Question

The time period of a simple pendulum of length l as measured in an elevator descending with accelaration g/3 is

Answer 1

Answer:  

$2\pi \sqrt\frac{3l}{g}$

Answer 2

The time period of elevator is $2\pi \sqrt{\dfrac{3l}{2g}}$  .

Given :

Elevator descending with acceleration of g/3 .

We know , time period a simple pendulum of length l is :

$T=2\pi\sqrt{\dfrac{l}{g_{eff}}}$    .......( 1 ).

Here , it system i.e the elevator is acceleration .

Therefore , the system is not inertial .

To convert it into inertial we need to apply the same acceleration in opposite direction in every equation .

Therefore ,

$g_{eff}=g-\dfrac{g}{3}\\\\g_{eff}=\dfrac{2g}{3}$

Here , g is acceleration due to gravity .

Putting value of $g_{eff}$ in equation 1 we get ,

We get ,

$T=2\pi\sqrt{\dfrac{l}{\dfrac{2g}{3}}}\\\\T=2\pi \sqrt{\dfrac{3l}{2g}}$

Hence , this is the required solution .

Learn More :

Time Period

https://brainly.in/question/11010150