Question

A pendulum is suspended in a lift and its period of oscillation is t when the lift is stationary. What will be the period t of oscillation of pendulum , if the lift begins to accelerate downward with acceleration equal to 3g/4

Answer 1

Answer:

Given:

Time period of pendulum in a stationary lift = t

The lift now accelerates downwards with 3g/4

To find:

New time period

Concept:

When the lift goes downwards with an Acceleration, it experiences an pseudo-force . This force changes the time period of the pendulum.

Calculation:

Initial Case:

t = 2π√(l/g) .............(1)

Now net acceleration experienced by pendulum is :

= g - 3g/4

= g/4 ..........(2)

New time period

$= 2\pi \sqrt{ \frac{l}{ \frac{g}{4} } }$

$= 2\pi \sqrt{ \frac{4l}{g} }$

$= 2(2\pi \sqrt{ \frac{l}{g} } )$

$= 2t$

So the final answer is 2t

Answer 2

Answer:

The Initial case :

t2π√(l/g)........(equation 1).

Therefore net acceleration experienced by the pendulum :

= g - 3g = 4 .

= g/4 .......(equation 2).

Now,New time period :

$2\pi \sqrt{} \frac{l}{ \frac{g}{4} } .$

$2\pi \sqrt{ \frac{4l}{g} }$

$2(2\pi \sqrt{ \frac{l}{g} } ).$

$= 2t.$

Answer = 2t .

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