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13.Oscillations
easy
The ratio of frequencies of two pendulums are $2 : 3$, then their length are in ratio
A
$\sqrt {2/3} $
B
$\sqrt {3/2} $
C
$4/9$
D
$9/4$
Solution
(d) Frequency $n \propto \frac{1}{{\sqrt {\rm{l}} }}$
==> $\frac{{{n_1}}}{{{n_2}}} = \sqrt {\frac{{{l_2}}}{{{l_1}}}} $
==>$\frac{{{l_1}}}{{{l_2}}} = \frac{{n_2^2}}{{n_1^2}} = \frac{{{3^2}}}{{{2^2}}} = \frac{9}{4}$
Standard 11
Physics
