Experimental Physics Question 6
Question 6 - 2024 (31 Jan Shift 2)
The measured value of the length of a simple pendulum is $20 \mathrm{~cm}$ with $2 \mathrm{~mm}$ accuracy. The time for 50 oscillations was measured to be 40 seconds with 1 second resolution. From these measurements, the accuracy in the measurement of acceleration due to gravity is $\mathrm{N} %$. The value of $\mathrm{N}$ is:
(1) 4
(2) 8
(3) 6
(4) 5
Show Answer
Answer: (3)
Solution:
$$ \begin{aligned} \mathrm{T} & =2 \pi \sqrt{\frac{\ell}{\mathrm{g}}} \ \mathrm{g} & =\frac{4 \pi^{2} \ell}{\mathrm{T}^{2}} \ \frac{\Delta \mathrm{g}}{\mathrm{g}} & =\frac{\Delta \ell}{\ell}+\frac{2 \Delta \mathrm{T}}{\mathrm{T}} \ & =\frac{0.2}{20}+2\left(\frac{1}{40}\right) \ = & \frac{0.3}{20} \end{aligned} $$
Percentage change $=\frac{0.3}{20} \times 100=6 %$
