The period of a simple pendulum is given by \(T=2 \pi \sqrt{L / g}\) where is the length of the pendulum and is the acceleration due to gravity. Thus if and are measured, we can estimate with \(g=4 \pi^{2} L / T^{2}\). Assume that the period is known to be \(T=1.5 \mathrm{~s}\) with negligible uncertainty, and that is measured to be \(0.559 \pm 0.005 \mathrm{~m}\). Estimate , and find the uncertainty in the estimate.

Equation Transcription:

Text Transcription:

T=2{pi}sqrt{L/g}

g=4{pi}^{2}L/T2

T=1.5 s

0.559{+/-}0.005 m

Solution :

Step 1 of 1:

Let the period perod of a simple pendulum is given by

Where,

L = length of the pendulum and

g = acceleration due to gravity.

So m

Here L=0.559 and m

We assume that the the period is T=1.5 s.

Our goal is :

We need to estimate g and the uncertainty in the estimate.

Now we have to find the estimate of g and the uncertainty in the estimate.

The formula of the acceleration is given by

We know that L and T values.

Therefore the acceleration due to gravity is .

Now we have to find the estimate of the uncertainty.

Consider,

Differentiate above equation with respect to “L” we get

Here we substitute and T values.

Hence .

Now,

The estimate of uncertainty is given by

Here m.

Hence .

Therefore, The estimate of uncertainty is g = 9.80810.0087