## Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (3rd Edition)

Published by Pearson

# Chapter 14 - Oscillations - Exercises and Problems: 54

#### Answer

For a simple pendulum of a mass on a string: $\omega = \sqrt{\frac{mgd}{I}} = \sqrt{\frac{g}{L}}$

#### Work Step by Step

For a mass on a string of length $L$, the moment of inertia is $I = mL^2$. The distance $d$ from the center of mass to the pivot is $L$. We can find an expression for the angular frequency as: $\omega = \sqrt{\frac{mgd}{I}}$ $\omega = \sqrt{\frac{mgL}{mL^2}}$ $\omega = \sqrt{\frac{g}{L}}$

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