Question

The period of oscillation of a simple pendulum is $T=2\pi \sqrt{\frac{L}{g}}.$ Length $L$ is about $10cm$ and is known to $1mm$ accuracy. The period of oscillation is about $0.5s$. The time of $100$ oscillations is measured with wristwatch of $1s$ time period. What is error in the determination of $g$ ?

• A. $2.5\%$
• B. $5\%$
• C. $7.5\%$
• D. $10\%$

Answer 1

Answer: (B)

$5\%$

Here, $T=2\pi \sqrt{\frac{L}{g}}$ or $T^2=4{\pi }^2\left(L/g\right)$

So, $g=\frac{4{\pi }^{2}L}{T^2}$

Thus, $\frac{\Delta g}{g}=\frac{\Delta L}{L}+2\frac{\Delta T}{T}$

$\%$ error in $g$ $=\frac{\Delta g}{g}\times 100$

$=(\frac{\Delta L}{L}+2\frac{\Delta T}{T})\times 100$

$=(\frac{\left(1/10\right)}{10}+2\frac{\left(1/100\right)}{0.5})\times 100=5\%$