# Damped oscillator given odd initial conditions

• oddjobmj
In summary, the damped oscillator is described by the equation mx''+bx'+kx=0 with the condition for critical damping expressed as b=W0. The position x(t) for t>0 can be determined by the general solution, x(t)=(C1+C2t)e-βt, with the initial conditions x(0)=0 and v(0)=v0. Solving for the constants, C1=0 and C2=v0, and substituting the given values of k/m=(2*pi rad/s)^2 and v0=10 m/s, the plot of x(t) can be accurately graphed using an appropriate range for t.
oddjobmj

## Homework Statement

(A) The damped oscillator is described by the equation mx''+bx'+kx=0. What is the condition for critical damping expressed in terms of m,b,k. Assume this is satisfied.

(B) For t<0 the mass is at rest (x=0). This mass is set in motion at t=0 by a sharp impulsive force so that the velocity is v0 at time t=0. Determin the position x(t) for t>0.

(C) Suppose k/m=(2*pi rad/s)2 and v0=10 m/s. Plot an accurate graph of x(t) using an appropriate range for t.

## Homework Equations

For critical damping; B=W0

General solution for critically damped oscillator:

x(t)=(C1+C2t)e-βt

## The Attempt at a Solution

I am running into issues at (B) where I'm not entirely sure how to find the values of the two constants in the general solution.

(A) I have shown through the given relations that b2=4km which I believe is the correct answer.

(B) Given the general solution for a critically damped oscillator I can take its derivative to find v(t):

v(t)=e-βt(C2-C2βt-C1β)

I know that at t=0 v(0)=v0 so I can solve the solution for v(t) for a constant:

C2=v0+C1β

I'm just not sure about how to solve for x(t). Can I consider x(0)=0 since it is set in motion at t=0 but since no time has elapsed it has not moved?

If not, how do I proceed in solving for the constants? I believe the rest of the problem will fall into place without much issue after that point.

Thank you for your time and help!

oddjobmj said:
I'm just not sure about how to solve for x(t). Can I consider x(0)=0 since it is set in motion at t=0 but since no time has elapsed it has not moved?

If not, how do I proceed in solving for the constants? I believe the rest of the problem will fall into place without much issue after that point.

Thank you for your time and help!

The initial conditions are x(0)=0 and v(0)=vo. Impulsive force means that the mass is set into motion in an infinitesimally short time, so the displacement during that time is negligible.

ehild

1 person
ehild said:
The initial conditions are x(0)=0 and v(0)=vo. Impulsive force means that the mass is set into motion in an infinitesimally short time, so the displacement during that time is negligible.

ehild

Ahh, fantastic. That definitely makes sense, thank you.

So, C1=0 and C2=v0 making the general solution:

x(t)=v0te-βt

(C) Since k/m=(2*π rad/s)2 and w02=k/m and w0

x(t)=10te-2πt

The plot:
http://www.wolframalpha.com/input/?i=Plot(x(t)=10*t*e^(-2*pi*t)),+(t,+0,+1)

Thanks again!

## 1. What is a damped oscillator?

A damped oscillator is a physical system that exhibits oscillatory behavior, where the amplitude of the oscillations decreases over time due to the presence of a damping force or resistance.

## 2. What are odd initial conditions for a damped oscillator?

Odd initial conditions refer to the initial conditions of a damped oscillator that do not follow the usual pattern of starting at equilibrium or with zero velocity. This could include starting at a maximum or minimum displacement, or with a nonzero initial velocity.

## 3. How do odd initial conditions affect the behavior of a damped oscillator?

Odd initial conditions can significantly alter the behavior of a damped oscillator. They can result in different amplitudes and frequencies of oscillation, and can also affect the rate of decay of the oscillations.

## 4. What is the equation of motion for a damped oscillator with odd initial conditions?

The equation of motion for a damped oscillator with odd initial conditions is a second-order differential equation that includes terms for the damping force, the spring force, and the initial conditions.

## 5. How can the behavior of a damped oscillator with odd initial conditions be predicted?

The behavior of a damped oscillator with odd initial conditions can be predicted by solving the equation of motion using mathematical techniques such as differential equations or numerical methods. The specific behavior will depend on the values of the damping coefficient, the mass, and the spring constant, as well as the initial conditions.

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