SHM - 1

Classical Mechanics Level pending

A pendulum with a bob of mass $m$ and a string of length $L$ is displaced from its equilibrium position $O$ by a small angle and then released. At the same time, a bob of mass $M$ is dropped and falls vertically downwards through a distance $L$ . A point $P$ is directly below bob of mass $M$ and it happens to be at the same horizontal level as $O$ . The dimensions of the two bobs are the same.

Which bob will arrive at its destination first – bob of mass $m$ reaching its equilibrium position $O$ or bob of mass $M$ arriving at the point $P$ ? (Ignore air resistance and you may wish to use: Period of pendulum, T = $\sqrt{2π(L/g)}$

Bob of mass m. Bob of mass M. The heavier ball will reach its destination first. Click to show/hide The two bobs arrive at their destinations at the same time

1 solution

Manmeswar Patnaik
Mar 5, 2016

The bob of mass m will take $t_{m}$ = $\frac{T}{4}$ s to reach its equilibrium position O.

Therefore,

$t_{m}$ = ¼ T

= $¼\sqrt{(2π(L/g)}$

= $1.6\sqrt{(L/g)}$

The bob of mass M experiences free-fall and the time, $t_{M}$ it takes to travel a vertical distance L to arrive at point P is

$t_{M} = \sqrt{(2L/g)}$

= $1.4\sqrt{(L/g)}$

Since $t_{m} > t_{M}$ , therefore bob of mass M will reach its destination first.

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