Question

The periof of oscilliation of a simple pendulum is $T=2\pi \sqrt{\frac{L}{g}}$, Measured value of L is 20.0 cm known to 1 mm accuracy and time for 100 osciliiation of the pendulum is found to be 90 s using wrist watch of 1 s resolution. The accuracy in the determination of g is:

• A. 3%
• B. 1%
• C. 5%
• D. 2%

Answer 1

Answer: (A)

3%

$\begin{array}{c}T=2\pi \sqrt{\frac{l}{g}}\\ \sqrt{g}=\frac{2\pi \sqrt{l}}{T}\\ g=\frac{4{\pi }^2}{T^2}l\\ \log g=\log 4x^2+\log l\\ -2\log T\end{array}$

$\begin{array}{c}\text{differentiating both sides:}\\ \frac{dg}{g}=\frac{dl}{l}-\frac{2dT}{T}\\ \text{but, error is always}\\ \text{added,}\\ \therefore \text{.}\\ \frac{dg}{g}=\frac{dl}{l}+\frac{2dT}{T}\\ \text{for \% error,}\\ \frac{dg}{g}\times 100\%=\frac{dl}{l}\times 100\%+\frac{2dT}{T}\times 100\%\end{array}$

$\begin{array}{c}\text{also,}\\ T=t/m\end{array}$

$\begin{array}{c}\log T=\log t-\log n\\ \text{differentiating both sides,}\\ \frac{dT}{T}=\frac{dt}{t}\\ \frac{dT}{T}\times 100\%=\frac{dt}{t}\times 100\%\end{array}$

$\text{now}$

$\begin{array}{cc}\frac{dg}{g}\times 100\%=& \frac{dl}{l}\times 100\%\\ & +\frac{2dt}{t}\times 100\%\\ \frac{dg\times 100\%}{9}=& \frac{0.1}{20}\times 100\\ & +\frac{2\times 1\times 100}{90}\\ & =2.22+0.5\\ & \\ & =2.72\%\end{array}$

$\approx 3\%$