Let's consider the formula for the period of a simple
pendulum of length L is formula is T period.
Is equal to 2π?
Times the square root of L over G&L is the length of
the pendulum.
Now on Earth, we tend to regard G's being fixed it.
There is a little with altitude, but we tend to
think of it as being fixed.
Just supposing we had a pendulum of a fixed length L, and we took
it somewhere else. Let's say the moon or Mars.
Somewhere out of the solar
system. Gravity there would not be the same and we might want to
measure gravity, so one of the ways will be to take our
pendulum, set it swinging.
And measure the time.
Once we measured the time, we could then use that to Calculate
G, but before we could do it, we would need to know what G was in
terms of the rest of the symbols in this formula. So we need to
rearrange this formula so that it said G.
Equals.
Now it's what?
Goes instead of this question mark that we're going to have a
look at in this video. We're going to be looking at how we
transform formally, how we move from an expression like this.
To another expression.
Involving exactly the same variables, but one that helps us
answer the questions that were after in terms of a different
variable. To do this?
Transformation of formula will need all the techniques that we
had in terms of solving equations, so the video solving
linear equations in one variable might be very useful to have a
look at, but let's just recap by having a look.
A simple linear equation.
So the one going to take three X +5.
Equals 6 - 3 * 5
- 2 X.
Now we've got this equation we want to solve it for X, so our
aim is to end up with X equals.
So one of the first things we would do is multiply out
the bracket.
So we have minus three times by 5 is minus 15.
Minus three times by minus 2X is plus 6X.
Next, we can simplify this bit. Here three X +5
equals 6X, and now six takeaway 15 is minus 9.
Now, one of the things we would want to do is to try and get all
the ex is together. So we were three X here under 6X there, so
will take 3X away from each side, so that's three X
takeaway. Three X +5 equals 6X takeaway, 3X minus 9.
3X takeaway 3X the X is go. We've known left and then we've
got this. Five is equal to six X takeaway 3X. That's three X
again takeaway 9.
Now we need to get the constant terms, the numbers together. So
to do that we would add 9 to each side. So we 5 + 9 is equal
to three X minus 9 + 9.
So again, we're doing to the same thing to both sides. Here
we took three X away from each side. Here we're adding nine to
each side, so five and nine is 14 equals 3X, and we minus 9 +
9.0. Now we need to divide both sides by three.
So we have 3X over three an 14 over 3, dividing both sides by
three, and so those three canceled and we're left with X
equals 14 over 3. Now I've written this out very fully.
I've written out every step that I've said, but we wouldn't
normally expect to see all of that written down.
Starting from here.
The next line we'd expect to see is this one because
we say Take 3X away from both sides and we'd expect
to see that as the result.
Then we'd say add 9 to both sides, and So what we would
expect to see having done that would be that.
And then we, say, divide both sides by three.
And so we see that So what we would see written down in our
exercise book or on a piece of paper that carried the
solution to this equation will be something like this.
OK, we've been through the steps now of solving an
equation. These techniques, particularly idea of a balance
of keeping the same on both sides by doing the same thing
to both sides is what we're going to do when we look at
the transformation of formally.
So let's begin with our first formula V equals U
plus AT. And the variable that we're going to try and
find is TI.
T is the variable we're going to find what is T expressed?
In terms of the rest of the letters or symbols that there
are in this formula well, in order to model this, what I'm
going to do is I'm going to write down an equation which
looks like this, but it just got
numbers in. With a T there as the unknown. So what we might
have is something like this. Seven for V5 for you.
Let's say a tool for A and then T.
Now what would we do to solve this equation? The first thing
we try and do is get the TS on their own and so to do that,
we take 5 away from each side. So this would be 2. Over here
equals 2 T, so having done that there, let's do it here.
V minus U taking you away from each side.
Now we would divide both sides by two over here.
In order to end up with just T on its own, so let's do the
same. Here. A is what multiplies by T, so we need to divide
everything on this side V minus U over a equals T. That's it.
Notice everything here is over a, not just one part of it, but
both the whole expression.
The minus U over a.
Let's take another one. V squared equals
U squared plus 2A S.
And this time, let's say the symbol that I'm going to try and
find all the variables that I'm going to try and find in terms
of all the others, is you.
So that's the case. Again, I'm going to write down an
equation over here which looks like this one.
So let's have.
25
equals U squared.
+9.
Weather 9 is in place of this lump here of algebra.
So what would we do here? Our first step would be to take
nine away from both sides. So let's do that. 25 - 9 equals
U squared and that gets us EU squared on its own. So let's
do that over here. Let's take this lump away from both
sides. 20 squared minus 2A S is equal to U squared.
