Classical mechanics equation of motion

[shyta]

Homework Statement

A point mass m moving along the z axis experiences a time dependent force and a fricitional force. Solve the equation of motion

m\ddot{z} = -m\gamma\dot{z} + F(t)

to find v(t) = \dot{z}(t) for the initial velocity \dot{z}(0) = v_0
Hint: what is the time derivative of e^{\gamma t}v(t)

The Attempt at a Solution

So I made use of the hint and got e^{\gamma t} (\ddot{z}(t) + \gamma\dot{z}(t) )

Manipulating the equation of motion, I got e^{\gamma t} (\ddot{z}(t) + \gamma\dot{z}(t) ) = e^{\gamma t} 1/m F(t)

Subbing in the hint and integrating: \dot{z}(t) = e^{-\gamma t}/m \int e^{\gamma t} F(t) dt

Just wondering if this is correct? and how do I make use of the initial condition v_0?

 

[I like Serena]

Hi again shyta!

Yes, that is correct. Good!

As for the initial condition.
What do you get if you substitute t=0 in your final formula?

 

[shyta]

Omg hi iloveserena again hahaha

For \dot{z}(t) = e^{-\gamma t}/m \int e^{\gamma t} F(t) dt

v_0 = 1/m \int e^{\gamma t} F(t) dt

This is the part I'm stuck at, I'm not sure what to do with the integration function :(

 

[I like Serena]

Well, let me rewrite if for you again:
\dot z(t) = {e^{−γt} \over m} \int_0^t e^{γT} F(T) dT + C

I've also added the integration constant C that vanishes when you take the derivative.
Can you substitute t=0 in this?

 

[shyta]

\dot z(t) = {e^{−γt} \over m} \int_0^t e^{γT} F(T) dT + C

mmm..

\dot z(0) = {1 \over m} \int_0^0 e^{γT} F(T) dT + C


I really have no clue on this part :S

 

[I like Serena]

Did you know that the integral of a function corresponds to the area of the surface enclosed by the function, the x-axis and 2 vertical lines at each limit?

In other words, suppose F(x) is the anti-derivative of f(x), what is:
\int_a^a f(x) dx

 

[shyta]

Hey wait! integration of 0 to 0 for any function is 0 right? so v_0 = C :O

 

[shyta]

hahah yes! omg how could I not see that!

 

[I like Serena]

Yes, that's exactly it!

 

[shyta]

thanks once again :D