Solution Found!
The period T of a simple pendulum is given by where L is
Chapter 3, Problem 8E(choose chapter or problem)
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
Questions & Answers
QUESTION:
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
ANSWER:
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.