A mass is suspended separately by two springs of spring constants k1 and k2 in successive order. The time periods of oscillations in the two cases are T1 and T2, respectively. If the same mass be suspended by connecting the two springs in parallel (as shown in figure), then the time period of oscillations is T. The correct relation is

T=2πmk or T∝1kor k=αT2Now, ke=k1 + k2∴ αT2=αT12+αT22or T-2 = T1-2 + T2-2

A mass is suspended separately by two springs of spring constants k1 and k2 in successive order. The time periods of oscillations in the two cases are T1 and T2, respectively. If the same mass be suspended by connecting the two springs in parallel (as shown in figure), then the time period of oscillations is T. The correct relation is

T=2πmk or T∝1kor k=αT2Now, ke=k1 + k2∴ αT2=αT12+αT22or T-2 = T1-2 + T2-2

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A mass is suspended separately by two springs of spring constants k₁ and k₂ in successive order. The time periods of oscillations in the two cases are T₁and T₂, respectively. If the same mass be suspended by connecting the two springs in parallel (as shown in figure), then the time period of oscillations is T. The correct relation is

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

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

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

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

answer is B.

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Detailed Solution

$T = 2 π \sqrt{\frac{m}{k}}$ or $T \propto \frac{1}{\sqrt{k}}$ or $k = \frac{α}{T^{2}}$

Now, $k_{e} = k_{1} + k_{2}$

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

or $T^{- 2} = T_{1}^{- 2} + T_{2}^{- 2}$

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