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Question

(A) For a body of mass $m$ executing $SHM$ under a force $F = -kx$, the time period is $T = 2\pi \sqrt{\frac{m}{k}}$.Since $F = kx$, we have $k = \fra

Vedclass · Accepted Answer

$\frac{12}{25}$

Vedclass · Answer

(A) For a body of mass $m$ executing $SHM$ under a force $F = -kx$, the time period is $T = 2\pi \sqrt{\frac{m}{k}}$.Since $F = kx$, we have $k = \frac{F}{x}$, so $T = 2\pi \sqrt{\frac{mx}{F}}$.This implies $T^2 \propto \frac{1}{F}$, or $F \propto \frac{1}{T^2}$.Let $F_1 = \frac{c}{T_1^2}$ and $F_2 = \frac{c}{T_2^2}$, where $c$ is a constant.When both forces act simultaneously in the same direction, the net force is $F_{net} = F_1 + F_2$.The new time period $T$ satisfies $F_{net} = \frac{c}{T^2}$.Substituting the expressions: $\frac{c}{T^2} = \frac{c}{T_1^2} + \frac{c}{T_2^2}$.$\frac{1}{T^2} = \frac{1}{T_1^2} + \frac{1}{T_2^2} = \frac{T_1^2 + T_2^2}{T_1^2 T_2^2}$.$T = \frac{T_1 T_2}{\sqrt{T_1^2 + T_2^2}} = \frac{(4/5) 	imes (3/5)}{\sqrt{(4/5)^2 + (3/5)^2}} = \frac{12/25}{\sqrt{16/25 + 9/25}} = \frac{12/25}{\sqrt{25/25}} = \frac{12}{25} \ s$.

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