$T_{1}$ , $T_{2}$ are time periods of oscillation of a block individually suspended to spring of force constants $K_{1}$ , $K_{2}$ respectively. If same block is suspended to parallel combination of same two springs, its time period is

T_{1} + T_{2}

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

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

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\frac{T_{1} T_{2}}{\sqrt{{T_{1}}^{2} + {T_{2}}^{2}}}

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Solution

The correct option is D $\frac{T_{1} T_{2}}{\sqrt{{T_{1}}^{2} + {T_{2}}^{2}}}$

for a parallel combination $k = k_{1} + k_{2}$
$T = 2 π \sqrt{\frac{m}{k}}$

$= 2 π \sqrt{\frac{m}{(k_{1} + k_{2})}}$

$T_{1} = 2 π \sqrt{\frac{m}{k_{1}}}$

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

$k_{2} = (2 π)^{2} \frac{m}{T_{2}^{2}}$

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

$∴ \frac{T_{1} T_{2}}{\sqrt{T_{1} + T_{2}}} = T$

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T1, T2 are time periods of oscillation of a block individually suspended to spring of force constants K1 ,K2 respectively. If same block is suspended to parallel combination of same two springs, its time period is

The correct option is D T1T2√T12+T22 for a parallel combination k=k1+k2T=2π√mk =2π√m(k1+k2) T1=2π√mk1 ⇒k1=(2π)2mT21 k2=(2π)2mT22 k1+k2=(2π)2(mT21+mT22) ∴T1T2√T1+T2=T

$T_{1}$ , $T_{2}$ are time periods of oscillation of a block individually suspended to spring of force constants $K_{1}$ , $K_{2}$ respectively. If same block is suspended to parallel combination of same two springs, its time period is