A body executes simple harmonic motion under | vedclass

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(A) For a body of mass $m$ executing $SHM$,the restoring force is $F = m \omega^2 x$,where $\omega = \frac{2\pi}{T}$.Thus,$F = m \left( \frac{2\pi}{T}

Vedclass · Accepted Answer

$12/25$

Vedclass · Answer

(A) For a body of mass $m$ executing $SHM$,the restoring force is $F = m \omega^2 x$,where $\omega = \frac{2\pi}{T}$.Thus,$F = m \left( \frac{2\pi}{T} \right)^2 x = \frac{4\pi^2 m}{T^2} x$.This implies $F \propto \frac{1}{T^2}$,or $F = kx$ where $k = \frac{4\pi^2 m}{T^2}$.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 $k_{eff} = k_1 + k_2$.Therefore,$\frac{4\pi^2 m}{T^2} = \frac{4\pi^2 m}{T_1^2} + \frac{4\pi^2 m}{T_2^2}$.This simplifies to $\frac{1}{T^2} = \frac{1}{T_1^2} + \frac{1}{T_2^2}$.Given $T_1 = 4/5 \ s$ and $T_2 = 3/5 \ s$,we have $\frac{1}{T^2} = \frac{1}{(4/5)^2} + \frac{1}{(3/5)^2} = \frac{25}{16} + \frac{25}{9}$.$\frac{1}{T^2} = 25 \left( \frac{9 + 16}{144} \right) = 25 \left( \frac{25}{144} \right) = \frac{625}{144}$.Taking the square root,$\frac{1}{T} = \frac{25}{12}$,so $T = \frac{12}{25} \ s$.

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