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An experiment to measure the acceleration due to gravity g of a pendulum is done. The...
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.
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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
Posted by william1941 on October 5, 2010 at 3:17 AM (Answer #1)
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.
Posted by mike-krupp on October 5, 2010 at 7:52 AM (Answer #2)
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