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(B) For a spring-mass system,the time period is given by $T = 2\pi \sqrt{\frac{m}{k}}$.In the first case,for spring $k_1$,the time period is $T_1 = 2\

Vedclass · Accepted Answer

$T^{-2} = T_1^{-2} + T_2^{-2}$

Vedclass · Answer

(B) For a spring-mass system,the time period is given by $T = 2\pi \sqrt{\frac{m}{k}}$.In the first case,for spring $k_1$,the time period is $T_1 = 2\pi \sqrt{\frac{m}{k_1}}$. Squaring both sides,we get $T_1^2 = 4\pi^2 \frac{m}{k_1}$,which implies $\frac{1}{T_1^2} = \frac{k_1}{4\pi^2 m}$.In the second case,for spring $k_2$,the time period is $T_2 = 2\pi \sqrt{\frac{m}{k_2}}$. Squaring both sides,we get $T_2^2 = 4\pi^2 \frac{m}{k_2}$,which implies $\frac{1}{T_2^2} = \frac{k_2}{4\pi^2 m}$.When the two springs are connected in parallel,the effective spring constant is $k_{eff} = k_1 + k_2$. The time period $T$ is given by $T = 2\pi \sqrt{\frac{m}{k_1 + k_2}}$.Squaring both sides,we get $T^2 = 4\pi^2 \frac{m}{k_1 + k_2}$,which implies $\frac{1}{T^2} = \frac{k_1 + k_2}{4\pi^2 m}$.Substituting the expressions for $\frac{k_1}{4\pi^2 m}$ and $\frac{k_2}{4\pi^2 m}$,we get $\frac{1}{T^2} = \frac{1}{T_1^2} + \frac{1}{T_2^2}$.This can be written as $T^{-2} = T_1^{-2} + T_2^{-2}$.

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