College Physics (4th Edition)

Published by McGraw-Hill Education

Chapter 10 - Problems - Page 400: 45

Answer

The new value of $\omega$ is $5.0~rad/s$

Work Step by Step

Let the equivalent spring constant of the two springs be $k$. Let $M$ be the mass of the cart. We can find an expression for the original value of $\omega$: $\omega = \sqrt{\frac{k}{M}}$ Let $M' = 4.0~M$. We can find an expression for the new value of the angular frequency $\omega'$: $\omega' = \sqrt{\frac{k}{M'}}$ $\omega' = \sqrt{\frac{k}{4.0~M}}$ $\omega' = \frac{1}{2.0}\times \sqrt{\frac{k}{M}}$ $\omega' = \frac{1}{2.0}\times \omega$ $\omega' = \frac{1}{2.0}\times 10.0~rad/s$ $\omega' = 5.0~rad/s$ The new value of $\omega$ is $5.0~rad/s$.

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