# Homework Help: Velocity of a particle large and small times

1. Oct 14, 2007

### rock.freak667

1. The problem statement, all variables and given/known data
The velocity,v, of a particle moving on the x-axis varies with the time,t,according to the differential equation $\frac{dv}{dt}+2v=sint$. Find v in terms of t given that v=0 when t=0. Deduce that when t is large;the velocity of the particle varies between -0.45 and 0.45. Show that is t is very small,then $v\approx\frac{1}{2}t^2$

2. Relevant equations

$$\frac{d}{dx}(\frac{e^{2t}}{5}(2sint-cost))=e^{2t}sint$$

3. The attempt at a solution

$$\frac{dv}{dt}+2v=sint$$

$$X e^{2t}$$

$$e^{2t}\frac{dv}{dt}+2e^{2t}v=e^{2t}sint$$

$$\int...dt$$

$$ve^{2t}=\frac{e^{2t}}{5}(2sint-cost)+C$$

using the initial values they gave me $C=\frac{1}{5}$

And so
$$v=\frac{1}{5}(2sint-cost)+\frac{e^{-2t}}{5}$$

As $t\rightarrow\infty(large),\frac{e^{-2t}}{5}\rightarrow0$

but $\frac{1}{5}(2sint-cost)$ does not approach any limit as both cost and sint oscillate between 1 and -1.

2. Oct 14, 2007

### real10

no not between -1 and 1 that is for individual terms.You have to combine them into 1 cosine term and the new amplitude is ur answer.
use the below formula:
A sint + B cost = (A^2+B^2 ) ^(1/2) * Cos(t-arctan(b/A))

3. Oct 14, 2007

### rock.freak667

but wouldn't Cos(t-arctan(b/A)) not approach a limit? Or would t tending to infinity mean that t would eventually be equal toarctan(B/A) and so it would be the cos0=1?

4. Oct 14, 2007

### bob1182006

the final limit will include a function that oscillates between some numbers.
since that function will oscillate it means that the velocity will vary.
Between [-0.45,0.45] in this case.
You will have to plug in the numbers that lead to the maximum value (0.45) and the minimum (-0.45). and that will be it for t->infinity.

5. Oct 14, 2007

### rock.freak667

Oh....thank you...shall try the next part now

6. Oct 14, 2007

### real10

yes it is just the amplitude as I said [( 2/5)^2 + (1/5)^2 ] ^1/2 = +-0.45