Question 122: Oscillations

Question

A simple pendulum is being used to determine
th value of gravitational acceleration g at a
certain place. Th length of the pendulum is
25.0 cm and a stop watch with 1s resolution
measures the time taken for 40 oscillations to
be 50 s. The accuracy in g is :

Accepted Answer

T = $$2\pi \sqrt {{l \over g}} $$
$$ \Rightarrow $$ $$g = {{4{\pi ^2}l} \over {{T^2}}}$$
$$ \Rightarrow $$ $${{\Delta g} \over g} = {{\Delta l} \over l} + 2{{\Delta T} \over T}$$
= $${{0.1} \over {25}} + 2{1 \over {50}}$$ = $${{11} \over {250}}$$
$$ 	herefore $$  % accuracy = $${{11} \over {250}}$$ $$ 	imes $$ 100% = 4.40 %

Answer

4.40%

Answer

3.40%

Answer

2.40%

Answer

5.40%

Oscillations

A simple pendulum is being used to determine th value of gravitational acceleration g at a certain place. Th length of the pendulum is 25.0 cm and a stop watch with 1s resolution measures the time taken for 40 oscillations to be 50 s. The accuracy in g is :

Choose the correct answer: