Question

The period of oscillation of a simple pendulum is T = $2\pi \sqrt{\frac{L}{g}}$. Measured value of L is 10 cm known to 1 mm accuracy and time for 100 oscillations of the pendulum is found to be 50 s using a wrist watch of 1 s resolution. What is the accuracy in the determination of g?

• A. 2%
• B. 3%
• C. 4%
• D. 5%

Answer 1

The correct option is D 5%

Here, $T=2\pi \sqrt{\frac{L}{g}}$
Squaring both sides, we get, $T^2=\frac{4{\pi }^{2}L}{g}org=\frac{4{\pi }^{2}L}{T^2}$
The relative error in g is, $\frac{\Delta g}{g}=\frac{\Delta L}{L}+2\frac{\Delta T}{T}$
Here, $T=\frac{t}{n}$ and $\Delta T=\frac{\Delta t}{n}\therefore \frac{\Delta T}{T}=\frac{\Delta t}{t}$
The errors in both L and t are the least count errors.
$\therefore \frac{\Delta g}{g}=\frac{0.1}{10}+2\left(\frac{1}{50}\right)=0.01+0.04=0.05$
The percentage error in g is
$\frac{\Delta g}{g}\times 100=\frac{\Delta L}{L}\times 100+2\left(\frac{\Delta T}{T}\right)\times 100=[\frac{\Delta L}{L}+2\left(\frac{\Delta T}{T}\right)]\times 100=0.05\times 100=5$