that pulling on either end
of this little snippet of string are tension forces. This force is equal on either end,
and we’ve called it capital 𝑇. And compared to this tension force,
the magnitude of the weight force acting on this tiny little segment of string is
small enough that we can neglect it. And let’s say that our objective
is, based on the forces that are acting on each little segment of string, we wanna
solve for an equation of motion for the 𝑦-position of this string at any time.
To do that, we’ll start with
Newton’s second law of motion, which says that the net force in the 𝑦-direction on
this little segment of string is equal to the mass of that segment times its
acceleration in that 𝑦-direction. When we consider this string that
our wave is propagating along, we can say that the string overall has a linear mass
density. We symbolize this with the letter
𝜇. And the linear mass density is the
total mass of our string divided by its total length.
And going further, say we took our
linear mass density and we multiplied it by an infinitesimal displacement in the
𝑥-direction, that is along the length of the string 𝐿. If for our linear mass density 𝜇,
instead of considering the overall length of string, we consider just the tiny
little segment we’ve highlighted here, then the overall length of that segment is
approximately equal to 𝑑𝑥, an infinitesimal displacement in the horizontal
direction. Which means that 𝑑𝑥 and 𝐿 will
cancel one another out when we consider this tiny segment of string.
Going back to our Newton’s second
law expression, that means we can replace the mass in this expression with 𝑑𝑥
times 𝜇. And then, going a step further, we
can replace the acceleration of our string segment in the 𝑦-direction with the
second derivative of the position of the string with respect to time. And we write this derivative as a
partial derivative because the 𝑦-position of our string doesn’t depend just on
time, we also see it depends on 𝑥. That tells us then that 𝑦 is not
just a function of 𝑡, but also of 𝑥.
At this point, we now have an
expression according with Newton’s second law of motion for the net force in the
𝑦-direction on this small string segment. We can move a step closer to our
equation of motion for our string if we look at the tension forces that act on the
end of this segment. After all they are the source of
force in the 𝑦-direction on this string.
Say we draw in horizontal reference
lines off each end of our small string segment, and then we name the angles that are
between those lines and our tension force vectors 𝜃 one and 𝜃 two. If we consider just the vertical ,
that is the 𝑦-component of each of these tension poles, we can say that the net
force in the 𝑦-direction 𝐹 sub 𝑦 is equal to 𝑇 times the sin of 𝜃 two minus 𝑇
times the sin of 𝜃 one. This is promising to us because we
see it connects with our Newton’s second law expression we have developed.
At this point, we’re going to make
an approximation. We’re going to approximate the
angles 𝜃 one and 𝜃 two as small angles. Here’s what that approximation
means. Say that we have a right triangle,
where one of the angles in our triangle is 𝜃. If we made no approximation for our
angle 𝜃, then we could say that the sin of 𝜃 is equal to 𝑦 over 𝑅, as we’ve
written our triangle. When 𝜃 is small though, as we’re
treating it in this case, then the sine of this angle is approximately equal to the
tangent of this angle. In other words, 𝑥 is approximately
equal to 𝑅.
We can visualise this if we imagine
the triangle shrinking down with smaller and smaller 𝜃 values, that indeed 𝑥 would
approach 𝑅. The tangent of the angle 𝜃 is
equal to 𝑦 divided by 𝑥, which for infinitesimal displacements 𝑦 and 𝑥, we can
write as 𝑑𝑦 over 𝑑𝑥. Here’s what all this means for our
equation for 𝐹 sub 𝑦. Because the sine of our two angles
is approximately equal to the tangent of those two angles and because the tangent
can be written as the partial derivative of 𝑦 with respect to 𝑥, we’re going to
replace the sin of 𝜃 sub two with 𝑑𝑦 𝑑𝑥 two and the sin of 𝜃 sub one with 𝑑𝑦
𝑑𝑥 one.
It’s at this point that we can
bring these two equations for 𝐹 sub 𝑦 together and set them equal one to
another. And as we do, keep in mind that our
goal is to find a solution to the position of our string at any given time 𝑡 or
position 𝑥. With these two equations joined up,
let’s now divide both sides of our equation by 𝜇 times 𝑑𝑥, so dividing the left
and the right side by that term. When we do, we have the
acceleration of our string segment on the left. And on the right, when we divide
our partial derivatives with respect to 𝑥 by another derivative with respect to 𝑥,
that simplifies that expression to be a second derivative of 𝑦 with respect to
position 𝑥.
