How do I find the frequency of oscillation for a damped harmonic oscillator?

Click For Summary
SUMMARY

The discussion focuses on deriving the frequency of oscillation for a damped harmonic oscillator subjected to linear air resistance and gravitational force. The equation of motion is established as x'' + (c/m)x' + (k/m)x - mg = 0, where c represents the damping coefficient, k is the spring constant, and m is the mass. The participants clarify that the gravitational term does not affect the frequency of oscillation and suggest expressing k in terms of a (spring stretch), m, and g (acceleration due to gravity). The final frequency is derived using the characteristic equation and the quadratic formula, leading to r = -g/2v_t ± sqrt(g^2/(4v_t^2) + g/a).

PREREQUISITES
  • Understanding of second-order linear differential equations
  • Familiarity with concepts of damping in oscillatory systems
  • Knowledge of spring mechanics and Hooke's Law
  • Basic proficiency in solving characteristic equations
NEXT STEPS
  • Study the derivation of the characteristic equation for second-order linear differential equations
  • Learn about under-damped, critically damped, and over-damped systems
  • Explore the relationship between angular frequency and frequency of oscillation
  • Investigate the effects of damping on the amplitude and frequency of oscillations
USEFUL FOR

Students studying physics, particularly those focusing on mechanics and oscillatory motion, as well as educators and tutors looking to clarify concepts related to damped harmonic oscillators.

Slightly
Messages
29
Reaction score
0

Homework Statement


The terminal speed of a freely falling object is v_t (assume a linear form of air resistance). When the object is suspended by a spring, the spring stretches by an amount a. Find the formula of the frequency of oscillation in terms of g, v_t, and a.

Homework Equations



the linearly damped force = -cx'

The Attempt at a Solution



I started by writing the equation,

x'' + cx'/m + kx/m - mg = 0

Is this the correct way to start the problem and where would I go to next? I know that undamped frequency is equal to sqrt(k/m) but I'm not sure what to do with the equation. I'm stuck!
 
Physics news on Phys.org
Slightly said:

Homework Statement


The terminal speed of a freely falling object is v_t (assume a linear form of air resistance). When the object is suspended by a spring, the spring stretches by an amount a. Find the formula of the frequency of oscillation in terms of g, v_t, and a.

Homework Equations



the linearly damped force = -cx'

The Attempt at a Solution



I started by writing the equation,

x'' + cx'/m + kx/m - mg = 0

Is this the correct way to start the problem
Almost. In your "mg" term, you forgot to divide by m like you did in your other terms.

Perhaps more importantly though, this mg is a uniform force across all x and it does not depend on x or any of its derivatives.

I think it might be fine to simply remove the "mg" term altogether. It doesn't affect the frequency of oscillation, in this problem. It will affect the center of oscillation, but you're not asked to find that. So if I were you I'd just get rid of it.

Can you express k in terms of a, m and g?

and where would I go to next? I know that undamped frequency is equal to sqrt(k/m) but I'm not sure what to do with the equation. I'm stuck!

So at the moment, you have a second order, homogeneous, linear, differential equation with constant coefficients. Single solutions of this type of equation come in the form x(t) = Ae^{\lambda t} The general solution may involve the linear sum of two, single solutions. (Here, \lambda can be, and generally is, a complex number.)

The first step in solving that is to write down its characteristic equation. Then find the roots of the characteristic equation.

You might wish to consult your textbook/course material on how to solve such differential equations. If not, even a quick Google search on "second order, homogeneous, linear, differential equation with constant coefficients" should produce all you need.
 
collinsmark said:
. (Here, \lambda can be, and generally is, a complex number.)

Can you explain to me why you came to this conclusion? How is this system under-damped?
 
Slightly said:
Can you explain to me why you came to this conclusion? How is this system under-damped?
I never said it was necessarily an imaginary number. Just a complex number (all real numbers are also complex).
 
That's funny.

So when I get the solution to the DE, where will the frequency come from?
 
Slightly said:
That's funny.

So when I get the solution to the DE, where will the frequency come from?
The fact that "frequency of oscillation" was even mentioned, I take that as implying the system is under-damped. I mean, if the system were critically damped or over-damped, there wouldn't be any oscillations at all.

An under-damped system has a sinusoidal function associated with it (like sine or cos). And in the case of a damped harmonic oscillator, it involves the sinusoidal function multiplied by an exponentiation function. It might be helpful to recall \cos{ \left( \omega_1 t \right)} = \frac{1}{2} \left( e^{i \omega_1 t} + e^{-i \omega_1 t} \right). Here, \omega_1 is an angular frequency (having units such as RAD/s).

So you might want to put the general solution in a form like x(t) = A e^{- \alpha t} \left( e^{ i \omega_1 t} + e^{- i \omega_1 t} \right)
 
I went through the DE and was able to obtain the characteristic equation to be

r^2+cr/m + k/m = 0

I said k = - mg/a

and c = mg/v_t

so... r^2+gr/v_t - g/a = o

In this case, can I assume r is the frequency?

so, using the quadratic formula,

r = -g/2v_t +/- sqrt(g^2/(4v^2)+ g/a) is this right?
 
I was able to figure it out! Thanks.
 
Slightly said:
I was able to figure it out! Thanks.
Great! :smile:
 

Similar threads

  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 8 ·
Replies
8
Views
3K
Replies
1
Views
2K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 3 ·
Replies
3
Views
6K
Replies
9
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K