Câu 112:
\(\eqalign{
& \left\{ \matrix{
{T_1} = 2\pi \sqrt {{{{l_1}} \over g}} \hfill \cr
{T_2} = 2\pi \sqrt {{{{l_1}} \over g}} \hfill \cr
T = 2\pi \sqrt {{{{l_1} + {l_2}} \over g}} \hfill \cr} \right. \Rightarrow {T^2} = T_1^2 + T_2^2 \cr
& \Rightarrow {T_2} = \sqrt {{T^2} - T_1^2} = \sqrt {{1^2} - {{0,8}^2}} = 0,6s \cr} \)
Câu 113:
Chu kì: \(T = 2\pi \sqrt {{l \over g}} = 2\pi \sqrt {{{0,8} \over {9,81}}} \approx 1,8s\)
Câu 114:
Ta có: \(T = 2\pi \sqrt {{l \over g}} \Rightarrow l = {{{T^2}.g} \over {4{\pi ^2}}} = {{{{1,5}^2}.9,8} \over {4{\pi ^2}}} = 0,56m = 56cm\)
Câu 115:
Ta có:
\(\eqalign{
& \left\{ \matrix{
{T_1} = 2\pi \sqrt {{1 \over g}} = 2s \hfill \cr
{T_2} = 2\pi \sqrt {{3 \over g}} \hfill \cr} \right. \Rightarrow {{{T_1}} \over {{T_2}}} = \sqrt {{1 \over 3}} \cr
& \Rightarrow {T_2} = \sqrt 3 {T_1} = 2\sqrt 3 = 3,46s \cr} \)
