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(B) The time period of a spring-mass system is given by $t = 2\pi \sqrt{\frac{m}{k}}$.Squaring both sides,we get $t^2 = 4\pi^2 \frac{m}{k}$,which impl

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$\frac{t_1 t_2}{\sqrt{t_1^2 + t_2^2}}$

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(B) The time period of a spring-mass system is given by $t = 2\pi \sqrt{\frac{m}{k}}$.Squaring both sides,we get $t^2 = 4\pi^2 \frac{m}{k}$,which implies $k = \frac{4\pi^2 m}{t^2}$.For the two springs individually,the spring constants are $k_1 = \frac{4\pi^2 m}{t_1^2}$ and $k_2 = \frac{4\pi^2 m}{t_2^2}$.When the springs are connected in parallel,the effective spring constant is $k_{eq} = k_1 + k_2$.The new time period $t$ is given by $t = 2\pi \sqrt{\frac{m}{k_{eq}}} = 2\pi \sqrt{\frac{m}{k_1 + k_2}}$.Substituting the values of $k_1$ and $k_2$:$t = 2\pi \sqrt{\frac{m}{\frac{4\pi^2 m}{t_1^2} + \frac{4\pi^2 m}{t_2^2}}} = 2\pi \sqrt{\frac{m}{4\pi^2 m (\frac{1}{t_1^2} + \frac{1}{t_2^2})}}$.$t = \frac{1}{\sqrt{\frac{1}{t_1^2} + \frac{1}{t_2^2}}} = \frac{1}{\sqrt{\frac{t_2^2 + t_1^2}{t_1^2 t_2^2}}} = \frac{t_1 t_2}{\sqrt{t_1^2 + t_2^2}}$.

$A$ block of mass $m$ is suspended separately by two different springs having time periods $t_1$ and $t_2$. If the same mass is connected to a parallel combination of both springs,then its time period will be

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$A$ mass of $200 \text{ g}$ is suspended from a spring with a force constant of $80 \text{ N/m}$. What is the time period of the oscillation in seconds?

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