Question

A simple pendulum is mounted in an elevator. Whathappens to the time period of the pendulum (does it increase, decrease,or remain the same) if the elevator

(a) accelerates upward at$\mathit{a}\mathbf{=}\mathbf{5}\mathbf{.}\mathbf{0}\mathit{m}\mathbf{/}{\mathit{s}}^{\mathbf{2}}$

(b) moves upward at a steady$\mathit{v}\mathbf{=}\mathbf{5}\mathbf{.}\mathbf{0}\mathit{m}\mathbf{/}\mathit{s}$

(c) acceleratesdownward at$\mathit{a}\mathbf{=}\mathbf{5}\mathbf{.}\mathbf{0}\mathit{m}\mathbf{/}{\mathit{s}}^{\mathbf{2}}$

(d) accelerates downward at$\mathit{a}\mathbf{=}\mathbf{9}\mathbf{.}\mathbf{8}\mathit{m}\mathbf{/}{\mathit{s}}^{\mathbf{2}}$

Justify your answers.

Short answer

1. T will decrease.
2. T will remain same.
3. T will increase.
4. T will be infinity.

Step-by-step solution

Definition of time period of a vertical mass-spring

Time period of a vertical mass-spring is given by

$T=2\pi \sqrt{\frac{L}{g_{net}}}$

Where L is the length of the pendulum and $g_{net}$is the net acceleration due to gravity.

Since, the motion of the object is affected by the net acceleration of the objects i.e.${\mathit{g}}_{\mathbf{n}\mathbf{e}\mathbf{t}}\mathbf{=}\mathit{g}\mathbf{\pm }\mathit{a}$

Consider downward g as positive.

Calculation when acceleration is upwards

(a) When$a=5.0m/s^2$ we have,

$g_{net}=g+a$

Hence, the time period T will decrease.

(b) When$v=5.0m/s$ we have,

$g_{net}=g$

Hence, the time period T will remain same.

Calculation when acceleration is downwards

(c) When$a=5.0m/s^2$ we have,

$g_{net}=g-a$

Hence, the time period T will increase.

(d) When$a=9.8m/s^2$ we have,

$g_{net}=g-a$

Hence, the time period T be infinity.