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  • Answer please! A simple pendulum hanging from ceiling of oscillation of a stationary lift has a time period T1. When the lift moves downward with constant velocity, the time period is T2. Then

A simple pendulum hanging from ceiling of oscillation of a stationary lift has a time period T1. When the lift moves downward with constant velocity, the time period is T2 . Then

 

 

  • Option 1)

    T2 is infinty.

  • Option 2)

    T2 > T1

  • Option 3)

    T2 < T1

  • Option 4)

    T2 = T1

 

Answers (7)

best_answer

T\;\alpha\;\frac{1}{\sqrt{g}} and g is same in both cases so time period remains same

 

Time period of simple pendulum accelerating downward -

T= 2\pi \sqrt{\frac{l}{g-a}}

- wherein

l= length of pendulum

g= acceleration due to gravity.

a= acceleration of pendulum.

 

 

 


Option 1)

T2 is infinty.

This is incorrect.

Option 2)

T2 > T1

This is incorrect.

Option 3)

T2 < T1

This is incorrect.

Option 4)

T2 = T1

This is correct.

Posted by

Avinash

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2

Posted by

Suchitra Nair

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As per your explanation giving answer should be 4 T2=T1

Posted by

sanjay Gajjar

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If possible pl plarify and update me

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Posted by

sanjay Gajjar

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3)T1>T2

Posted by

Kacho Ahmad Khan

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4 option

 

Posted by

Raju Khapre

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Option 4). T2=T1

Posted by

Mahaveer

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