Will the time period pendulum increase or decrease when the experiment is done in an elevator moving down with constant acceleration?

Answer

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Hint: Here the varying quantity is acceleration. According to the simple pendulum concept, the time period of the simple pendulum depends only on the acceleration and length of the pendulum. Since the acceleration is in the denominator of the equation of time period, when the acceleration increases the time period decreases.Formula used: Time period of a simple pendulum is, \[T=2\pi \sqrt{\dfrac{L}{g}}\], where L is the length of the simple pendulum and g is the acceleration due to gravity.Complete step by step answer: To find the new time period of the simple pendulum, we can redraw the figure given in the question. We can consider the length of the simple pendulum as L. At the rest, the only acceleration acting on the body is an acceleration due to gravity. That is, g. But the lift starts to move upwards with an acceleration of \[\dfrac{g}{4}\]. Thus the total acceleration of the pendulum will be the sum of acceleration due to gravity and the acceleration of the lift.

The time period of a simple pendulum at rest will be,\[T=2\pi \sqrt{\dfrac{L}{g}}\], where L is the length of the simple pendulum and g is the acceleration due to gravity.When the lift starts to move upwards, the acceleration of the lift becomes \[\dfrac{g}{4}\]. So we have to add this acceleration also with the acceleration due to gravity to get the total acceleration of the simple pendulum.Total acceleration will be,\[a=g+\dfrac{g}{4}\]\[a=\dfrac{5g}{4}\]We can plug this acceleration in the equation of time period to find the new time period.\[{{T}_{n}}=2\pi \sqrt{\dfrac{L}{a}}\]\[{{T}_{n}}=2\pi \sqrt{\dfrac{4L}{5g}}\]We can compare the new time period with the initial time period.\[{{T}_{n}}=\sqrt{\dfrac{4}{5}}T\]It can simplify as, \[{{T}_{n}}=\dfrac{2}{\sqrt{5}}T\], Hence the correct option is C.Note: Motion in a lift is always a special case. Here the acceleration of the lift will bring great impact to the motion. Here we are adding the acceleration of the lift to the acceleration due to gravity since the lift moves upwards. If the lift is moving downward, we have to reduce the acceleration of the lift from the acceleration due to gravity.The normal force acting on the body from the lift at various conditions are;1. \[N=mg\], when the lift is at rest or moving with a constant velocity.2. \[N=mg+ma\], when the lift moves upward with an acceleration.3. \[N=mg-ma\], when the lift moves downward with an acceleration.

The normal force is equal to your weight. So, you will feel heavier when the lift moves upward with an acceleration a and lighter when the lift moves downward with an acceleration a.

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Imagine that you are in a stationary elevator and you drop an object.
Using a distance measuring scale attached to the elevator and a timing device you measure the acceleration of free fall relative to the elevator as $g$.

Now repeat the experiment in an elevator which is moving up or down at constant velocity relative to the Earth.
Again using the scale attached to the elevator and a timing device you measure the acceleration of free fall relative to the elevator as $g$.

Next repeat the experiment in an elevator accelerating upwards at an acceleration $a$ relative to the Earth and you find that the acceleration of free fall measured relative to the elevator is $g+a$. To show that this is reasonable repeat the experiment with the elevator in free fall relative to the Earth ie accelerating downwards with an acceleration of $g$ relative to the Earth.

If you "drop" an object in such a situation and you are in the elevator you will not see the object moving relative to you or the elevator as you, the elevator and the object are all accelerating downwards at $g$ relative to the earth.

What does all this tell you? It tells you that as far as an observer in an accelerating, relative to the Earth, elevator is concerned the acceleration of free fall is $g + (\pm a)$ where $\pm a$ is the acceleration of the elevator relative to the Earth with the positive sign for an upward acceleration relative to the Earth and the negative sign for a downward acceleration relative to the Earth.

Note that if the downward acceleration of the elevator relative to the Earth is greater than $g$ you have a situation where a "dropped" object actually accelerates upwards relative to the elevator.

Now consider your question as to whether one should add or subtract $a$ from $g$ in a situation where the elevator is in free fall relative to the Earth. You, standing in the elevator, pull the pendulum bob from its equilibrium position and release it. The pendulum bob does not move relative to you or the elevator. The period of the bob is "infinite" because the tension in the string which is connecting the bob to the elevator is zero.

Accelerate the elevator at $2g$ downwards relative to the Earth and you will measure the same period for the pendulum as in a stationary elevator with the pendulum oscillating "up side down" ie the point of suspension will be below the pendulum bob.

In your equation for the period of the pendulum the $g$ is the value of the acceleration of free fall of a body relative to the elevator. and when the elevator is accelerating upwards with an acceleration of $a$ relative to the Earth the acceleration of free fall relative to the elevator is $g \; { \Large +} \;a$.

You might find it interesting to read about the Equivalence Principle?

Text Solution

Solution : i) When the lift moves up with uniform velocity i.e. `a=0`. There would be no change in the time period of a simple pendulum. ii) When the lift moves down with uniform velocity i.e. `a=0`, there would be no change in the time period of a simple pendulum. iii) When lift is moving up with acceleration 'a' then relative acceleration `=g+a` `therefore` Time period, `T=2pisqrt((l)/(g+a))` so when lift is moving up with uniform acceleration time period of pendulum in it decreases. iv) When lift is moving down with acceleration 'a' time period, `T=2pisqrt((l)/(g-a))` (`g-a=` relative acceleration of pendulum) So time period of pendulum in the lift decreases. v) If the falls freely, `a=g` then the time period of a simple pendulum becomes infinite.