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  • Help me solve this - Oscillations and Waves - JEE Main-5

A man measures the period of a simple pendulum inside a stationary lift and finds it to be T s. If the lift accelerates upwards with an accelleration \frac{g}{}4, then the period of the pendulum will be 

  • Option 1)

    T

  • Option 2)

    \frac{T}{}4

  • Option 3)

    \frac{2T}{\sqrt{5}}

  • Option 4)

    2T\sqrt{5}

 

Answers (1)

best_answer

In stationary lift T = 2\pi\sqrt{\frac{l}{g}}

In upward moving lift T' = 2\pi\sqrt{\frac{l}{g+a}}

(a = acceleration of lift)

\Rightarrow \frac{T'}{T} = \sqrt{\frac{g}{g+\frac{g}{4}}} = \sqrt{\frac{4}{5}} \Rightarrow T' = \frac{2T}{\sqrt{5}}

 

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)

T

This is incorrect.

Option 2)

\frac{T}{}4

This is incorrect.

Option 3)

\frac{2T}{\sqrt{5}}

This is correct.

Option 4)

2T\sqrt{5}

This is incorrect.

Posted by

divya.saini

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