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Question

(D) For a body of mass $m$ executing $SHM$ with force constant $k$,the time period is $T = 2\pi \sqrt{\frac{m}{k}}$,which implies $k = \frac{4\pi^2 m}

Vedclass · Accepted Answer

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

Vedclass · Answer

(D) For a body of mass $m$ executing $SHM$ with force constant $k$,the time period is $T = 2\pi \sqrt{\frac{m}{k}}$,which implies $k = \frac{4\pi^2 m}{T^2}$.When the first force acts,$k_1 = \frac{4\pi^2 m}{T_1^2}$.When the second force acts,$k_2 = \frac{4\pi^2 m}{T_2^2}$.When both forces act simultaneously in the same direction,the effective force constant is $k_{eff} = k_1 + k_2$.Substituting the expressions for $k$,we get $\frac{4\pi^2 m}{T^2} = \frac{4\pi^2 m}{T_1^2} + \frac{4\pi^2 m}{T_2^2}$.Dividing by $4\pi^2 m$,we obtain $\frac{1}{T^2} = \frac{1}{T_1^2} + \frac{1}{T_2^2}$.$\frac{1}{T^2} = \frac{T_1^2 + T_2^2}{T_1^2 T_2^2}$.Taking the reciprocal and square root,$T = \frac{T_1 T_2}{\sqrt{T_1^2 + T_2^2}} \, s$.

$A$ body executes $SHM$ under the influence of one force and has a period $T_1 \, s$. The same body executes $SHM$ with period $T_2 \, s$ when under the influence of another force. When both forces act simultaneously and in the same direction,the time period of the same body is:

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