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  • How to solve this problem- - Oscillations and Waves - JEE Main-4

The time period of a simple pendulum of length L as measured in an elevator desending with acceleration \frac{g}{}3 is

  • Option 1)

    2\pi \sqrt{\frac{3L}{g}}

  • Option 2)

    \pi \sqrt{\frac{3L}{g}}

  • Option 3)

    2\pi \sqrt{\frac{3L}{2g}}

  • Option 4)

    2\pi \sqrt{\frac{2L}{3g}}

 

Answers (1)

best_answer

The effective acceleration in a lift descending with acceleration \frac{g}{3} is g_{eff}= g - \frac{g}{3} =\frac{2g}{3} \\* \Rightarrow T = 2\pi\sqrt{\frac{L}{g_{eff}}} = 2\pi\sqrt{\frac{3L}{2g}}

 

Time period of simple pendulum accelerating horizontally -

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

- wherein

l= length of pendulum

g= acceleration due to gravity.

a= acceleration of pendulum.

 

 

 


Option 1)

2\pi \sqrt{\frac{3L}{g}}

This is incorrect.

Option 2)

\pi \sqrt{\frac{3L}{g}}

This is incorrect.

Option 3)

2\pi \sqrt{\frac{3L}{2g}}

This is correct.

Option 4)

2\pi \sqrt{\frac{2L}{3g}}

This is incorrect.

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