Now what I would want to do now is to take the square root of
both sides 'cause I want just you and here I've got you
squared. So I need the square root of 25 - 9 and I want the
square root of all of that not square root of 25 minus the
square root of 9. I want the square root of 25 - 9, so again
I've got to do the same over here I want the square root of.
All. Love it.
Let's just check why I need the square root of all of this.
25 - 9 is 16, so that 16 equals U squared. Take the square root
of both sides. Four is equal to you and we're happy. That's the
answer. But what if I do the square root of 25 minus the
square root of nine? Well, that's 5 - 3 is 2, which is not
the answer very definitely not the answer, so we need to take
the square root of everything, not just each little piece. And
so this over here has to be the square root of all of that.
I've been working so far with the formula that are for uniform
acceleration, so let's continue with that and take.
Another one of them S equals UT plus 1/2 AT squared, and
this time it's a.
Then I'm going to be looking for can I rearrange this formula?
Can I transform it so it says a
equals? Again, let me model it with an equation. Let's say that
S is 21, but U times by T is 15 + 1/2 of a
Times by 9.
And it's this a the time after first of all. Let's isolate the
term with the unknown in it. So that's the ater. So let's take
15 away from both sides. So I'm 21 - 15 is equal to 1/2 of
a times by 9. So again, let's do that here. Let's take this lump
away. S minus Utah equals 1/2.
AT squared, so again, my first steps are to try and isolate the
variable. The term that I'm
looking for. Now this one looks a bit complicated. We
got a half. There would be nice to get rid of the half,
so let's multiply both sides by two. If we do it at this
side, that's just a times by 9. We do it this side 2 * 21
- 15 and it multiplies all of it. So let's do the same
here. Multiply both sides by 2, two times S minus UT.
Equals AT squared.
Going back to the equation we got equals a Times by 9. I just
want a on its own and so I must divide both sides by 9.
Here the thing that's doing the multiplying it's T
squared. So let's divide both sides by T squared and in the
same way as I divided everything by 9, I've got to
divide everything here by T squared.
And that gives me a.
Notice I haven't made any effort to cancel because T
is not a common factor. It is not a common factor.
No, we've been working with the equations that are to do with
uniform acceleration. But there are some other kinds of
equations that we need to gain practice at and in order to
develop these skills, I'm going to start with some made up
equations that don't have physical applications in the
real world, but we will come back once we've developed those
skills to looking at some real equations that do contain
physical applications that we will be able to transform, but
we need to develop some skills to begin with. So first of all.
Let's take this expression, let's call it rather than a
formula.
So then we have Y times 2X plus one equals X Plus One and the
thing that we're going to be after is X. Can we rearrange it
so it says X equals?
Well gain, let me try and model this.
With an equation. So all I need to do to model
this as an equation is replaced the Y by three.
And if I was to solve this inequation, my first step would
be to multiply out the bracket. So let's do that. 6X plus three
is equal to X plus one.
So let's do that over here. Multiply out this bracket. So
that's two XY Plus Y.
Is equal to X plus one.
Our next step with the equation will be to get all the excess
together. So I would take X away from both sides, so that would
be 6X takeaway X +3, and taking the X away from this side just
leaves me with one. So let's do that here. Let's take this X
away from both sides, so have two XY minus X Plus Y.
Is equal to 1.
Now I would naturally want to combine these two in some way.
6X minus X is just 5X.
I can't really do that at this side, but what I can do is
gather together the terms in X by taking out X as a common
factor. So we take out X from this as a common factor. The
other factor is 2 Y minus.
And taking X out of there is one.
Plus Y equals 1.
What now? Well now I've got my ex is together over here. Let's
write this as 5X plus 3 equals 1. My next step will be to leave
the X term on its own by taking three away from each side.
So let's do that here. Lead the Exterm on its own by
taking the other term. That's why away from both sides.
Now I'm just going to simplify this. Five X equals minus two,
and in order to solve this to get a value of XI need to divide
both sides by this number 5. So that's X equals minus two over
5. So if I come back to this this term in the bracket, two Y
minus one is the term that's multiplying the X, and so I need
to divide. Both sides by so X is equal to 1 minus Y over
2Y plus. Sorry made a mistake there. Two Y minus one.
I just look again at what we've done here. We've mimic the
solving of an equation. First, we multiplied out the brackets.
Then we got the terms together
in X. The variable that we were after.
Having got those terms together.
We then isolated them so that they were on their own.
So we need to bear that in mind and follow it through.
Let's take another expression Y over Y plus X.
+5 is equal to X.
