Due to some force $F_1$ a body oscillates with period $4/5\, sec$ and due to other force $F_2$ oscillates with period $3/5\, sec$. If both forces act simultaneously, the new period will be .... $\sec$
Diffcult
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(c) Under the influence of one force ${F_1} = m\omega _1^2y$ and under the action of another force, ${F_2} = m\omega _2^2y$.
Under the action of both the forces $F = {F_1} + {F_2}$
$ \Rightarrow m{\omega ^2}y = m\omega _1^2y + m{\omega ^2}y$
Under the action of both the forces $F = {F_1} + {F_2}$
$ \Rightarrow m{\omega ^2}y = m\omega _1^2y + m{\omega ^2}y$
$ \Rightarrow \omega _1^2 + \omega _2^2$
$ \Rightarrow {\left( {\frac{{2\pi }}{T}} \right)^2} = {\left( {\frac{{2\pi }}{{{T_1}}}} \right)^2} + {\left( {\frac{{2\pi }}{{{T_2}}}} \right)^2}$
$ \Rightarrow T = \sqrt {\frac{{T_1^2T_2^2}}{{T_1^2 + T_2^2}}} $
$ = \sqrt {\frac{{{{\left( {\frac{4}{5}} \right)}^2}{{\left( {\frac{3}{5}} \right)}^2}}}{{{{\left( {\frac{4}{5}} \right)}^2} + {{\left( {\frac{3}{5}} \right)}^2}}}} = 0.48\,sec $
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