Question

In an experiment, on the measurement of g using a simple pendulum the time period was measured with an accuracy of 0.2 % while the length was measured with accuracy of 0.5%. Calculate the percentage error in the value of g.
A. 0.7%
B. 0.1%
C. 0.2%
D. 0.9%

Answer 1

This problem can be solved with the concept of Relative Error and Measurement using Time Period formula. Using simple rules of error handling on multiplication and addition we can easily solve this problem. This is just a question of addition no in-depth knowledge of Harmonic Motion or time period calculation is required.

Complete step-by-step solution:
Time period formula using pendulum
$T = 2\pi \sqrt {\dfrac{L}{g}}$
Or
$g = L \times {\left( {\dfrac{{2\pi }}{T}} \right)^2}$ -----(1)
And we are asked to calculate the relative accuracy in g i.e.
$\dfrac{{\Delta g}}{g}$
putting the given values in the equation number (1)
we get:
$g = 0.5L \times {\left( {\dfrac{{2\pi }}{{0.2T}}} \right)^2}$

And we know that Relative Errors on multiplication and division get added.
And Relative Errors on exponential get multiplied.

Error in measuring length is $\dfrac{{\Delta l}}{l}$ = 0.5
Error in measuring time period is $\dfrac{{\Delta T}}{T} = 0.2$
Therefore:
$\dfrac{{\Delta g}}{g} = \dfrac{{\Delta l}}{l} + \dfrac{{2\Delta T}}{T}$
$= 0.5 + 2 \times 0.4$ $= 0.9\%$
Thus the percentage error in the value of g is 0.9%
Hence option (D) is correct.

Note:-
Relative Errors on multiplication and division get added like If error in measuring a is x and error in measuring b is y then:
error in measuring $\dfrac{a}{b}$ is $x + y$
error in measuring a$\times$b is $x + y$
error in measuring $a^3$ is $x \times 3$