Question

A uniform meter stick is suspended from the ceiling of an elevator on one end and is swinging back and forth. The elevator is accelerating upwards with acceleration a.

The period of this meter stick, in terms of its length L, g and a is ______.

Answer 1

`T = 2\pi\cdot \sqrt{(l)/(g + a) }`

Step-by-step explanation:

It is known that stick is experimenting a Simple Harmonic Movement and, to be exactly, can be modelled as a simple pendulum. The period of oscilation of the stick is:

`T = (2\pi)/(\omega)`

The pendulum is modelled by the Newton's Laws. The Free Body Diagram is presented below:

`\Sigma F_(r) = T - m\cdot g \cdot \cos \theta = m\cdot (\omega^(2)\cdot l + a\cdot \cos \theta)`

`\Sigma F_(t) = m\cdot g \cdot \sin \theta = m\cdot (l\cdot \alpha - a \cdot \sin \theta)`

Let assume that pendulum is just experimenting small oscillations, so that:

`\theta \approx \sin \theta`

Then:

`m\cdot g \cdot \theta = m\cdot (l\cdot \alpha - a\cdot \theta)`

`g\cdot \theta = l\cdot \alpha - a\cdot \theta`

`(g + a)\cdot \theta = l\cdot \alpha`

`\alpha = (g+a)/(l)\cdot \theta`

Where
`\omega =\sqrt{(g + a)/(l) }`.

Finally, the period is:

`T = 2\pi\cdot \sqrt{(l)/(g + a) }`