A system pendulum is oscillating in a lift.If the lift is going down with constant velocity,the time period of the simple pendulum is $T_{1}$ .If the lift is going down with some retardation its time period is $T_{2}$ then

T_{1} > T_{2}

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T_{1} < T_{2}

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T_{1} = T_{2}

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Depends upon the mass of the pendulum bob

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Solution

The correct option is A $T_{1} > T_{2}$

Time period of pendulum with constant velocity is given by
$T_{1} = 2 π \sqrt{\frac{l}{g}}$
while the time period of pendulum in a lift moving downwards with constant retardation is given by
$T_{2} = 2 π \sqrt{\frac{l}{g_{e f f}}} = 2 π \sqrt{\frac{l}{g - (- a)}} = 2 π \sqrt{\frac{l}{g + a}}$
its clear from above two equations that $T_{1} > T_{2}$ , Since denominator of the second equation is larger than that of first equation.

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A system pendulum is oscillating in a lift.If the lift is going down with constant velocity,the time period of the simple pendulum is T1.If the lift is going down with some retardation its time period is T2 then

A system pendulum is oscillating in a lift.If the lift is going down with constant velocity,the time period of the simple pendulum is $T_{1}$ .If the lift is going down with some retardation its time period is $T_{2}$ then