Question

A student conducted an experiment to measure the acceleration due to gravity,g, of a simple pendulum. The data obtained are tabulated:
LENGTH O THREAD,L/M 0.35 0.65 1.00 1.45 1.95
TIME FOR 20 24.1 32.4 40.1 47.5 56.3
OSCILLATIONS,T/S
Given that the relation between the periodic time,T,=2 pi square root of L/G, find the value of g using a graphical approach.

Answer 1

Answer on Question #39214, Physics, Other

A student conducted an experiment to measure the acceleration due to gravity, $g$, of a simple pendulum.

The data obtained are tabulated:

LENGTH O THREAD,L/M 0.35 0.65 1.00 1.45 1.95

TIME FOR 20 OSCILLATIONS,T/S 24.1 32.4 40.1 47.5 56.3

Given that the relation between the periodic time,T,=2 pi square root of L/G, find the value of g using a graphical approach.

Solution:

A simple pendulum is one which can be considered to be a point mass suspended from a string or rod of negligible mass. For small amplitudes, the period of such a pendulum can be approximated by:

$$T = 2 \pi \sqrt {\frac {L}{g}}$$

For $g$ we have:

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

The data obtained are tabulated:

LENGTH O THREAD,L/M 0.35 0.65 1.00 1.45 1.95

TIME FOR 20 OSCILLATIONS,T/S 24.1 32.4 40.1 47.5 56.3

Period of pendulum, the time of one oscillation is $\frac{\text{TIME FOR 20 OSCILLATIONS}}{20}$:

We draw a graph of $L$ with respect to $T^2$. We get a straight line passing through origin. The slope of this line is: $g / (4\pi^2)$.

For example, the slope of line is $\tan (\alpha) = 2 / 8 = 1 / 4$.

Thus,

$$\frac {g}{4 \pi^ {2}} \approx \frac {1}{4}$$ $$g \approx \pi^ {2} = 3.14 ^ {2} = 9.86$$

Answer. Finding graphically $g = 9.86 \, \text{m/s}^2$ .