- #1

Pochen Liu

- 52

- 2

- Homework Statement:
- I have an equation which I used a double derivative to solve, now I need to find values for the variables such that when external force is applied resonance occurs

- Relevant Equations:
- *Attached

I have the equation $$\frac{d^2y}{dt^2} + 5y = 0$$

where I've worked out $$y = Acos(\sqrt5t) + Bsin(\sqrt 5 t)$$

$$y'' = -5Bsin(\sqrt 5 t)

$$

using $$y = e^{\lambda x}$$ and using y(0) = 0 (the spring is released from equilibrium)

so an external force $$Acos(\omega(t - \phi))$$ is applied so we therefore

$$\frac{d^2y}{dt^2} + 5y = -4Bsin(\sqrt 5 t) = Acos(\omega(t - \phi))$$

$$-4Bsin(\sqrt 5 t) = -4Bcos(\frac{\pi}{2} - \sqrt 5 t)$$

so $$\frac{\pi}{2} - \sqrt 5 t = \omega(t - \phi)$$

$$-\sqrt 5(t-\frac{\pi}{2 \sqrt 5}) = \omega(t - \phi)$$

we can conclude that these are the values to create resonance

$$ \omega = - \sqrt 5 $$

$$ \phi = \frac{\pi}{2\sqrt5}$$

A = any positive integer as only they only need to be in phase, so any amplitude is added

Are these the right conclusions?

where I've worked out $$y = Acos(\sqrt5t) + Bsin(\sqrt 5 t)$$

$$y'' = -5Bsin(\sqrt 5 t)

$$

using $$y = e^{\lambda x}$$ and using y(0) = 0 (the spring is released from equilibrium)

so an external force $$Acos(\omega(t - \phi))$$ is applied so we therefore

$$\frac{d^2y}{dt^2} + 5y = -4Bsin(\sqrt 5 t) = Acos(\omega(t - \phi))$$

$$-4Bsin(\sqrt 5 t) = -4Bcos(\frac{\pi}{2} - \sqrt 5 t)$$

so $$\frac{\pi}{2} - \sqrt 5 t = \omega(t - \phi)$$

$$-\sqrt 5(t-\frac{\pi}{2 \sqrt 5}) = \omega(t - \phi)$$

we can conclude that these are the values to create resonance

$$ \omega = - \sqrt 5 $$

$$ \phi = \frac{\pi}{2\sqrt5}$$

A = any positive integer as only they only need to be in phase, so any amplitude is added

Are these the right conclusions?