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(C) For a body performing $S.H.M.$ under a restoring force $F = kx$,the time period is given by $T = 2\pi \sqrt{\frac{m}{k}}$,which implies $k = \frac

Vedclass · Accepted Answer

$\frac{T_1 T_2}{\sqrt{T_1^2+T_2^2}}$

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(C) For a body performing $S.H.M.$ under a restoring force $F = kx$,the time period is given by $T = 2\pi \sqrt{\frac{m}{k}}$,which implies $k = \frac{4\pi^2 m}{T^2}$.Given $F_1 = k_1 x$ and $F_2 = k_2 x$,we have $k_1 = \frac{4\pi^2 m}{T_1^2}$ and $k_2 = \frac{4\pi^2 m}{T_2^2}$.When both forces act simultaneously in the same direction,the effective force constant is $k_{eff} = k_1 + k_2$.The new time period $T$ is given by $T = 2\pi \sqrt{\frac{m}{k_1 + k_2}}$.Squaring both sides,$T^2 = 4\pi^2 \frac{m}{k_1 + k_2}$,so $\frac{1}{T^2} = \frac{k_1 + k_2}{4\pi^2 m} = \frac{k_1}{4\pi^2 m} + \frac{k_2}{4\pi^2 m}$.Substituting the expressions for $k_1$ and $k_2$,we get $\frac{1}{T^2} = \frac{1}{T_1^2} + \frac{1}{T_2^2} = \frac{T_1^2 + T_2^2}{T_1^2 T_2^2}$.Therefore,$T = \frac{T_1 T_2}{\sqrt{T_1^2 + T_2^2}}$.

$A$ body performs $S.H.M.$ under the action of force $F_1$ with period $T_1$ seconds. If the force is changed to $F_2$,it performs $S.H.M.$ with period $T_2$ seconds. If both forces $F_1$ and $F_2$ act simultaneously in the same direction on the body,the period in seconds will be

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