# An experiment to measure the acceleration due to gravity g of a pendulum is done. The length of threads l/m are 0.35, 0.65, 1.00, 1.45, 1.95 & ...time for 20 oscillation t/s are 24.1, 32.4,...

An experiment to measure the acceleration due to gravity g of a pendulum is done. The length of threads l/m are 0.35, 0.65, 1.00, 1.45, 1.95 & ...

time for 20 oscillation t/s are 24.1, 32.4, 40.1, 47.5, 56.3 respectively. Given that the relation between the period T, the length of the pendulum l and the acceleration due to gravity g is T=2pi x sq. rt. of l/g. Find the value of g.

### 2 Answers | Add Yours

The time period T of a pendulum is given as T = 2*pi*sqrt [ l / g ], where l is the length of the pendulum and g is the acceleration due to gravity.

Now for the pendulum of length:

0.35 m, T= 24.1 / 20 = 1.205

0.65 m, T = 32.4 / 20 =1.62

1.00 m, T= 40.1/20 = 2.00

1.45 m , T = 47.5 / 20 = 2.375

1.95 m , T = 56.3 /20 = 2.815

Now T = 2 pi sqrt [ l /g ]

So 2 = 2 pi sqrt [ 1 / g]

=> pi sqrt [ 1/g ] = 1

=> 1/g = (1/ pi )^2

=> g = pi ^2 m/s^2

**Therefore the value of g is pi^2 m/s^2**

I like to do this a different way, although we get the same result.

Start with T = 2 * pi * sqrt(L/g)

Solve for g: g = (4 * pi^2 * L)/(T/20)^2

I hope you can read my notation. The factor of T/20 is because we are given the period of 20 oscillations.

Substituting our data for L and T, we get

L T g

0.35 24.1 9.52

0.65 32.4 9.78

1.00 40.1 9.82

1.45 47.5 10.15

1.95 56.3 9.71

The value of 10.15 is probably an outlier that we could discard, but what the hey. The average value for g is

g = 9.80 m/s^2, in good agreement with the textbooks.

Comparing with the previous response:

His value for g = pi^2 = 9.87. I think his answer is different because he simplified, considering only the data for L= 1 and T/20 = 2.

Incidentally, I did all the arithmetic with Excel, the poor man's mathematical analyser.