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