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

Published by Pearson

Chapter 14 - Oscillations - Exercises and Problems - Page 406: 73

Answer

$f = \sqrt{f_1^2+f_2^2}$

Work Step by Step

If the mass is a distance of $x$ from the equilibrium point, the force on the mass is $k_1x+k_2x$ which is $(k_1+k_2)x$. Thus, the equivalent spring constant is $k_1+k_2$. We can find an expression for the frequency using: $f = \frac{1}{2\pi}~\sqrt{\frac{k_1+k_2}{m}}$ We can find an expression for $\sqrt{f_1^2+f_2^2}$ using the formula above: $\sqrt{f_1^2+f_2^2}$ $= \sqrt{(\frac{1}{2\pi}~\sqrt{\frac{k_1}{m}})^2+ (\frac{1}{2\pi}~\sqrt{\frac{k_2}{m}})^2}$ $= \sqrt{(\frac{1}{2\pi})^2~\frac{k_1}{m}+ (\frac{1}{2\pi})^2~\frac{k_1}{m}}$ $= \sqrt{(\frac{1}{2\pi})^2~\frac{k_1+k_2}{m}}$ $= \frac{1}{2\pi}~\sqrt{\frac{k_1+k_2}{m}}$ Therefore, $f = \sqrt{f_1^2+f_2^2}$.

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.