Question

n an experiment of a simple pendulum, measurements show that the pendulum has length m, mass kg, and period s. Take m/s2 . i. Use the measured length to predict the theoretical pendulum period with a range of error (use the error propagation method you learned in Lab 1). ii. Compute the percentage difference (as defined in Lab 1) between the measured value and the predicted value .

Answer 1

Answer:

The answer is "$(1.265 \pm 0.010) \ s \ and \ 0.709 \%$"

Explanation:

In point i:

$T_{theo}= 2\pi \sqrt{\frac{l}{g}}$

        $=2\pi\sqrt{\frac{0.397}{9.8}}\\\\= 1.265 \ s$

If  error in the theoretical time period :

$\frac{\Delta T_{theo}}{T_theo} = \frac{1}{2} \frac{\Delta l }{l}\\\\\Delta T_{theo} = 1.265 \times \frac{1}{2} \times \frac{0.006}{0.397}$

           $= 0.010 \ s$

 $T_{theo} = (1.265 \pm 0.010) \ s$

In point ii:

$\% \ difference = \frac{|T_{exp} -T_{theo}|}{\frac{T_{exp}+T_{theo}}{2}} \times 100$

                    $= \frac{1.274 -1.265}{\frac{1.274+1.265}{2}} \times 100\\\\=0.709 \%$