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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.
Solution
It is given that:
Length of the simple pendulum, l = 1 feet
Time period of simple pendulum, T = \[\frac{\pi}{3} s\] Acceleration due to gravity, g = 32 ft/s2
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,
\[T = 2\pi\sqrt{\frac{l}{g + a}}\]
On substituting the respective values in the above formula, we get:
\[\frac{\pi}{3} = 2\pi\sqrt{\frac{1}{32 + a}}\]
\[\frac{1}{9} = 4\left( \frac{1}{32 + a} \right)\]
\[ \Rightarrow 32 + a = 36\]
\[ \Rightarrow a = 36 - 32 = 4 ft/ s^2\]
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