when two waves of the same frequency Collide on a string the result is called a standing wave we
see that the resulting superposition of waves no longer appears to travel in any particular
direction instead simply repeating the same oscillation in place there are spots on the
string that never move and these are called nodes and there are spots on the string that move with
maximum amplitude and these are called antinodes in this video we're interested in how a standing
wave can arise on a string that's fixed at both ends as a result of rightward traveling waves
interfering constructively with their leftward traveling Reflections so the first thing to
work out is just the math of how to combine a rightward moving wave with its leftward moving
reflection so a wave traveling to the right can be represented by the wave function y r of x and t is
equal to a cosine KX minus Omega T and I'll post a link to the video where this was first derived
similarly a wave traveling to the left is given by y L of x and t is equal to a cosine KX plus
Omega T but we have to be careful about Reflections we've learned previously that
Reflections at a fixed end are inverted and are leftward traveling waves are reflections of y r
we can deal with this mathematically by giving the reflected wave a minus sign after making this
replacement we can add the two waves y r and Y L so this is a superposition of the rightward wave
and Its leftward Reflection so we have an A cosine KX minus Omega T minus a cosine KX plus Omega T
and the way to get a handle on what we're looking at is to apply the trig identities for the cosine
of a sum and a difference so before we apply those identities we're going to factor an a out of the
entire thing and we're left with a times this hideous expression the cosine of a difference is
the cosine of the first piece cosine of the second piece plus sine of the first piece sine of the
second piece and then when we apply the identity for the cosine of a sum of two terms and remember
we still have a minus sign in front of this piece we have the cosine of the first piece cosine of
the second minus the sine of the first sign of the second we notice when we distribute the
minus sign into the second term that the cosine terms are going to cancel each other out we have
one that's positive out in front here and one is negative here also when we distribute that minus
sign the sine terms in that second piece become positive so we end up with 2 times sine KX sine
Omega T and we can move that 2 out in front to obtain the wave function for the standing wave
Ys of x and t is 2A sine KX sine Omega T now it's useful to rearrange this expression with
the time dependence put out in front because now we can interpret this as an oscillating amplitude
sitting in front of a fixed sine function sine k x so sine KX is just an ordinary sine wave in
space and then the 2A sine Omega T out in front tells you we have an amplitude that's oscillating
between 2A and negative 2A so we have an ordinary sine function with its amplitude growing to 2A it
then shrinks to zero it then goes to negative 2A so it flips upside down and so on so this form of
the equation helps us understand what we're seeing in the animation of the standing wave now we need
to figure out what standing waves are allowed to vibrate on a string of length L clamped at both
ends and again our wave function has this form 2A sine KX sine Omega T remember a was the amplitude
of the original waves that combine together and the standing wave oscillates with twice
that original amplitude for the individual waves now we see that for x equals zero we automatically
get a node out of this wave equation because sine of zero is zero so sine of KX is zero when X is
zero now we just have to enforce that our wave function is zero at x equals L as well and this
requires that the sine of k l is equal to zero well this occurs when KL is an integer multiple
of Pi so this allows us to say what the values of K are for the standing waves that satisfy the
boundary conditions and that's n pi over L where n is 1 2 3 and so on alternatively because the
wavelength is 2 pi over K we can say that Lambda is 2 pi times the reciprocal of K which is L Over
N pi and our Pi's are going to cancel out and we get that Lambda is 2L Over N where n is 1 2 3 and
so on so these are the wavelengths corresponding to standing waves on a string of length L that's
clamped at both ends and this one is worth putting in a box now we have to be really careful here to
note that every allowed wavelength corresponds to a different frequency recall that the wave
speed formula V equals F Lambda relates speed frequency and wavelength of a wave on a string
and V is a constant here V is the same for every standing wave because we're holding the tension in
linear Den density of the string constant so with each choice of n we get a wavelength of
Lambda n and a corresponding frequency of FN note that as the Lambda ends get smaller the
corresponding frequencies FN those have to get bigger to hold the wave speed constant so when
we're writing down the explicit wave functions for these standing waves we can write each
omega as 2 pi FN and we'll get into the details on how to calculate the FNS in a moment for now we
want to visualize the first several standing waves on this string so for n equals one we get Lambda 1
is equal to 2L divided by one or just 2L so that's a wavelength of twice the length of the string and
that means one half of a wavelength fits on the string and here's the explicit wave function for
this standing wave we just subbed in that K is 2 pi over Lambda 1 and Omega is 2 pi times F1
and this standing wave is called the fundamental or the first harmonic for the string so here's is
what it looks like next we Sub in n equals 2 and that gives us Lambda 2 is 2L over 2 or simply the
length of the string so the wavelength is equal to the length of the string and again our explicit
wave function for the standing wave is given by just subbing in n equals two for the wavelength
and the frequency and this is called the second harmonic for the string and here's what that one
looks like notice that that second harmonic is vibrating much faster we have a shorter
wavelength in fact it's half as big so we have a larger frequency and that's actually twice as much
as the first harmonic subbing in n equals three we find a wavelength of 2L over three and here's
what that one looks like and finally subbing in n equals four we get a wavelength of 2L over four
in other words L over two so we get a complete wave every half length of the string another way
to say that is that two waves will fit on the string finally it's not that hard to work out
the frequencies of all these harmonics so recall our wave speed formula V b equals F Lambda and
again every time we choose an N we end up with a corresponding wavelength of Lambda n and frequency
of FN so we can sub in our little formula for the nth harmonic wavelength that's 2L over n and then
go ahead and solve for the nth harmonic frequency and that's a useful formula we get that the nth
harmonic frequency is n times the wave velocity over twice the length but what I want to do
with this is sub in n equals 1 and we find that the fundamental frequency or the first harmonic
is just V over 2L and we can do a really sneaky little trick here we can replace V over 2L in that
previous line with F1 and we find out that the nth harmonic frequency is n times the fundamental
frequency or n times the first harmonic so every single resonant frequency on the string is just an
integer multiple of the lowest resonant frequency on the string and this is one of the most powerful
formulas that we have for problem solving in standing waves if you enjoyed this video or at
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