# How is this equation dimensionally homeogenous?Watch

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#1

can you please explain very detailed.... I'm so lost...
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#2

How is this equation dimensionally homogeneous on both sides? step-by-step if possible please x
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7 months ago
#3
(Original post by tnk191)

How is this equation dimensionally homogeneous on both sides? step-by-step if possible please x
Have a go yourself first. What do you think??

Also, tell us what each variable represents. That would be a good way to start looking at what the dimensions are.
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#4
(Original post by RDKGames)
Have a go yourself first. What do you think??

Also, tell us what each variable represents. That would be a good way to start looking at what the dimensions are.
well,

LHS: the dimensionis just LT no?

RHS: bit messy but

x0 is distance so L?
e being resistance is (ML^2)/T^3.I^2
k being the spring constant in N/M is therefore (M.L)/T^2 * 1/M which is just L/T^2...
c being the viscosity in kg/s is M/T

I can isolate the base dimensions, but don't see how that will get me to L.T which is the LHS?
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7 months ago
#5
Rewrite each variable as its unit. So replace each M by kg, and M^2 is going to be (kg)^2 etc. You can rewrite the constants as '1' if necessary.

Ideally everything cancels out so you have the same units on each side of the equation.
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7 months ago
#6
Worth noting is that if you have a transcendental function (i.e. exp or sin or cos), the argument to that function must be dimensionless for things to make sense.
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7 months ago
#7
(Original post by tnk191)
well,

LHS: the dimensionis just LT no?

RHS: bit messy but

x0 is distance so L?
e being resistance is (ML^2)/T^3.I^2
k being the spring constant in N/M is therefore (M.L)/T^2 * 1/M which is just L/T^2...
c being the viscosity in kg/s is M/T

I can isolate the base dimensions, but don't see how that will get me to L.T which is the LHS?
I'd agree when you say but it doesn't make sense for you to then say since it's the same quantity. So the LHS has dimensions of .

Then I don't think even refers to the resistance. By the looks of it, I believe it refers to (Euler's number)

Then if we say is the spring constant, then indeed.

And of course .

Now then... take it bit by bit. One thing to note is that when you take cosine or exponential of a quantity, that quantity must be dimensionless.

So let's check that we are taking the exponential of a dimensionless quantity.

so indeed we are good here. This expression has no dimension. (I assume is mass kg but to avoid new notation just take )

Now try the same thing for the expression inside cosine.
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