What are the values of the mass $m$ such that $mx''+25x = 12 \cos(36\pi t)$ will exhibit resonance?

Asked Apr 01 '21 at 09:15

Active Apr 01 '21 at 10:44

Viewed 34 times

$mx''+25x = 12 cos(36πt)$

my take on this:

${\omega_0} = \sqrt{k \over m} = \sqrt{25 \over m}$

$x_h = c_1cos({\omega_0}t) + c_2sin({\omega_0}t)$

$x_h = c_1cos(\sqrt{25 \over m} t) + c_2sin(\sqrt{25 \over m}t)$

$x_p = {{12 \over {m({\omega_0^2 - \omega^2})}}cos(36\pi t)} = {{12 \over {25-1296m^2}}cos(36\pi t)}$

but I am stuck here. am I on the wrong road?

edited Apr 01 '21 at 10:44

David C. Ullrich

asked Apr 01 '21 at 09:15

2

Does this answer your question? Differential equation with resonance. Same question, different numbers and a $\sin$ instead of a $\cos$. – Toby Mak Apr 01 '21 at 09:18
I assume you're finding the general solution since there are no initial conditions. Is that right? – Toby Mak Apr 01 '21 at 09:19
@TobyMak thank you! mostly yes, because I didn't know in which direction to go. :) – just1frustredstudent Apr 01 '21 at 09:21
Remember that you can always modify the particular solution with terms from the complimentary one. So you could also use a modified $$x_p=A\frac{\cos(ω_0t)-\cos(ωt)}{ω_0^2-ω^2}$$ with a now obvious difference quotient for $ω_0-ω\to 0$. – Lutz Lehmann Apr 01 '21 at 09:45

0 Answers0