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Change in Amplitude with respect to Driven Frequencies

  • Thread starter Azer
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  • #1
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Homework Statement



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

Homework Equations



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

(2) [tex] \omega^2_0=\frac{25b^2}{m^2} [/tex]

The Attempt at a Solution



Plugging in 1.1 for [tex] \omega_d [/tex] in equation 1 (since it is 10% more than [tex] \omega_0 [/tex] ) and using equation 2 to substitute [tex] \omega^2_0 [/tex] for [tex] \frac{25b^2}{m^2} [/tex] gives an amplitude of [tex] \frac{mF_0}{b^2}*1/7.25 [/tex] . Plugging in 0 for the driving force yields an amplitude of [tex] \frac{mF_0}{b^2}*1/5 [/tex] . 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.
 
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Answers and Replies

  • #3
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[tex] \omega_0 [/tex] should not change whether the system is driven or not. It is the natural frequency. [tex] \omega_d [/tex] however, will be changing in each of the three calculations of amplitude.
 
  • #4
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