Question

A simple pendulum of length 1 feet suspended from the ceiling of an elevator takes π/3 seconds to complete one oscillation. Find the acceleration of the elevator.

Answer 1

It is given that:
Length of the simple pendulum, l = 1 feet
Time period of simple pendulum, T = $\frac{\mathrm{\pi }}{3}\mathrm{s}$
Acceleration due to gravity, g = 32 ft/s^2
​
​Let a be the acceleration of the elevator while moving upwards.

Driving force$\left(f\right)$ is given by,
f = m(g + a)sinθ

Comparing the above equation with the expression, f = ma, we get:
Acceleration, a = (g + a)sinθ = (g +a)θ (For small angle θ, sin θ → θ)
$=\frac{\left(g+a\right)x}{l}={\omega }^{2}x$ (From the diagram $\theta =\frac{x}{l}$)
$\Rightarrow \omega =\sqrt{\frac{\left(g+a\right)}{l}}$
Time period $\left(T\right)$ is given as,
$T=2\mathrm{\pi }\sqrt{\frac{l}{g+a}}$
On substituting the respective values in the above formula, we get:
$$\begin{gathered} \frac{\mathrm{\pi }}{3}=2\mathrm{\pi }\sqrt{\frac{1}{32+a}} \\ \frac{1}{9}=4\left(\frac{1}{32+a}\right) \\ \Rightarrow 32+a=36 \\ \Rightarrow a=36-32=4\mathrm{ft}/{\mathrm{s}}^2 \end{gathered}$$