This time. We're going to be finding why we're going to be
getting Y equals, and we want some lump of numbers and X is
over this sigh.
So let's model this with an equation Y over Y plus 3
+ 5 is equal to 3.
Look at this side. Here we've got an algebraic fraction.
Y over Y plus 3Y plus three is the denominator. It's in the
bottom of the fraction.
So in order to get rid of that, we need to multiply everything
in this equation by this denominator. So I'm going to
write that out in full Y over Y, plus three. Put that in a
bracket times Y plus 3.
Plus five times Y.
Three is equal to three times Y plus three. Let me just
emphasize we have to do the same thing to both sides, so here
I've had to multiply everything on both sides of the equation by
this term Y plus 3.
Here those will cancel out, just leaving me with why, so
let's do that at this site. Let's multiply everything by
this term Y plus X so we know this first one. When we've
multiplied it by wiper sex is just going to give us Why.
Plus five times by Y plus X is equal to
X times by Y plus X.
Coming back to the equation here, faced with a lot of
brackets, we know the first thing we would do is multiply
out the brackets. So let's do that, why?
Plus five times by Y is 5 Y five times by three is 15
is equal 2 three times by Y is 3 Y and three times by
three is 9.
Do the same here, why?
+5 Y. Five times by X Plus 5X is
equal to X times by Y is XY&X times by X is X.
Square.
On this side now we would like to get all of our wise together.
So we've Y plus
5Y6Y. Take away 3 Y so we've six. Why
already and we're going to take away 3Y. Plus
15 is equal to 9, so let's gather all the
wise together Y +5. Y is 6 Y.
I need to bring this term over, so I'll take XY away from both
sides minus XY.
And then let's just write down the other terms.
Now here I simplify this. I'd have six Y takeaway 3 Y and that
would just leave me with three Y plus 15 equals 9. I can't do
that here, but the thing that I can do is bring them much closer
together by taking out why.
The thing that I'm after as a common factor. So let's take why
out of these two terms. Why brackets? Why times by 6 Y means
I must have a 6 in their minus XY means I must have a minus X
there plus 5X equals X squared.
Now over here I'd isolate this term in why by taking 15 away
from both sides 3 Y is equal to. Now this is 9 - 15,
so I've got to do the same here. Take 5X away from each
side, isolating this term in YY times 6 minus X is equal to
X squared minus.
5X.
This side now I've got 3 Yi, just need Y so I would divide
everything on this side by three. So I've got to do the
same with this. I've got to divide everything on this side
by this term 6 minus X so we have X squared minus 5X.
Divided by 6 minus X.
Put it all in a bracket to keep it together. Notice I'm
attempting no canceling 'cause there is no common factor in
this numerator and this denominator. They do not share a
common factor, so that's my answer and I finished. I've got
Y equals a lump of algebra involving X is and numbers.
We started with a real formula. We started with the formula for
a simple pendulum.
Of length Lt equals 2π square root.
L over G and the problem that we posed was what was G in terms of
T? So G.
After G equals.
Let's have a look at a simple equation. Let's say it's 10
equals 2π and I'll keep the two Pike's. Part is just a number,
so 2π is just a number.
Square root of 3 over G.
Now as an equation, this one is a bit tricky because the G is
trapped inside this square root sign. We need to bring it out
and the way to get a square root sign to disappear, so to speak,
is to square both sides, reverse the operation if you like, so
will square. Both sides of this equation so 10 multiplied by
itself. I'm going to write it as 10 squared equals 2π multiplied
by itself, 'cause we're squaring the whole of both sides times by
a nice square this.
Three over G.
Now we need to do the same here, square the whole of both sides.
So that will be T squared.
2π all squared L over G.
Coming back to the equation, jeez, in the denominator I don't
want it there. I want it to say G equals I want G upstairs, so
to speak, so I have to get rid of it out of the denominator and
so to do that I must multiply both sides by G.
So I've got 10 squared times. G is equal to 2π squared
times three, so let's do that here. Multiply both sides by G.
So I have T squared times by G is equal to 2π
all squared times by L.
Gee, I want on its own. I want to say G equal, so I must divide
by this thing 10 squared. So G is equal to 2π all squared times
3, all over 10 squared.
And I can work that out. Let's do the same with this G equals
2π all squared times by L, and I'm getting rid of this T
squared by dividing everything by T squared.
And we can leave it like that.
We can make it look a little bit nicer if we want to buy
spotting that we've got a square here under Square
there, and it might be nice to combine those two.
Squaring process is by writing 2π over T all
squared times by L.
Basically this is the answer that we finished with.
Let's take another couple of.
