# Homework Help: At what angular frequencies is the amplitude half of its resonant value?

1. Oct 13, 2014

### Ange

1. The problem statement, all variables and given/known data

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?

2. Relevant 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).

3. 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 Im 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.

2. Oct 18, 2014