- #1
Piamedes
- 41
- 0
Homework Statement
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.
Homework Equations
[tex]
\sum F = ma
[/tex]
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.
[tex]
a = (4 \frac {1}{s^2}) x
[/tex]
[tex]
a = \frac {d^2x} {dt^2}
[/tex]
[tex]
\frac {d^2x} {dt^2} - (4 \frac {1}{s^2}) x = 0
[/tex]
Which is just a 2nd order differential equation, whose general solution is:
[tex]
x = C_1 e^(\lambda_1 t) + C_2 e^(\lambda_2 t)
[/tex]
Where lambda is the solution to the auxiliary equation,
[tex]
\lambda^2 - 4\lambda = 0
[/tex]
Therefore
[tex]
x = C_1 e^(2 t) + C_2 e^(-2 t)
[/tex]
and v(t) is
[tex]
v = 2 C_1 e^(2 t) - 2 C_2 e^(-2 t)
[/tex]
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
[tex]
3 = C_1 + C_2
[/tex]
[tex]
0 = 2 C_1 - 2 C_2
[/tex]
[tex]
C_1 = \frac {3}{2} and C_2 = \frac {-3}{2}
[/tex]
So v(t) becomes
[tex]
v = (3 \frac {m}{s}) ( e^(2 t) + e^(-2 t) )
[/tex]
Then I wanted to make it look nicer so I rewrote it as:
[tex]
v = (6 \frac {m}{s}) \cosh {(2 \frac {1}{s} t)}
[/tex]
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.