Examples of real formula that we might want to be able to
manipulate. So let's have a look at the lens Formula One over F.
Is equal to one over U.
This one over V.
Let's say it's you.
That we're after we want you
equals. Well, let me write down an equation.
Similar.
One over 3 equals 1 over U plus one over 5.
I want to isolate this term first in one over you.
To do that, I'm going to take this away from both sides. 1/3
takeaway 150 is equal to one
over you. So let's do that. Isolate this term in you.
And so to do that, we take away one over V from both sides, so
we want over F.
Take away one over V is equal to one over you.
Now, faced with problems like this, it's very, very
tempting to simply turn everything upside down.
Well, I just have a think about that. This says 1/3 - 1/5. Now a
third is bigger than 1/5.
So 1/3 - 1/5 is a positive number, so you at the very least
has got to be positive.
Watch what happens if I just turn everything upside down.
3 - 5 equals U, which tells me that minus two is equal to you.
But we just agreed that you had to be positive and not negative.
You can't do that.
These are fractions, and so because their fractions we have
to combine them as fractions, which means we have to find a
common denominator and a common denominator for three and five
is a number that both three and five will divide into.
Easiest number is 3 times by 5.
So three Zing to three times by 5, which is 15 goes five times.
So I found multiplied 3 by 5. I must multiply by this one by 5,
so that's 5 minus.
I've multiplied 5 by three, so I multiply this one by three
equals one over. You know if I did that here, I've got to do it
over here. So I want to common denominator.
Here the common denominator I talk was three times by 5, so
let's take F times by vis.
I've multiplied F.
By the so I must multiply the
one. By faith.
Minus.
I've multiplied the V by F.
So I must multiply the one by F.
Now if I come back over here, let me just simplify this. This
is 2 over 15 is one over you, and now I've got a complete
fraction on both sides. I can turn it upside down and say
that's what you is.
Now there was a lot of calculation went on here. I
can't do that calculation here, but I have got a complete
fraction here so I can turn it.
Upside down.
To give me you.
So let's take as our final example the time dilation
formula from relativity.
The formula is T equals T, not.
Divided by.
1. Minus. V squared over C squared to the
power. 1/2 or square root.
So let's take some numbers and in fact what we're going to be
after is we're going to be looking for this expression
here, V over C. So I just write that down there. That's
what we're going to be looking for. Can we get V over C in
terms of T&T Nord?
So let's have 6 equals 5 over 1
minus X squared.
To the half. Now solving this.
What would we do? Well, we've got a square root. Let's Square
both sides in order to get rid of that square root. So that
would be 6 squared equals 5 squared all over 1 minus X
squared. So let's do that here. Square both sides.
T squared. Equals TN squared all over 1
minus V squared over C squared.
Now, well here the term I want in X is in the denominator. I
need it upstairs, so let's multiply both sides of this
equation by one minus X squared. That would be 6
squared times 1 minus X squared is equal to 5 squared.
So let's do that here.
T squared times, 1 minus V squared over
C squared is equal to T Nord squared.
Now I want the X squared bit.
And it would be nice perhaps to multiply out the bracket, but
there's a slightly quicker way. I can divide both sides by 6
squared, and that's nice. 'cause it keeps the square bits
together, so to speak. So let me do that one minus X squared is
equal to 5 squared over 6 squared. So I'm going to do that
here. Divide both sides by T squared, 1 minus V squared over
C squared. Is equal to TN squared over T squared?
Now.
Running out of paper here. So what I'm going to do is
turn over and write this one down at the top of the next
page, and this one as well.
So we've got 1 minus V squared over. C squared is
equal to T not squared over T squared.
And here, with one minus X squared is equal to 5 squared
over 6 squared.
It's The X squared that we're after so I can add X squared
both sides, so this is one equals 5 squared over 6 squared
plus X squared. So let me add V squared over C squared to each
side. One is equal to T not squared over T squared plus B
squared over C squared.
Now I want the X squared on it so.
So take this away from both sides. 1 - 5 squared over
6 squared is equal to X squared, so 1 minus T, not
squared over T squared is equal to V squared over C squared.
Taking this away from both
sides. Almost there now it's X that I want not X squared
selects. Take the square root of
both sides. So do the same here. The oversee that I
want, so let's take the square root of both sides.
And that's it. We have found the ratio of the velocity to
the speed of light in terms of the two times.
So. That concludes this video on transformation of formula thing
to do is to treat it always as though it was an equation that
you were trying to solve.
In terms of the variable that you want, X equals G equals and
you just perform the steps as though you were solving an
equation. If you keep that in mind, then most of your problems
should be taken care of.