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  • I have a doubt, kindly clarify. The time period of a simple pendulum in a lift desending with constant acceleration g is

The time period of a simple pendulum in a lift desending with constant acceleration g is

  • Option 1)

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

  • Option 2)

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

  • Option 3)

    zero

  • Option 4)

    infinte

 

Answers (1)

This is the case of freely falling lift and in free fall of lift effective g for pendulum will be zero. So T= 2\pi \sqrt{\frac{l}{0}} = \infty

 

Time period of simple pendulum accelerating up -

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

- wherein

l= length of pendulum

g= acceleration due to gravity.

a= acceleration of pendulum.

 

 

 


Option 1)

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

This is incorrect.

Option 2)

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

This is incorrect.

Option 3)

zero

This is incorrect.

Option 4)

infinte

This is correct.

Posted by

Vakul

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