We now have a nice-looking
second-order partial differential equation. And we want to know what 𝑦-value,
specifically what 𝑦 function, will be a solution to this equation. If we had to guess a solution,
since we know the wave on our string is a transverse wave, we might say, as a guess,
that our solution would be 𝑦 as a function of 𝑥 and 𝑡 is equal to 𝐴 times the
sin of 𝑘𝑥, where 𝑘 is the wavenumber, minus 𝜔𝑡, where 𝜔 is the wave’s angular
frequency.
If we were to take this function
and insert it into our differential equation, we would indeed find that it is a
solution to this equation. That’s great news, but there is a
condition to this function being a solution, and here it is. This function works as a solution
if, and only if, the quantity 𝑇 over 𝜇 leading off the right side of our
expression is equal to the square of 𝜔 over 𝑘. In other words, there is a
necessary connection between the 𝑘 and the 𝜔 in our proposed solution and the 𝑇
and the 𝜇 in our equation of motion.
And with that, we’ve done it. We’ve found a solution for the
vertical, or 𝑦-displacement, of a point on our string at any 𝑥-location or at any
time value. Beyond finding this solution, we
can draw an interesting and helpful conclusion from this equality of 𝑇 over 𝜇 to
𝜔 over 𝑘 squared. If we recall that wave speed 𝑣 is
equal to wavelength times wave frequency, we also know that wavelength 𝜆 is equal
to two 𝜋 over wavenumber and that frequency is equal to angular frequency 𝜔 over
two 𝜋. Put together like this, we see that
the factors of two 𝜋 cancel out and that wave speed 𝑣 is equal to 𝜔 over 𝑘.
Let’s look back over here. If we take the square root of both
sides of this equation, we see that 𝜔 over 𝑘 is also equal to the square root of
the tension in our string 𝑇 over 𝜇, its linear mass density. Bringing these two results
together, we can say that wave speed 𝑣, where our wave is moving along a string, is
equal to the square root of the tension in that string divided by its linear mass
density.
This result, which is like a bonus
result beyond the one that we wanted in the first place, helps us answer the
question we raised at the start. To speed up communication mediated
by waves moving on strings, we could increase the tension of those strings. Let’s get a bit of practice with
this relationship for wave speed through an example.
Transverse waves are sent along a
6.00-meter-long string with a speed of 25.00 meters per second. The string is under a tension of
12.00 newtons. What is the mass of the string?
We can label the string mass
𝑚. And we’ll start on our solution by
recalling the relationship between wave speed, string tension, and string linear
mass density. The speed of a transverse wave
moving along a string can be determined by knowing the string’s tension and its
linear mass density, taking their ratio in the square root. The linear mass density 𝜇 of a
string is equal to its overall mass divided by its overall length.
If we combine these two equations,
we can say that the speed of the wave is equal to the square root of its tension
times its length all divided by its mass. Rearranging this equation to solve
for mass, we find it’s is equal to 𝑇 times 𝐿 over 𝑣 squared. In the problem statement, we’re
given 𝐿, 𝑣, and 𝑇, so we’re ready to plug in and solve for 𝑚. Entering this expression on our
calculator, to three significant figures, 𝑚 is 115 grams. That’s the mass of the string in
this example.
Now let’s summarize what we’ve
learned about propagation of waves along a string. In this segment, we saw that
transverse waves moving along a string satisfied this equation. The acceleration of the string in
the 𝑦-direction is equal to the tension in the string divided by its linear mass
density all multiplied by the second derivative of the height of the string with
respect to its horizontal position.
We also saw that the equation 𝑦 as
a function of 𝑥 and 𝑡 is equal to the amplitude times the sin of 𝑘𝑥, where 𝑘 is
the wavenumber, minus 𝜔𝑡, where 𝜔 is the angular frequency of a wave, is a
solution to this differential equation. Meaning that, this equation
describes the height of waves on a string at any position 𝑥 and at any time 𝑡. And finally, we saw that this
solution requires that wave speed 𝑣 be equal to the square root of string tension
divided by string linear mass density.
