Question

The period of a pendulum T is assumed to depend only on the mass m, the length of the pendulum `, the acceleration due to gravity g, and the angle of swing θ. By means of dimensional analysis, simplify this problem and express this dependence in non dimensional terms.

Answer 1

The expression is shown in the explanation below:

Step-by-step explanation:

Thinking process:

Let the time period of a simple pendulum be given by the expression:

`T = \pi \sqrt{(l)/(g) }`

Let the fundamental units be mass= M, time = t, length = L

Then the equation will be in the form

`T = M^(a)l^(b)g^(c)`

`T = KM^(a)l^(b)g^(c)`

where k is the constant of proportionality.

Now putting the dimensional formula:

`T = KM^(a)L^(b) [LT^(-) ^(2)]^(c)`

`M^(0)L^(0)T^(1) = KM^(a)L^(b+c)`

Equating the powers gives:

a = 0

b + c = 0

2c = 1, c = -1/2

b = 1/2

so;

a = 0 , b = 1/2 , c = -1/2

Therefore:

`T = KM^(0)l^{(1)/(2) } g^{(1)/(2) }`

T =
`2\pi \sqrt{(l)/(g) }`

where k =
`2\pi`