A body at the end of a spring executes S.H.M. | vedclass

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(B) The period of oscillation for a spring-mass system is given by $T = 2\pi \sqrt{\frac{M}{k}}$.For the first spring,$t_1 = 2\pi \sqrt{\frac{M}{k_1}}

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$T^2 = t_1^2 + t_2^2$

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(B) The period of oscillation for a spring-mass system is given by $T = 2\pi \sqrt{\frac{M}{k}}$.For the first spring,$t_1 = 2\pi \sqrt{\frac{M}{k_1}} \implies t_1^2 = 4\pi^2 \frac{M}{k_1} \implies k_1 = \frac{4\pi^2 M}{t_1^2}$.For the second spring,$t_2 = 2\pi \sqrt{\frac{M}{k_2}} \implies t_2^2 = 4\pi^2 \frac{M}{k_2} \implies k_2 = \frac{4\pi^2 M}{t_2^2}$.When two springs are connected in series,the effective spring constant $k_{eq}$ is given by $\frac{1}{k_{eq}} = \frac{1}{k_1} + \frac{1}{k_2} = \frac{k_1 + k_2}{k_1 k_2}$,so $k_{eq} = \frac{k_1 k_2}{k_1 + k_2}$.The period of oscillation for the series combination is $T = 2\pi \sqrt{\frac{M}{k_{eq}}}$.Squaring both sides,$T^2 = 4\pi^2 \frac{M}{k_{eq}} = 4\pi^2 M \left( \frac{k_1 + k_2}{k_1 k_2} \right) = 4\pi^2 M \left( \frac{1}{k_2} + \frac{1}{k_1} \right)$.Substituting the expressions for $k_1$ and $k_2$,we get $T^2 = 4\pi^2 M \left( \frac{t_2^2}{4\pi^2 M} + \frac{t_1^2}{4\pi^2 M} \right) = t_1^2 + t_2^2$.

$A$ body at the end of a spring executes $S.H.M.$ with a period $t_1$,while the corresponding period for another spring is $t_2$. If the period of oscillation with the two springs in series is $T$,then

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