Question

A man enters a tall tower, needing to know its height. He notes that a long pendulum extends from the ceiling almost to the floor and that its period is 26.5 s. (a) How tall is the tower? m (b) If this pendulum is taken to the Moon, where the free-fall acceleration is 1.67 m/s2, what is the period there?

Answer 1

Explanation:

It is given that,

Time period of the pendulum, T = 26.5 s

1. The time period of the pendulum is given by :

$T=2\pi \sqrt{\dfrac{l}{g}}$, l is the length of the pendulum

$l=\dfrac{T^2g}{4\pi ^2}$

$l=\dfrac{(26.5)^2\times 9.8}{4\pi ^2}$

l = 174.32 meters

2. On the surface of moon, the acceleration due to gravity, $g'=1.67\ m/s^2$

Time period is given by :

$T=2\pi \sqrt{\dfrac{174.32\ m}{1.67\ m/s^2}}$

T = 64.19 seconds

Hence, this is the required solution.