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