Question

A simple pendulum is suspended from the ceiling of a lift. When the lift is at rest its time period is T. With that acceleration should the lift be accelerated upwards in order to reduce its period to T/2? (g is the acceleration due to gravity)
a)2g b)3g c)4g d)g
[This question is from 12th MH Board chapter Oscillation and Waves]
Friends, please help me with the above question!!

Answer 1

ANSWER:

• The upward acceleration of the lift = 3g m/s².

GIVEN:

• A simple pendulum is suspended from the ceiling of a lift.

• When the lift is at rest its time period is T.

• New time period of the pendulum = T/2.

TO FIND:

• The upward acceleration of the lift.

EXPLANATION:

$\pink\bigstar\green{ \boxed{\large{\bold{\orange{T = 2\pi\sqrt{\dfrac{l}{g}}}}}}} \\ \\ \\ \sf T' = 2 \pi \sqrt{\dfrac{l}{a} } \\ \\ \\ \sf W.K.T. \ T' = \dfrac{T}{2} \\ \\ \\ \sf \dfrac{T}{2} = 2 \pi \sqrt{\dfrac{l}{a} } \\ \\ \\ \sf \dfrac{ 2 \pi}{2} \sqrt{\dfrac{l}{g} } = 2 \pi \sqrt{\dfrac{l}{a} } \\ \\ \\ \sf \dfrac{ 1}{2} \sqrt{\dfrac{1}{g} } = \sqrt{\dfrac{1}{a} } \\ \\ \\ \sf \dfrac{ 1}{4g} = \dfrac{1}{a} \\ \\ \\ \sf a_{net} = 4g \\ \\ \\ \sf a_{net} = g + a_{upwards} \\ \\ \\ \sf 4g = g + a_{upwards} \\ \\ \\ \sf a_{upwards} = 3g$

•°• The upward acceleration of the lift = 3g m/s².

Answer 2

Question :

A simple pendulum is suspended from the ceiling of a lift. When the lift is at rest its time period is T. With that acceleration should the lift be accelerated upwards in order to reduce its period to T/2? (g is the acceleration due to gravity)

a) 2g b) 3g c) 4g d) g

Required Answer :

• The upward acceleration of the lift = 3g m/s²

Step by step explanation :

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Using formula :

$\qquad\red\bigstar\:{\underline{\bf{\green{T = 2\pi \sqrt{\dfrac{1}{g}}}}}}$

Also,

$\qquad\red\bigstar\:{\underline{\bf{\green{T' = 2\pi \sqrt{\dfrac{1}{a}}}}}}$

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$:\Rightarrow\qquad\tt W \:K\:T\:T' = \dfrac{T}{2}$

$:\Rightarrow\qquad\tt \dfrac{T}{2} = 2\pi \sqrt{\dfrac{1}{a}}$

$:\Rightarrow\qquad\tt \dfrac{2\pi}{2}\sqrt{\dfrac{1}{g}} = 2\pi \sqrt{\dfrac{1}{a}}$

$:\Rightarrow\qquad\tt \dfrac{\cancel{2\pi}}{2}\sqrt{\dfrac{1}{g}} = \cancel{2\pi} \sqrt{\dfrac{1}{a}}$

$:\Rightarrow\qquad\tt \dfrac{1}{2}\sqrt{\dfrac{1}{g}} = \sqrt{\dfrac{1}{a}}$

Squaring both sides :

$:\Rightarrow\qquad\tt \bigg(\dfrac{1}{2}\sqrt{\dfrac{1}{g}}\bigg)^2 = \bigg(\sqrt{\dfrac{1}{a}}\bigg)^2$

$:\Rightarrow\qquad\tt \dfrac{1}{4}\:\times\:\dfrac{1}{g} = \dfrac{1}{a}$

$:\Rightarrow\qquad\tt \dfrac{1}{4g} = \dfrac{1}{a}$

$:\Rightarrow\qquad{\boxed{\frak{\pink{a\:=\:4g}}}}\:\purple\bigstar$

Using formula :

$\qquad\red\bigstar\:{\underline{\bf{\green{a_{(net)} = g + a_{(upwards)}}}}}$

Putting all known values :

$:\Rightarrow\qquad\tt 4g = g + a_{(upwards)}$

$:\Rightarrow\qquad\tt a_{(upwards)} = 4g - g$

$:\Rightarrow\qquad{\boxed{\frak{\pink{a_{(upwards)}\:=\:3g}}}}\:\purple\bigstar$

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$\therefore\:{\underline{\frak{Option\:b)\:3g\:is\:correct}}}$