Change in Amplitude with respect to Driven Frequencies

[Azer]

Homework Statement

A mass-spring system has b/m = \omega_0/5 , where b is the damping constant and \omega_0 the natural frequency. How does its amplitude when driven at frequencies 10% above \omega_0 compare with its amplitude at \omega _0 ? How does its amplitude when driven at frequencies 10% below \omega_0 compare with its amplitude at \omega _0 ?

Homework Equations

(1) A(\omega) = \frac{F_0}{m\sqrt{(\omega^2_d - \omega^2_0)^2 + \frac{b^2\omega^2_0}{m^2}}}

(2) \omega^2_0=\frac{25b^2}{m^2}

The Attempt at a Solution

Plugging in 1.1 for \omega_d in equation 1 (since it is 10% more than \omega_0 ) and using equation 2 to substitute \omega^2_0 for \frac{25b^2}{m^2} gives an amplitude of \frac{mF_0}{b^2}*1/7.25 . Plugging in 0 for the driving force yields an amplitude of \frac{mF_0}{b^2}*1/5 . The amplitude when driven over the amplitude given by the natural frequency should be 5/7.25*100%. Rounded to two sig figs, as my online homework demands, should yield 69%. It is telling me that I have made a rounding error, but I cannot find where.

 

[Redbelly98]

Welcome to Physics Forums.

According to this, the final ω₀ under the √ should actually be ω:

http://hyperphysics.phy-astr.gsu.edu/Hbase/oscdr.html#c1

 

[Azer]

\omega_0 should not change whether the system is driven or not. It is the natural frequency. \omega_d however, will be changing in each of the three calculations of amplitude.

 

[Redbelly98]

Yes, most definitely.

Um, I meant it should be ω_d, not ω. Sorry. Here's a more direct link:

http://hyperphysics.phy-astr.gsu.edu/Hbase/oscdr.html#c2

Look at the term called A. Also, their ω is your ω_d