# Velocity function of a variable force

1. Jul 6, 2008

### Piamedes

1. The problem statement, all variables and given/known data
A particle with a mass of 3kg is at rest at x = 3m, and then a force = (12 N/m)x is applied to it. What is the acceleration of the particle when it reaches x = 5m? Determine the v(t) for the particle.

2. Relevant equations
$$\sum F = ma$$

3. The attempt at a solution

The first part was simple. I just solved for acceleration and got the same answer as the book, 20 m/s^2. However there was no answer for the second part so I want to know if I solved for v(t) properly.

$$a = (4 \frac {1}{s^2}) x$$

$$a = \frac {d^2x} {dt^2}$$

$$\frac {d^2x} {dt^2} - (4 \frac {1}{s^2}) x = 0$$

Which is just a 2nd order differential equation, whose general solution is:
$$x = C_1 e^(\lambda_1 t) + C_2 e^(\lambda_2 t)$$

Where lambda is the solution to the auxiliary equation,
$$\lambda^2 - 4\lambda = 0$$

Therefore
$$x = C_1 e^(2 t) + C_2 e^(-2 t)$$

and v(t) is
$$v = 2 C_1 e^(2 t) - 2 C_2 e^(-2 t)$$

The problem stipulates that at x=3m, v=0, so when if I set t=0 at x=3m, then I can solve for C1 and C2
$$3 = C_1 + C_2$$

$$0 = 2 C_1 - 2 C_2$$

$$C_1 = \frac {3}{2} and C_2 = \frac {-3}{2}$$

So v(t) becomes
$$v = (3 \frac {m}{s}) ( e^(2 t) + e^(-2 t) )$$

Then I wanted to make it look nicer so I rewrote it as:
$$v = (6 \frac {m}{s}) \cosh {(2 \frac {1}{s} t)}$$

Is this the proper solution to the question?
If not could someone please explain to me how to get the correct one. Thank you for the help.

2. Jul 6, 2008

### alphysicist

Hi Piamedes,

Since the problem involves v and x, I would write

$$\frac {d^2x} {dt^2}$$

in a form that does not have t in it:

$$\frac {d^2x} {dt^2} =v \frac{dv}{dx}$$

Then you can integrate over v and x directly. (But you might also want to drop the (1/s^2) that is inside your equation.)

3. Jul 6, 2008

### Piamedes

I tried doing that at first, but since the problem required v as a function of time when I integrated again to solve for x as a function of t it came out to
$$\ln [x + \sqrt{x^2 - 9}] = 2 t$$

So when I tried solving for x it I lost the +/- portion of the square. I went back and used the general solution because it seemed easier than trying to figure out a mistake in my page of algebra work.

Does doing it different ways yield different answers?

4. Jul 6, 2008

### alphysicist

I'm not sure I understand what happened, but different way should definitely give the same answer.

Looking over your original post, I think your values for C1 and C2 are incorrect. Your second equation for them is:

$$0=2 C_1 - 2 C_2$$
so they must be equal.

5. Jul 7, 2008

### Piamedes

oops. I solved it all the way through to the end and messed up some algebra.
Its should be
$$C_1 = C_2 = \frac {3}{2}$$

Which means that
$$x = (\frac {3}{2} \frac {m}{s}) (e^(2 t) + e^(-2 t) ) = (\frac {3}{2} \frac {m}{s}) \cosh {(2t)}$$

$$x = (3 \frac{m}{s}) (e^(2 t) - e^(-2 t) ) = (3 \frac {m}{s}) \sinh {(2t)}$$

$$a = (6 \frac {m}{s}) (e^(2 t) + e^(-2 t) ) = (6 \frac {m}{s}) \cosh {(2t)}$$

And if I check from the original equation
$$a = 4x$$

$$a = 4 ( (\frac {3}{2} \frac {m}{s}) \cosh {(2t)} ) = 4x$$

By showing that my solution fits the given parameters, is that proof that the answer is correct or am I still missing something.
Thanks for all the help so far.

6. Jul 8, 2008

### alphysicist

You have a small error here. Notice that the exponential form gives x=3 at t=0, while the cosh from doesn't.

7. Jul 8, 2008

### Piamedes

Thanks for catching that error. I must have forgotten to factor out the 1/2 when I transformed it into cosh.

It should be
$$x = (3 \frac {m}{s}) \cosh {(2t)}$$

Thanks for the help.

8. Jul 8, 2008

### alphysicist

Sure, glad to help! (And the way you put all the details of your work in your posts definitely made it easy to read through.)

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