How is this equation dimensionally homeogenous? Watch

tnk191
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can you please explain very detailed.... I'm so lost...
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tnk191
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How is this equation dimensionally homogeneous on both sides? step-by-step if possible please x
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RDKGames
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(Original post by tnk191)
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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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tnk191
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(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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ThomH97
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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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DFranklin
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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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RDKGames
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(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 [x_0] = L but it doesn't make sense for you to then say [x(t)] = LT since it's the same quantity. So the LHS has dimensions of L.

Then I don't think e even refers to the resistance. By the looks of it, I believe it refers to e= 2.718... (Euler's number)

Then if we say k is the spring constant, then [k] = LT^{-2} indeed.

And of course [c] = MT^{-1}.

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.

\left[ -\dfrac{c}{2M}\cdot t \right] = \dfrac{[-1][c]}{[2M]} \cdot [t]= \dfrac{MT^{-1}}{M} \cdot T = 1 so indeed we are good here. This expression has no dimension. (I assume M is mass kg but to avoid new notation just take [M] = M)

Now try the same thing for the expression inside cosine.
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