Amplitude of a Damped, Driven Pendulum

[AJKing]

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_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}}

The Attempt at a Solution

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

Am I solving for ω or ω₀?

 

[darkxponent]

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_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}}

The Attempt at a Solution

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

Am I solving for ω or ω₀?

Definitely ω, ω₀ is the undamped natural frequency.

 

[Pqpolalk357]

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

 

[darkxponent]

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

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

 

[Pqpolalk357]

Do you know by any chance how to proceed ?

 

[darkxponent]

Do you know by any chance how to proceed ?

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

 

[Pqpolalk357]

Please explain to me how to proceed.

 

[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 = g^0.5

We must keep out eyes for clever approximations.