A mass is suspended separately by two springs of spring constant $k_{1}$ and $k_{2}$ in successive order. The time periods of oscillations in the two cases are $T_{1}$ and $T_{2}$ 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^{2} = T_{1}^{2} + T_{2}^{2}

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

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

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T = T_{1} + T_{2}

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Solution

The correct option is A $T^{- 2} = T_{1}^{- 2} + T_{2}^{- 2}$
$T = 2 π \sqrt{\frac{M}{k}}$
$∴$ $k = \frac{4 π^{2} m}{T^{2}}$
In the given situation
$k = k_{1} + k_{2}$
$∴$ $\frac{4 π^{2} m}{T^{2}} = \frac{4 π^{2} m}{T_{1}^{2}} + \frac{4 π^{2} m}{T_{2}^{2}}$
$∴$ $T^{- 2} = T_{1}^{- 2} + T_{2}^{- 2}$

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The correct option is A T−2=T−21+T−22T=2π√Mk∴ k=4π2mT2In the given situationk=k1+k2∴ 4π2mT2=4π2mT21+4π2mT22∴ T−2=T−21+T−22