At what angular frequencies is the amplitude half of its resonant value?

[Ange]

Homework Statement

A simple pendulum has a length of 1m. In free vibration the amplitude of its swings falls off by a factor of e in 50 swings. The pendulum is set into forced vibration by moving its point of suspension horizontally in SHM with an amplitude of 1 mm.

a) Built Differential

b) Found Amplitude at exact resonance = 0.1576m

c) At what angular frequencies is the amplitude half of its resonant value?

Homework Equations

A(wo)=0.1576m

wo/gamma=50π

A(w)=(F/m)/((wo^2−w^2)2+(γω)^2)^0.5
A(w)=(wo^2)Eo / ((wo^2−w^2)2+(γω)^2)^0.5 , where Eo is the horizontal displacement from the support(0.001m).

The Attempt at a Solution

A(w)=(F/m)/((wo^2−w^2)2+(γw)^2)^0.5
A(w)=0.5A(wo)=0.1576m/2

Solution is : ω0 ± 0.017 /s

I'm stuck in solving for w, not sure how to go about it I feel like I am missing information but something must cancel out somewhere. I tried expanding out the bottom, that gets me nowhere( thought I may use quadratic formula). Not sure where to go.

 

[Greg Bernhardt]

Thanks for the post! Sorry you aren't generating responses at the moment. Do you have any further information, come to any new conclusions or is it possible to reword the post?