Question 148: Oscillations

Question

The period of oscillation of a simple pendulum is $$T = 2\pi \sqrt {{L \over g}} $$.  Measured value of L is 20.0 cm known to 1 mm accuracy and time for 100 oscillations of the pendulum is found to be 90 s using wrist watch of 1 s resolution. The accuracy in the determination of g is:

Accepted Answer

Given $$T = 2\pi \sqrt {{L \over g}} $$
$$ \Rightarrow g = {{4{\pi ^2}L} \over {{T^2}}}$$
$$ \Rightarrow g = {{4{\pi ^2}L{n^2}} \over {{t^2}}}$$
[ as $$T = {t \over n}$$ ]
So, percentage error in $$g$$ =
$${{\Delta g} \over g} 	imes 100 = {{\Delta L} \over L} 	imes 100 + 2{{\Delta t} \over t} 	imes 100$$
= $${{0.1} \over {20.0}} 	imes 100 + 2 	imes {1 \over {90}} 	imes 100$$
= 2.72 % = 3 %

Answer

1 %

Answer

5 %

Answer

2 %

Answer

3 %

Oscillations

The period of oscillation of a simple pendulum is $T = 2\pi \sqrt {{L \over g}} $. Measured value of L is 20.0 cm known to 1 mm accuracy and time for 100 oscillations of the pendulum is found to be 90 s using wrist watch of 1 s resolution. The accuracy in the determination of g is:

Choose the correct answer: