Question

A simple pendulum is being used to determine the value of gravitational acceleration$g~$at a certain place. The length of the pendulum is $25.0cm~$and a stopwatch with $1s$ resolution measures the time taken for $40$ oscillation to be $50s$. The accuracy in$g~$is:
(a) $2.40$
(b) $3.40$
(c) $5.40$
(d) $4.40$

Answer 1

In order to solve this question, we need to use the formula of simple pendulum. The period of a simple pendulum is$T=2\pi \sqrt{\dfrac{L}{g}}$, where $L$is the length of the string and $g$ is the acceleration due to gravity.
For simple pendulum,
$T=2\pi \sqrt{\dfrac{L}{g}}$
So, $\dfrac{\Delta T}{T}=\dfrac{1}{2}\left( \dfrac{\Delta L}{L}+\dfrac{\Delta g}{g} \right)$
Given that,
$\Delta T=1\sec \text{ and }T=50\sec$
$\Delta L=0.1cm\text{ and }L=25cm$
$\dfrac{\Delta g}{g}=\dfrac{2\Delta T}{T}+\dfrac{\Delta L}{L}$
$\dfrac{\Delta g}{g}=2\times \dfrac{1}{50}+\dfrac{0.1}{25}=\dfrac{1.1}{25}$
$\dfrac{\Delta g}{g}=\dfrac{1.1}{25}$
$$

Therefore, the correct answer for this question is option (d).

Additional Information: A pendulum is essentially a weight that is hung from a fixed point. It is placed in such a way that it allows the device to swing freely to and fro. The pendulum bob of a simple pendulum is treated as a point mass. Further, the string from which it’s hanging is of negligible mass.f you look at it from the perspective of physics, you will find these simple pendulums quite intriguing. This is so because they serve as a great example of simple harmonic motion, which is much similar to rubber bands or springs.

Note: While solving this question, we need to have good knowledge about the simple pendulum and its motion. The formula used to solve this question must be used carefully and all the values for the variables should be kept in place. This question may be solved using different methods as per the convenience.