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 and acceleration a₀ (b) is going down with an acceleration a₀ and (c) is moving with a uniform velocity.

Answer 1

The length of the simple pendulum is l.
​Let x be the displacement of the simple pendulum..
(a)

From the diagram, the driving forces f is given by,
f = m(g + a₀)sinθ ...(1)
Acceleration (a) of the elevator is given by,
$$\begin{gathered} a=\frac{f}{m} \\ =\left(g+a_0\right)\sin \theta \\ =\left(g+a_0\right)\frac{x}{l}{\left(\mathrm{From}\mathrm{the}\mathrm{diagram}\sin \theta =\frac{x}{l}\right)}_{} \end{gathered}$$
[ when θ is very small, sin θ → θ = x/l]

$\therefore a=\left(\frac{g+a_0}{l}\right)x$ ...(2)
As the acceleration is directly proportional to displacement, the pendulum executes S.H.M.
Comparing equation (2) with the expression a =${\omega }^{2}x$, we get:
${\omega }^2=\frac{g+a_0}{l}$

Thus, time period of small oscillations when elevator is going upward(T) will be:
$T=2\mathrm{\pi }\sqrt{\frac{l}{g+a_0}}$

(b)

When the elevator moves downwards with acceleration a₀,
Driving force (F) is given by,
F = m(g − a₀)sinθ
On comparing the above equation with the expression, F = ma,
$$\begin{gathered} \mathrm{Acceleration},a=\left(g-a_0\right)\mathrm{sin\theta }=\frac{\left(g-a_0\right)x}{l}={\omega }^{2}x \\ \mathrm{Time}\mathrm{period}\mathrm{of}\mathrm{elevator}\mathrm{when}\mathrm{it}\mathrm{is}\mathrm{moving}\mathrm{downward}\left(T\text{'}\right)\mathrm{is}\mathrm{given}\mathrm{by}, \\ T\text{'}=\frac{2\mathrm{\pi }}{\mathrm{\omega }}=2\mathrm{\pi }\sqrt{\frac{l}{g-a_0}} \end{gathered}$$
(c) When the elevator moves with uniform velocity, i.e. a₀ = 0,
For a simple pendulum, the driving force $\left(F\right)$ is given by,
$$\begin{gathered} F=\frac{mgx}{l} \\ \mathrm{Comparing}\mathrm{the}\mathrm{above}\mathrm{equation}\mathrm{with}\mathrm{the}\mathrm{expression},F=ma,\mathrm{we}\mathrm{get}: \\ a=\frac{gx}{l} \\ \Rightarrow \frac{x}{a}=\frac{l}{g} \\ \mathrm{T}=2\mathrm{\pi }\sqrt{\frac{\mathrm{displacement}}{\mathrm{Acceleration}}} \\ =2\mathrm{\pi }\sqrt{\frac{l}{g}} \end{gathered}$$