A simple pendulum suspended from the ceiling of a stationary lift has period $T_{0}$ . When the lift descends at steady speed, the period is $T_{1}$ , and when it descends with constant downward acceleration, the period is $T_{2}$ . Which one of the following is true?

T_{0} = T_{1} = T_{2}

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T_{0} = T_{1} < T_{2}

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T_{0} = T_{1} > T_{2}

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T_{0} < T_{1} < T_{2}

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Solution

The correct option is B $T_{0} = T_{1} < T_{2}$

The formula for time period is $T = 2 π \sqrt{\frac{L}{T^{'} / m}}$ where $T^{'} = t e n s i o n$ in string

For no acceleration of lift, $T = m g$

$⟹ T_{o} = T_{1} = 2 π \sqrt{\frac{L}{g}}$

When lift accelerates downward-

$m g - T e n s i o n = m a ⟹ \frac{T e n s i o n}{m} = g - a$

$⟹ T_{2} = 2 π \sqrt{\frac{L}{g - a}}$

$⟹ T_{0} = T_{1} < T_{2}$

Answer-(B)

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A simple pendulum suspended from the ceiling of a stationary lift has period T0. When the lift descends at steady speed, the period is T1, and when it descends with constant downward acceleration, the period is T2. Which one of the following is true?

The correct option is B T0=T1<T2The formula for time period is T=2π√LT′/m where T′=tension in stringFor no acceleration of lift, T=mg⟹To=T1=2π√LgWhen lift accelerates downward-mg−Tension=ma⟹Tensionm=g−a⟹T2=2π√Lg−a⟹T0=T1<T2Answer-(B)

A simple pendulum suspended from the ceiling of a stationary lift has period $T_{0}$ . When the lift descends at steady speed, the period is $T_{1}$ , and when it descends with constant downward acceleration, the period is $T_{2}$ . Which one of the following is true?