Question

A pendulum consisting of a 0.5 kg mass tied to a 0.3 m string is set into oscillation at the same moment that a stone is dropped from a 44.1-m-tall building. how many cycles of oscillation will the pendulum go through before the stone hits the ground?

Answer 1

Answer:

2.72 cycles

Explanation:

First of all, let's find the time that the stone takes to reaches the ground. The stone moves by uniform accelerated motion with constant acceleration g=9.8 m/s^2, and it covers a distance of S=44.1 m, so the time taken is

$S=\frac{1}{2}at^2\\t=\sqrt{\frac{2S}{a}}=\sqrt{\frac{2(44.1m)}{9.8 m/s^2}}=3 s$

The period of the pendulum instead is given by:

$T=2 \pi \sqrt{\frac{L}{g}}=2 \pi \sqrt{\frac{0.3 m}{9.8 m/s^2}}=1.10 s$

Therefore, the number of oscillations that the pendulum goes through before the stone hits the ground is given by the time the stone takes to hit the ground divided by the period of the pendulum:

$N=\frac{t}{T}=\frac{3 s}{1.10 s}=2.72$