Understanding resonance in a spring

  • #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?
 

Answers and Replies

  • #2
haruspex
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so an external force $$Acos(\omega(t - \phi))$$ is applied so we therefore
$$\frac{d^2y}{dt^2} + 5y = -4Bsin(\sqrt 5 t) = Acos(\omega(t - \phi))$$
The left hand side equals the right hand side, but the term in the middle makes no sense.
You seem to be assuming that the solution to the unforced case still applies, but it doesn't, and you have also made an arithmetic error. The arithmetic error gave you a 4 instead of 0.
 

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