# Amplitude of a Damped, Driven Pendulum

1. Jun 29, 2013

### AJKing

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

$A_m = 0.1576 m$

$F/m = \frac{gζ}{l}$

Let ζ = amplitude of driver.

$Q= 50 π$

$A(ω) = \frac{F/m}{((ω_0^2 - ω^2)^2 + (γω)^2)^{0.5}}$

3. The attempt at a solution

Solution is exactly stated as: "ω0 ± 0.017 sec-1"

Am I solving for ω or ω0?

2. Jun 29, 2013

### darkxponent

Definitely ω, ω0 is the undamped natural frequency.

3. Jun 30, 2013

### Pqpolalk357

Can you explain to me how did you solve this question ? Thank you

4. Jun 30, 2013

### darkxponent

I didn't solve any equation. I just told AJKing whether to solve for ω or ω0

5. Jun 30, 2013

### Pqpolalk357

Do you know by any chance how to proceed ?

6. Jun 30, 2013

### darkxponent

Yes i know how to proceed but let OP reply first. Highjacking a thread is not allowed here at PF!

7. Jun 30, 2013

### Pqpolalk357

Please explain to me how to proceed.

8. Jul 2, 2013

### AJKing

It would seem obvious how to proceed - solve for ω - but it is difficult to do directly.

Here's Wolfram Alpha's attempt at solving this:

$A(ω) = \frac{F/m}{((ω_0^2 - ω^2)^2 + (γω)^2)^{0.5}}$

Where ω0 = g0.5

We must keep out eyes for clever approximations.