Consider a damped driven harmonic oscillator, for which $\beta = \omega_0/4$ and the driving force is given by $F = F_0\cos\omega t$ ($\omega_0$ and $F_0$ represent initial condition of those variables). Our goal is to find the general solution, which is the sum of the complementary and particular solutions:

The differential equation is the following:

$$\ddot{x} + 2\beta\dot x + \omega_0^2x = F_0\cos\omega t$$

c) For the particular solution, try the following function:

$$x_p(t) = B_1\cos\omega t + B_2\sin\omega t = F_0\cos\omega t$$

Substitute the trial solution into your equation of motion, rearrange, and equate the coefficients to find expressions for $B_1$ and $B_2$.

First, the first derivative and second derivatives of the particular solution were taken, which is the following: \begin{align} x_p(t) &= B_1\cos\omega t + B_2\sin\omega t\\ \dot x_p(t) &= -\omega B_1\sin\omega t + \omega B_2\cos\omega t\\ \ddot x_p(t) &= -(\omega^2)B_1\cos\omega t - \omega^2B_2\sin\omega t \end{align} Plugging into the differential equation we get the following:

$$(-\omega^2B_1\cos\omega t - \omega^2B_2\sin\omega t) + 2\beta\omega(-B_1\sin\omega t + B_2\cos\omega t) + \omega_0^2(B_1\cos\omega t + B_2\sin\omega t) = F_0\cos\omega t$$

Then the terms were simplified into cos and sin:

$$[-\omega^2B_1 + 2\beta\omega B_2 + \omega_0^2B_1]\cos\omega t + [-\omega^2B_2 - 2\beta\omega B_1 + \omega_0^2B_2]\sin\omega t = F_0\cos\omega t$$

How do I obtain the equations for B(1) and B(2) when there is only one equation present?

Using G. Paily's suggestion, since there does not exist any constant such that csinωt = F(0)cosωt, then the sin terms must be equal to zero. Using this fact, we can create two equations: cos terms = F(o)cosωt and sin terms = 0. First use the sin terms to figure out what B(2) is, which is the following:

B(2) = 2βωB(1)/(ω(o)^2 -ω^2)

Taking B(2) and substituting into the cos terms after simplification, I we are left with:

B(1)cosωt(ω(o)^2 + 4(β^2)(ω^2)/(ω(o)^2-ω^2) - ω^2) = F(o)cosωt

After substituting for β and simplifying again, we derive the equation for B(1):

B(1) = F(o)/(ω(o)^2 + ω^2/[4(ω(o)^2-1)] -ω^2

The final answer is the following:

B(1) = F(o)/(ω(o)^2 + ω^2/[4(ω(o)^2-1)] -ω^2

B(2) = 2βωB(1)/(ω(o)^2 -ω^2)

Thank you for your help.


The fact that you can eliminate the $\sin$ term tells you that its coefficient $[-\omega^2 B_{2}-2\beta\omega B_{1}+\omega_{0}^{2}B_{2}]$ must be zero. This gives you another relation between $B_{1}$ and $B_{2}$, and with two equations and two unknowns, you can solve cleanly for $B_{1}$ and $B_{2}$.

| cite | improve this answer | |

First of all you have a mistake in calculus: you say "Plugging into the differential equation I get: ..." Do check, you improperly substituted x¨_p (t). You left a minus inside the round parentheses that should be plus. So, in fact you have, not minus.

−ω^2 B_1 cos(ωt) − ω^2 B_2 sin(ωt) +2βω B_2 cos(ωt) − 2βω B_1 sin(ωt) +(ω_0)^2 B_1 cos(ωt) + (ω_0)^2 B_2 sin(ωt) -F_0 cos(ωt) = 0

So, you can immediately get your equations for B_1 and B_2, by equating separately the coefficients of B_1, and those of B_2 ,

[(ω_0)^2 − ω^2]B_1 + 2βω B_2 = F_0, [(ω_0)^2 − ω^2]B_2 − 2βω B_1 = 0 .

Now, from the 2nd equality you get,

(1) B_2 = 2βω B_1/[(ω_0)^2 − ω^2] ,

and substituting in the 1st equation you get an equation exclusively for B_1:

B_1{[(ω_0)^2 − ω^2] + (2βω)^2 /[(ω_0)^2 − ω^2]} = F_0 ,

from which

(2) B_1 = F_0 / {[(ω_0)^2 − ω^2] + (2βω)^2 /[(ω_0)^2 − ω^2]} ,

Substituting in (1),

(3) B_2 = 2βω F_0/{[(ω_0)^2 − ω^2]^2 + (2βω)^2} .

You see, the interesting conclusion is that for ω = ω_0 you have an amplification of the sinusoidal part of the solution, B_2 = F_0/(2βω) , while the cosine part becomes zero (B_1 = 0). On the other hand, if β=0, the wave remains purely cosine, as the driving force, and what is more, if ω = ω_0, B_1 -> infinity.

Good luck !

| cite | improve this answer | |
  • $\begingroup$ You can use latex formatting. It helps in the understanding. Use \$ equation here \$. Also, you can write the Greek letters by writing: \beta, for instance, inside the \$. You can write nice formatted fractions the following way: \$\frac{numerator}{denominator}\$. Also, for a main equation use \$\$ equation here \$\$. It will be in the center. Take a read on this. =) $\endgroup$ – Physicist137 Nov 27 '14 at 2:54

Not the answer you're looking for? Browse other questions tagged or ask your own question.