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(B) The time period of a mass $m$ suspended by a spring of constant $k$ is given by $t = 2 \pi \sqrt{m/k}$.For the first spring with constant $k_1$,th

Vedclass · Accepted Answer

${t_0}^{-2} = {t_1}^{-2} + {t_2}^{-2}$

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(B) The time period of a mass $m$ suspended by a spring of constant $k$ is given by $t = 2 \pi \sqrt{m/k}$.For the first spring with constant $k_1$,the time period is $t_1 = 2 \pi \sqrt{m/k_1}$,which implies $k_1 = 4 \pi^2 m / t_1^2$.For the second spring with constant $k_2$,the time period is $t_2 = 2 \pi \sqrt{m/k_2}$,which implies $k_2 = 4 \pi^2 m / t_2^2$.When the two springs are connected in parallel,the effective spring constant is $k_{eff} = k_1 + k_2$.The time period of the system is $t_0 = 2 \pi \sqrt{m/k_{eff}} = 2 \pi \sqrt{m/(k_1 + k_2)}$.Squaring both sides,we get $t_0^2 = 4 \pi^2 m / (k_1 + k_2)$,so $1/t_0^2 = (k_1 + k_2) / (4 \pi^2 m)$.Substituting the expressions for $k_1$ and $k_2$:$1/t_0^2 = (4 \pi^2 m / t_1^2 + 4 \pi^2 m / t_2^2) / (4 \pi^2 m) = 1/t_1^2 + 1/t_2^2$.Thus,${t_0}^{-2} = {t_1}^{-2} + {t_2}^{-2}$.

$A$ mass $m$ is suspended separately by two different springs,and the time periods are $t_1$ and $t_2$ respectively. If it is connected by both springs in parallel as shown in the figure,then the time period is $t_0$. The correct relation is:

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