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Chapter 1: Q47E (page 462)

Question: A building in San Francisco has light fixtures consisting of small \({\bf{2}}.{\bf{35}} - {\bf{kg}}\) bulbs with shades hanging from the ceiling at the end of light, thin cords \({\bf{1}}.{\bf{50}}{\rm{ }}{\bf{m}}\) long. If a minor earthquake occurs, how many swings per second will these fixtures make?

Short Answer

Expert verified

The number of swings per second is \(0.406\,{\rm{swing/sec}}\)

Step by step solution

01

Identification of given data

Length of cord \(L = 1.50\,{\rm{m}}\)

Mass of the bulbs \(m = 2.35\,\,{\rm{kg}}\)

02

Significance of frequency of simple pendulum

It is defined as the number of waves that passes a point per unit time. It is expressed as,

\(f = \frac{1}{{2\pi }}\sqrt {\frac{g}{L}} \) 鈥�(i)

Where, \(L\) is the length of pendulum and \(g\) is the acceleration due to gravity (\(9.8\,{\rm{m/}}{{\rm{s}}^{\rm{2}}}\))

03

Determining number of swings per second

Using equation (i)

\(f = \frac{1}{{2\pi }}\sqrt {\frac{g}{L}} \)

Substituting all the values in above equation

\(\begin{array}{c}f = \frac{1}{{2\pi }}\sqrt {\frac{{9.8\,{\rm{m/}}{{\rm{s}}^{\rm{2}}}}}{{1.5\,{\rm{m}}}}} \\ = 0.406\,{\rm{Hz}}\end{array}\)

Hence the number of swings per second is \(0.406\,{\rm{swing/sec}}\)

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