Question

The time period of oscillation of a simple pendulum is $T = 2\pi \sqrt {\dfrac{l}{g}} $, measured value of the length is 20cm known to have 1mm accuracy and time for 100 oscillation of the pendulum is found to be 90 s using wristwatch of 1s resolution. What is the accuracy in determining the value of acceleration due to gravity?

Answer 1

Hint: The formula for the time period of oscillation of a simple pendulum has been already provided in the question itself. This is a hint in itself. However, the problem is not easy and it still needs to be solved. The problem should be treated as a derivation because we will need to perform some mathematical operations for deriving the formula that we actually need.

Complete step by step answer:
The length of the pendulum (l) given in the problem is 20 cm, $l = 20cm$.
The accuracy for the measured value of length of the pendulum is 1 mm, i.e. $\Delta l = 0.1cm$
As given in the problem the time period for single oscillation is T, now assume t to be the time period of 100 oscillations, it is given by,
$t = 100T$
Taking log of both the sides we get,
$\ln (t) = \ln (100) + \ln (T)$
Differentiating the above equation we get,
$\dfrac{{\Delta t}}{t} = \dfrac{{\Delta T}}{T} = \dfrac{1}{{90}}$ ……. (1)
Now the time period of oscillation of a simple pendulum is
$T = 2\pi \sqrt {\dfrac{l}{g}} $
Therefore, the acceleration due to gravity is given by,
$ \Rightarrow g = 4{\pi ^2}\dfrac{l}{{{T^2}}}$
Now, after taking log of both the side the above equation becomes,
$\ln (g) = ln(4{\pi ^2}) + \ln (l) - \ln ({T^{ - 2}})$
$ \Rightarrow \ln (g) = ln(4{\pi ^2}) + \ln (l) + 2\ln (T)$
Differentiating the above equation we get,
$\dfrac{{\Delta g}}{g} = \left( {\dfrac{{\Delta l}}{l} + 2\dfrac{{\Delta T}}{T}} \right)$ ……… (2)
Putting the value from equation (1) in equation (2) we get,
$ \Rightarrow \dfrac{{\Delta g}}{g} = \left( {\dfrac{{0.1}}{{20}} + 2\dfrac{1}{{90}}} \right)$
$\therefore \dfrac{{\Delta g}}{g} = 0.0272 = 2.72\% $
Since we are talking about the accuracy of measurement be take the value on both the sides, hence we have, $\dfrac{{\Delta g}}{g} = \pm 2.72\%.$

Hence, the accuracy of determining the value of acceleration due to gravity is found to be $ \pm 2.72\% $.

Note: The problem talks about acceleration due to gravity. It is a very commonly used term and hence we tend to overlook its actual meaning. Let us just quickly understand what acceleration due to gravity actually is. Consider an object being allowed to free fall under the action of gravitational force only so the acceleration that is achieved by this object is known as acceleration due to gravity.