Question

the period of oscillation of a simple pendulum is T=2pi√4g. measured value of l is 20.0 cm known to 1mm accuracy and time for 100 oscillations of the pendulum is found to be 90s using a wrist watch of 1s resolution. What is the determination of g.

Answer 1

Answer:

The accuracy in the determination of the value of g with the given data is 2.72%.

Explanation:

The period of oscillation of a simple pendulum is given by

$\rm T = 2\pi \sqrt{\dfrac{l}{g}}\ ......\ (1).$

where,

• l = length of the pendulum.
• T = time period of oscillation of the pendulum.
• g = acceleration due to gravity.

Given that:

$\rm l = 20.0\ cm=0.2\ m.\\\Delta l = 1\ mm = 1\times 10^{-3}\ m.\\T = 90\ s.\\\Delta T = 1\ s.$

From equation (1),

$\rm T^2=(2\pi)^2 \times \dfrac lg\\\Rightarrow g = \dfrac{4\pi ^2l}{T^2}.$

The uncertainty in the value of g is given by

$\rm \dfrac{\Delta g}{g}=\dfrac{\Delta l}{l}+\dfrac{\Delta (T^2)}{T}=\dfrac{\Delta l}{l}+\dfrac{2\Delta T}{T}\\\\\dfrac{\Delta g}{g}=\dfrac{1\times 10^{-3}}{0.2}+\dfrac{1}{90}=0.0272.$

Thus, the accuracy in the determination of the value of g is given by

$\rm Accuracy= \dfrac{\Delta g}{g}\times 100\%=0.0272\times 100\% = 2.72\%.$