A body executes SHM under the action of force | vedclass

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(D) For a body of mass $m$ executing $SHM$,the force is $F = kx$,where $k$ is the force constant. The time period is given by $T = 2\pi \sqrt{\frac{m}

Vedclass · Accepted Answer

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

Vedclass · Answer

(D) For a body of mass $m$ executing $SHM$,the force is $F = kx$,where $k$ is the force constant. The time period is given by $T = 2\pi \sqrt{\frac{m}{k}}$,which implies $k = \frac{4\pi^2 m}{T^2}$.For force $F_1$,$k_1 = \frac{4\pi^2 m}{T_1^2}$.For force $F_2$,$k_2 = \frac{4\pi^2 m}{T_2^2}$.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,$T^2 = \frac{4\pi^2 m}{k_1 + k_2}$.Taking the reciprocal,$\frac{1}{T^2} = \frac{k_1 + k_2}{4\pi^2 m} = \frac{k_1}{4\pi^2 m} + \frac{k_2}{4\pi^2 m}$.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 executes $SHM$ under the action of force $F_1$ with time period $T_1$. If the force is changed to $F_2$,it executes $SHM$ with time period $T_2$. If both the forces $F_1$ and $F_2$ act simultaneously in the same direction on the body,its time period is:

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