- [Instructor] We are told a transverse wave
travels to the right along a string,
they draw it right over here.
Two dots have been painted on the string.
In the diagrams below, those dots are labeled P and Q,
so that's these dots here.
The figure below shows the string at an instant in time.
At the instant shown, dot P has maximum displacement
and dot Q has zero displacement from equilibrium.
At each of the dots P and Q, draw an arrow
indicating the direction of the instantaneous
velocity of that dot.
If either dot has zero velocity,
write V equals zero next to the dot.
So pause this video and see if you can have a go
at it on your own before I work through it with you.
All right, now let's work through this together.
So, let's think about the instantaneous velocity
of this dot right over here.
And to help us think about it,
let's think about what's going on
on either side of this dot.
So, if we think about this point right over here,
what's gonna happen a little bit after
this moment in time we're looking at?
Well, the wave is traveling in that direction
so in a little while, the wave is gonna look,
in a moment after that, the wave is gonna look
something like this.
The wave is gonna look something like this.
So, this dot is in the process of moving down.
And what about this dot right over here?
Well, this dot a little bit later
is going to be higher, so it's moving up.
And it's this point where we're at our maximum,
when we're at our maximum displacement from equilibrium,
we are going from moving down to going up
right at that moment.
So right at this moment of maximum displacement,
we have zero velocity.
So let me do that in red so you can see it.
So here, at point P, we would write V is equal to zero.
Now what about at point Q?
Well, we just sketched what's going to happen
a moment later.
Our transverse wave is gonna look like this
and so point Q is going to be up here.
So point Q is moving upwards.
So if we want to draw its velocity vector,
it would look something like that.
Now let's do the next part.
The figure below shows the string at the same instant
as shown in part A-one.
At each of the dots P and Q, draw an arrow
indicating the direction of the instantaneous
acceleration of that dot.
If either dot has zero acceleration,
write A equals zero next to the dot.
So pause this video again and see
if you can have a go at it.
One way to think about it is all the points around P
and actually all the points in this area right over here
are going to have a net upward acceleration.
So even though this point for example right over here
has a velocity downwards, that downward velocity
is becoming less and less and less as you get closer
to P and then it flips around, becomes an upward velocity,
and becomes more and more upward as we get closer
and closer to this point.
So everything in this area right over here
has a net upward acceleration.
So P would have a net upward acceleration as well.
It has zero instantaneous velocity
but we're, at that moment, we're going from
a downward velocity to an upward velocity.
So we have an upward acceleration.
What about at point Q?
Well, one interesting thing to appreciate
is that everything in this region right over here
has an upward acceleration and everything over here
has a downward acceleration,
and then everything over here has an upward acceleration.
And so point Q is at that transition point
where we're going from a downward acceleration
to an upward acceleration.
And so here, right at that point at this moment,
you're going to have zero acceleration.
So here, we write acceleration is equal to zero.
And then it says, the figure below represents
the string at time t equals zero,
the same instant as shown in part A
where dot P is at its maximum displacement from equilibrium.
For simplicity, dot Q is not shown.
On the grid below, draw the string at a later time,
time t is equal to capital T divided by four,
where capital T is the period of the wave.
So pause this video again and see if you could do that,
you can even sketch it out if you don't have access
to this drawing right here.
To think about a full period,
we can look at a full wavelength of this wave
and so one way to do it is you start at a point
and then you see, well how far do you have to go
to get to this same point again?
So we went up, we go down,
you could say we're at the same point
but here, we're going up, and here we're going down.
So we wanna go back to when we're going back up.
So that is a full wavelength.
A full wavelength here is 24 centimeters.
So after one-fourth of a period,
this wave would've shifted one-fourth of this length.
And so, one-fourth would be one of these sections
right over here.
So all I will do is draw the same wave
but instead of starting here, I'm gonna start here.
One-fourth of a wave length to the right.
But then I'll draw the exact same wave.
It just shifted.
So notice, it has shifted one-fourth
of a wavelength to the right.
And so it will look something
like that.
And then if we look at the next parts of the question,
on your drawing above, draw a dot to indicate the position
of point P on the string at time t is equal
to the period divided by four and clearly label
the dot with the letter P.
So let's just remind ourselves of where P was before.
P is at this point that's halfway
between 12 and 24 centimeters.
So that's gonna be at 18 centimeters.
So it's right over here.
So at this point, P is going to be right over here.
So from T is equal to zero to T is equal
to the period divided by four,
P went from down here up to here.
And then last but not least, it says
now consider the wave at time T is equal to a full period.
Determine the distance traveled, not the displacement,
by dot P between the times t equals zero
and t is equal to capital T.
So pause this video and see if you can figure it out.
So we already saw at time t is equal to zero,
P is right over here.
And so what's gonna happen over a full period?
Well, over that time period, P is gonna move
all the way up to positive eight centimeters on our Y axis
and then all the way back down over a full period.
And so if you this displacement, you'd say it's zero,
you're back to where you started but the distance,
well it would've traveled eight, 16 centimeters up
and then 16 centimeters back down.
So the distance traveled would be 32 centimeters.
And we're done.