Two masses $m_{1}$ and $m_{2}$ $(m_{1} > m_{2})$ are suspended by two springs vertically and are in equilibrium, extensions in the springs were same. Both the masses are displaced in the vertical direction by same distance and released. In subsequence motion $T_{1}, T_{2}$ are their periods and $E_{1}, E_{2}$ are the energies of oscillations respectively then :

$T_{1} = T_{2}; E_{1} < E_{2}$

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$T_{1} > T_{2}; E_{1} > E_{2}$

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$T_{1} < T_{2}; E_{1} > E_{2}$

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$T_{1} = T_{2}; E_{1} > E_{2}$

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Solution

The correct option is D
$T_{1} = T_{2}; E_{1} > E_{2}$

$K_{1} = m_{1} \frac{g}{x}; K_{2} = m_{2} \frac{g}{x}$

Thus, using $T = 2 π \sqrt{\frac{m}{k}}$

$T_{1} = 2 π \sqrt{\frac{m_{1}}{(\frac{m_{1} g}{x})}}$

$T_{1} = 2 π \sqrt{\frac{m_{2}}{(\frac{m_{2} g}{x})}}$

$\Rightarrow T_{1} = T_{2}$

Energy of oscillation :

Since $ω$ and A are same for both and $m_{1} > m_{2} \Rightarrow E_{1} > E_{2}$

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The correct option is D T1=T2;E1>E2 K1=m1gx;K2=m2gx Thus, using T=2π√mk T1=2π√m1(m1gx) T1=2π√m2(m2gx) ⇒T1=T2 Energy of oscillation : Since ω and A are same for both and m1>m2⇒E1>E2