Question 279: Units , Dimensions and Error Analysis

Question

A student performs an experiment for determination of $g\left(=\frac{4 \pi^2 l}{\mathrm{~T}^2}\right), l=1 m$, and he commits an error of $\Delta l$. For T , he takes the time of $n$ oscillations with the stop watch of least count $\Delta \mathrm{T}$ and he commits a human error of 0.1 s . For which of the following data, the measurement of $g$ will be most accurate?

Accepted Answer

Given:
$ g = \frac{4 \pi^2 l}{T^2} $
Errors are in $l$ and $T$.
$ l = 1 	ext{ m} $
$ \Delta l $ is given.
The student records total time for $n$ oscillations.
So, time for one oscillation:
$ T = \frac{T_n}{n} $
Step 1: Relate fractional errors
For $ g = 4 \pi^2 \frac{l}{T^2} $,
$ \frac{\Delta g}{g} = \frac{\Delta l}{l} + 2 \frac{\Delta T}{T} $
But note $T$ is the period of
one
oscillation, and the student measures total time $T_n$ for $n$ oscillations.
So,
$ T = \frac{T_n}{n} \implies \frac{\Delta T}{T} = \frac{\Delta T_n}{T_n} $
Determine uncertainty in $T_n$.
Step 2: Error in measured total time $ T_n $
Total error in timing for $n$ oscillations (using a stopwatch of least count $\Delta T$):
$ \Delta T_n = \sqrt{(\Delta T)^2 + (0.1)^2} $
since least count and human error combine.
So,
$ \frac{\Delta T}{T} = \frac{\sqrt{(\Delta T)^2 + (0.1)^2}}{T_n} $
Step 3: Estimate the time for one oscillation
For $l = 1\,	ext{m}$,
$ T = 2\pi \sqrt{\frac{1}{9.8}} \approx 2.0\,	ext{s} $
Thus $T \approx 2\,	ext{s}$.
Hence $T_n = nT = 2n\,	ext{s}$.
Step 4: Compute fractional errors
$ \frac{\Delta g}{g} = \frac{\Delta l}{l} + 2 \frac{\Delta T_n}{T_n} $
Now compute for each case.
Option A
$ \Delta l = 5	ext{ mm} = 0.005	ext{ m}, \; \Delta T = 0.2	ext{ s}, n = 10 $
$ \frac{\Delta l}{l} = 0.005 $
$ T_n = 2n = 20\,	ext{s} $
$ \Delta T_n = \sqrt{(0.2)^2 + (0.1)^2} = \sqrt{0.05} \approx 0.224 $
$ \frac{\Delta T_n}{T_n} = 0.224/20 = 0.0112 $
$ \frac{\Delta g}{g} = 0.005 + 2(0.0112) = 0.0274 $
Option B
$ \Delta l = 0.005,\ \Delta T = 0.2,\ n=20 $
$ T_n = 40	ext{ s} $
$ \Delta T_n = 0.224 $
$ \frac{\Delta T_n}{T_n} = 0.224/40 = 0.0056 $
$ \frac{\Delta g}{g} = 0.005 + 2(0.0056)=0.0162 $
Option C
$ \Delta l = 0.005, \; \Delta T = 0.1, \; n=10 $
$ \Delta T_n = \sqrt{(0.1)^2+(0.1)^2}=\sqrt{0.02}=0.141 $
$ T_n = 20\,	ext{s} $
$ \frac{\Delta T_n}{T_n}=0.141/20 = 0.00705 $
$ \frac{\Delta g}{g}=0.005 + 2(0.00705)=0.0191 $
Option D
$ \Delta l = 1	ext{ mm} = 0.001\,	ext{m},\ \Delta T=0.1,\ n=50 $
$ \Delta T_n = 0.141 $
$ T_n = 100\,	ext{s} $
$ \frac{\Delta T_n}{T_n} = 0.141 / 100 = 0.00141 $
$ \frac{\Delta g}{g} = 0.001 + 2(0.00141) = 0.00382 $
Ã¢ÂÂ
Option D
has the
smallest fractional error
, i.e. the
most accurate measurement of $ g $
.
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Final Answer:
Option D

Answer

$$\begin{array}{l}	riangle\mathcal l\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;	riangle\mathrm T\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\mathrm n\5\;\mathrm{mm}\;\;\;\;\;\;\;\;\;\;\;\;0.2\;\sec\;\;\;\;\;\;\;\;\;\;\;\;10\end{array}$$

Answer

$$\begin{array}{l}	riangle\mathcal l\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;	riangle\mathrm T\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\mathrm n\5\;\mathrm{mm}\;\;\;\;\;\;\;\;\;\;\;\;0.2\;\sec\;\;\;\;\;\;\;\;\;\;\;\;20\end{array}$$

Answer

$$\begin{array}{l}	riangle\mathcal l\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;	riangle\mathrm T\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\mathrm n\5\;\mathrm{mm}\;\;\;\;\;\;\;\;\;\;\;\;0.1\;\sec\;\;\;\;\;\;\;\;\;\;\;\;10\end{array}$$

Answer

$$\begin{array}{l}	riangle\mathcal l\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;	riangle\mathrm T\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\mathrm n\1\;\mathrm{mm}\;\;\;\;\;\;\;\;\;\;\;\;0.1\;\sec\;\;\;\;\;\;\;\;\;\;\;\;50\end{array}$$

Units , Dimensions and Error Analysis

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