A simple pendulum in a lift moving up with a uniform acceleration ‘a’ has a time period $T_{1}$ and when the lift is moving down with the same acceleration has a time period $T_{2}$ . If the lift were stationary, find the time period of that pendulum?

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

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

No worries! We‘ve got your back. Try BYJU‘S free classes today!

T = \frac{\sqrt{2} T_{1} T_{2}}{\sqrt{T_{1}^{2} + T_{3}^{2}}}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

T = \frac{\sqrt{3} T_{1} T_{2}}{\sqrt{T_{1}^{2} + T_{2}^{2}}}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is A
The effective acceleration due to gravity when the lift moves up is (g+a) and when the lift moves down it is (g-a)
For a simple pendulum in a stationary lift $T = 2 π \sqrt{\frac{l}{g}} \Rightarrow g = \frac{4 π^{2} l}{T^{2}}$
When the lift moves up with uniform acceleration $g + a = \frac{4 π^{2} l}{T_{1}^{2}}$
When the lift moves down with uniform acceleration $g - a = \frac{4 π^{2} l}{T_{2}^{2}}$
Adding the two $2 g = \frac{4 π^{2} l (T_{1}^{2} + T_{2}^{2})}{T_{1}^{2} \times T_{2}^{2}},$ substituting $g = \frac{4 π^{2} l}{T^{2}}$ and solving $T = \frac{\sqrt{2} T_{1} T_{2}}{\sqrt{T_{1}^{2} + T_{2}^{2}}}$

Suggest Corrections

Similar questions

The time period of simple pendulum is T. When the length increases by 10 cm, its period is $T_{1}$ . When the length is decreased by 10 cm, its period is $T_{2}$ .Then the relation between T, $T_{1}$ and $T_{2}$ is

Q. Time period of a simple pendulum is

T

inside a lift when the lift is stationary. If the lift moves upward with an acceleration

g / 2

, the time period of the pendulum will be:

Q. A simple pendulum has time period '

t

'. Its time period in a lift which is moving upward with acceleration

3 m s^{- 2}

is :

Q. Time period of a simple pendulum inside a lift that is accelerating up at

2 {m s}^{2}

T_{1}

. If lift is retarding down at

6 m s

then time period of same pendulum is

T,

Then

T, / T_{2}

Q. A simple pendulum has period T inside a lift when it is stationary. The lift is accelerated upwards with constant acceleration a. The period

Join BYJU'S Learning Program

A simple pendulum in a lift moving up with a uniform acceleration ‘a’ has a time period T1 and when the lift is moving down with the same acceleration has a time period T2. If the lift were stationary, find the time period of that pendulum?

A simple pendulum in a lift moving up with a uniform acceleration ‘a’ has a time period $T_{1}$ and when the lift is moving down with the same acceleration has a time period $T_{2}$ . If the lift were stationary, find the time period of that pendulum?