What is the relationship between period and mass in a simple pendulum?

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1. A simple pendulum consists of a light inextensible string AB with length L, with the end A fixed, and a perticle of mass M attatched to B. The pendulum oscillates with period T.

It is suggested that T is proportional to the product of powers of M, L and g. Use dimensional analysis to find this relationship.



2. T = K (l/g)1/2



3. ?? Dont know where to start on this one. some sort of substitution to find the variables, but i don't know how.
 
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also, i am presuming that little g is the surface gravity, ie, 9.81. I think that's what that represents, but i better check that out first.
 
Yes, g stands for the acceleration due to gravity. What are its dimension?
 
g is in N/Kg i think? is that what you mean by its dimension?
 
Those are units. By dimensions I mean something independent of particular units. The fundamental dimensions are length (L), mass (M), and time (T). So, how would you express the dimension of g in terms of these quantities?

Here's a wiki page that might help you: http://en.wikipedia.org/wiki/Dimensional_analysis"
 
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Doc Al said:
Here's a wiki page that might help you: http://en.wikipedia.org/wiki/Dimensional_analysis"

Yep, sure did help, they pretty much has this exact question as an example on that page;

What is the period of oscillation T of a mass m attached to an ideal linear spring with spring constant k suspended in gravity of strength g? The four quantities have the following dimensions: T [T]; m [M]; k[M / T2]; and g[L / T2]. From these we can form only one dimensionless product of powers of our chosen variables, G1 = T2k / m. The dimensionless product of powers of variables is sometimes referred to as a dimensionless group of variables, but the group, G1, referred to means "collection" rather than mathematical group.

so, i basically substitiute dimensions, equate indices and solve.

T = kma Lb gc => T = Ma Lb (LT-2)c

a = 0

-2c = 1 => c = 0.5

b + c = 0 => b = 0.5,

...so T = K (L/G)1/2

I think that's right, cheers. :smile:
 
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