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
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Answer 1

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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

$\red{ \qquad \underline{ \pmb{{ \mathbb{ \maltese  \:  COMPLETE \: \: QUESTION \:   \maltese }}}}}$

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

$\huge\boxed{\fcolorbox{red}{ink}{SOLUTION:}}$

$\large \mathfrak{ \text{W}e \: \text{K}now }$

$T = 2\pi \sqrt{ \frac{T}{g} }$

• Thus, T is inversely proportional to square root of g.
• For, time period to become T/2, we have
• g eff= 4g

$\Large \orange{\qquad \underline{ \pmb{{ \mathbb{ \maltese  \:beacause \:  \maltese }}}}}$

$2\pi = \sqrt{ \frac{1}{4g} } = \frac{T}{2}$

$\purple{\qquad \qquad \underline{ \pmb{{ \mathbb{ \maltese  \:  REQUIRED \: \: INFO \:  \maltese }}}}}$

• Thus, to get g

eff =4g, lift should be moving upward with 3g, so that net acceleration of pendulum should be 3g(downward due to pseudo force)+g(downward due to gravity)=4g

$\Large\green{\qquad\underline{\pmb{{ \mathbb { \maltese  \:  Answer\:  \maltese }}}}}$

$\huge\boxed{\fcolorbox{red}{ink}{b)3g is correct option}}$