Question

A simple pendulum of length l is suspended through the ceiling of an elevator. Find the time period of small oscillations if the elevator (a) is going up with an acceleration $a_0$ (b) is going down with an acceleration $a_0$ and (c) is moving with a uniform velocity.

Answer 1

(a) Here driving forces,

$f=m(g+a_0)sin\theta ...\left(1\right)$

Acceleration,

$a=\frac{F}{m}$

$=(g+a_0)sin\theta$

$=(g+a_0)\frac{x}{l}$

[Because when q is small $sin\theta \to \theta =x/l$]

$\therefore a=\left(\frac{g+a_0}{l}\right)x$

$\therefore$ Acceleration is proportional to displacement, so the motion is SHM.

Now, ${\omega }^2=\frac{g-a_0}{l}$

$\therefore T=2\pi \sqrt{\frac{l}{g+a_0}}$

(b) When the elevator is going downwards with acceleration $a_0$

Driving force,

$F=m(g-a_0)sin\theta$

Acceleration = $(g-a_0)sin\theta \frac{(g-a)}{l}{\omega }^{2}x$

$T=\frac{2\pi }{\omega }2\pi \sqrt{\frac{l}{g-a_0}}$

(c) When moving with uniform velocity,

$a_0=0$

For the simple pendulum,

Driving force = $\frac{mgx}{l}$

$\Rightarrow a=\frac{gx}{l}$

$\Rightarrow \frac{x}{a}=\frac{l}{g}$

$T=2\pi \sqrt{\frac{\text{displacement}}{\text{Acceleration}}}$

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