A pendulum suspended from the roof of an elevator at rest has a time period $T_{1}$ . When the elevator moves up with an acceleration $‘ a^{'}$ , its time period becomes $T_{2}$ ; when the elevator moves down with an acceleration $‘ a^{'}$ ; its time period becomes $T_{3}$ , then

T_{1} = T_{2} = T_{3}

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

T_{2} > T_{3} > T_{1}

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

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

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

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

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

Open in App

Solution

The correct option is C $T_{1} = \frac{T_{2} T_{3} \sqrt{2}}{\sqrt{T_{2}^{2} + T_{3}^{2}}}$
When elevator is at rest, time period of pendulum is
$T_{1} = 2 π \sqrt{\frac{l}{g}}$
$\Rightarrow$ When elevator is moving up with acceleration
$a$ , Then $T_{2} = 2 π \sqrt{\frac{l}{g + a}}$
$\Rightarrow$ When elevator moves down with acceleration $a$ , then
$T_{3} = 2 π \sqrt{\frac{l}{g - a}}$
$\Rightarrow T_{3} > T_{2}$ and $T_{1}$ both
$\Rightarrow T_{1} = \frac{T_{2} T_{3} \sqrt{2}}{\sqrt{T_{2}^{2} + T_{3}^{2}}}$

Suggest Corrections

Similar questions

Q. A pendulum suspended from the roof of an elevator at rest has a time period

T_{1}

, when the elevator moves up with an acceleration a its time period becomes

T_{2}

, when the elevator moves down with an acceleration a, its time period becomes

T_{1}

, then.

Q. A simple pendulum of length

1 m

is freely suspended from the ceiling of an elevator. The time period of small oscillations as the elevator moves up with an acceleration of

2 m / s^{2}

is (use

g = 10 m / s^{2}

)

A simple pendulum of length l is suspended through the ceiling of an elevator. Find the time period of small oscillations if the elevator (a) is going up with an acceleration $a_{0}$ (b) is going down with an acceleration $a_{0}$ and (c) is moving with a uniform velocity.

Q. A pendulum bob is hanging from the roof of an elevator with the help of a light string. When the elevator moves up with a uniform acceleration

^{'} a^{'}

, the tension in the string is

T_{1}

. When the elevator moves down with the same acceleration, the tension in the string is

T_{2}

. If the elevator was stationary, the tension in the string would be

Q. A seconds pendulum is suspended from the roof of a lift. If the lift is moving up with an acceleration

9.8 m / s^{2}

, its time period is

Join BYJU'S Learning Program

A pendulum suspended from the roof of an elevator at rest has a time period T1. When the elevator moves up with an acceleration ‘a′, its time period becomes T2; when the elevator moves down with an acceleration ‘a′; its time period becomes T3, then

The correct option is C T1=T2T3√2√T22+T23When elevator is at rest, time period of pendulum is T1=2π√lg ⇒ When elevator is moving up with acceleration a, Then T2=2π√lg+a ⇒ When elevator moves down with acceleration a, then T3=2π√lg−a ⇒T3>T2 and T1 both ⇒T1=T2T3√2√T22+T23

A pendulum suspended from the roof of an elevator at rest has a time period $T_{1}$ . When the elevator moves up with an acceleration $‘ a^{'}$ , its time period becomes $T_{2}$ ; when the elevator moves down with an acceleration $‘ a^{'}$ ; its time period becomes $T_{3}$ , then