A pendulum is hanging from the ceiling of a cage. When the cage is moving up with certain acceleration and when it is moving down with the same acceleration, the tensions in the string are $T_{1} and T_{2}$ respectively. When the cage moves horizontally with the same acceleration, the tension in the string is,

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\sqrt{2 (T_{1}^{2} - T_{2}^{2})}

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

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

\sqrt{2 (T_{1}^{2} + T_{2}^{2})}

answer is B.

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Detailed Solution

$T_{1} - mg = ma$

$T_{1} = ma + mg ↑$

$= m (a + g) ↑$

$T_{2} = m (g - a) ↓$

$T_{1}^{2} + T_{2}^{2} = m^{2} g^{2} + m^{2} a^{2} + 2 mga + m^{2} g^{2} + m^{2} a^{2} - 2 mga$

$= 2 (m^{2} g^{2} + m^{2} a^{2})$

$T \cos θ = mg$

$Tsin θ = ma$

$T^{2} = {(mg)}^{2} + {(ma)}^{2}$

$T^{2} = m^{2} (a^{2} + g^{2})$

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

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

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

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

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