A body of mass $m$ has time period $T_{1}$ with one spring and has time period $T_{2}$ with another spring. if both the spring are connected in parallel and same mass is used, then new time period $T$ is given as

T^{2} = T_{1}^{2} + T_{2}^{2}

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T = T_{1} + T_{2}

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\frac{1}{T} = \frac{1}{T_{1}} + \frac{1}{T_{2}}

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\frac{1}{T^{2}} = \frac{1}{T_{1}^{2}} + \frac{1}{T_{2}^{2}}

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Solution

The correct option is D $\frac{1}{T^{2}} = \frac{1}{T_{1}^{2}} + \frac{1}{T_{2}^{2}}$
Angular frequency, $ω = \sqrt{\frac{k}{m}} \Rightarrow \frac{2 π}{T} = \sqrt{\frac{k}{m}}$

$\Rightarrow k = m {(\frac{2 π}{T})}^{2}$ where, $T$ is time period.

Net Spring constant when two spring is connected in parallel.

$k = k_{1} + k_{2}$

$\Rightarrow m {(\frac{2 π}{T})}^{2} = m {(\frac{2 π}{T_{1}})}^{2} + m {(\frac{2 π}{T_{2}})}^{2}$

$\Rightarrow \frac{1}{T^{2}} = \frac{1}{{T_{1}}^{2}} + \frac{1}{{T_{2}}^{2}}$

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A body of mass m has time period T1 with one spring and has time period T2 with another spring. if both the spring are connected in parallel and same mass is used, then new time period T is given as

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A body of mass $m$ has time period $T_{1}$ with one spring and has time period $T_{2}$ with another spring. if both the spring are connected in parallel and same mass is used, then new time period $T$ is given as