Forced oscillation, Equation of Motion and verifying solutions

[sm1t]

Hi, Been registered for a while here, but this is my first post, been using the forum as more of a resource before. I am going through some past papers, but I am faltering at 1 question.

Homework Statement

A particle moving along the x-axis with velocity v experiences a resistive force –bv, but no spring-like restoring force, where b is the friction coefficient.

(a) Write down the equation of motion. [3]
$$\ddot{x}= -(b/m)v$$

(b) Show that the equation of motion is satisfied by
$$x(t) = C - (v_{0}/\gamma)e^{-\gamma}$$

where , m is the mass of the particle, and C and $$\gamma=(b/m)$$ are free parameters. [3]

Not the quickest with latex so the answer was to just differentiate twice, you can then see they are equivalent.

(c) At the particle is at rest at t=0 x=0. At this instant a driving force is switched on $$F = F_{0}cos(\omega*t)$$ what is the equation of motion for t > 0 ? [2]


$$\ddot{x}= (F_{0}/m)cos(\omega*t) -(b/m)v$$

(d) Show that, when both forces are present, x(t)= A*cos(ωt−δ) is a solution to the equation of motion with appropriate choice of A and δ. Find A and δ .
We are also told that [cos(δ) = 1/(rootof 1 +tan^2(δ)] and sin(δ) = tan(δ)/(rootof 1 +tan^2(δ)]

Again I try to differentiate through but I come to a block, I use the 2 above identities but doesn't help me? I know I must be missing something simple or just not seeing it.


Any help much appreciated.

 

[Quantumpencil]

If you haven't already, try verifying it by using complex exponentials rather than trig functions rather than trig functions. It's usually much easier.

 

[thomate1]

Hello there, welcome to the forum! I will try to help you with this question.

First, let's review the concept of forced oscillation. When a particle experiences a resistive force, it may still oscillate due to a driving force that is applied. In this case, the equation of motion for the particle can be written as:

\ddot{x} = -\gamma\dot{x} + \frac{F}{m}

where \gamma = \frac{b}{m} is the damping coefficient, \dot{x} is the velocity of the particle, and F is the driving force.

Now, in part (c) of the question, we are asked to find the equation of motion for t > 0, when the driving force is switched on. As you have correctly written, the equation of motion in this case is:

\ddot{x} = \frac{F_{0}}{m}\cos{\omega t} - \gamma\dot{x}

To solve this equation, we can use the method of undetermined coefficients. We assume a solution of the form:

x(t) = A\cos(\omega t - \delta)

where A and \delta are constants to be determined. Substituting this into the equation of motion and simplifying, we get:

\ddot{x} + \gamma\dot{x} + \omega^{2}x = \frac{F_{0}}{m}\cos{\omega t}

Comparing this with the original equation, we can see that A = \frac{F_{0}}{m\omega^{2}} and \delta = 0. Therefore, the solution for x(t) is:

x(t) = \frac{F_{0}}{m\omega^{2}}\cos(\omega t)

Now, to show that this solution satisfies the equation of motion, we can differentiate it twice and substitute it into the equation. You should get the same result as the original equation, thus verifying the solution.

For part (d) of the question, we are asked to find the values of A and \delta when both forces are present. Using the identities provided, we can write:

A = \frac{F_{0}}{m\omega^{2}}\cos(\delta)

and

\sin(\delta) = \frac{\tan(\delta)}{\sqrt{1 + \tan^{2}(\delta)}}

Sub