How to solve a differential equation involving spring constants?
$begingroup$
A spring with a spring constant $k$ of $320$ Newtons per meter is loaded with a $5$-kg weight and allowed to reach equilibrium. It is then displaced $1$ meter downward and released. If the weight experiences a damping force in Newtons equal to $64$ times the velocity at every point, find the equation of motion.
$y(t)= ?$
where $t$ is time and $y(t)$ is displacement in time.
So I found that the characteristic equation would be;
$$5r^2 + 64r + 320 = 0$$
and I solved that equation for its roots which are:
$r = -(32/5) + (24/5)i$ and $-(32/5) - (24/5)i$
So the general equation would be:
$$c_1e^{-32t/5} cos(24t/5) + c_2e^{-32t/5} sin(24t/5) = y(t)$$
The only initial condition I can pull from the problem is $y(0) = 1$ which tells me that $c_1=1$ but I do not know how to find $c_2$.
Are there more initial conditions in the problem or is my method wrong?
calculus ordinary-differential-equations
edited Dec 13 '18 at 2:48
Andrei
12.4k21128
asked Dec 13 '18 at 2:31
S. SnakeS. Snake
485
$endgroup$
add a comment |
$begingroup$
A spring with a spring constant $k$ of $320$ Newtons per meter is loaded with a $5$-kg weight and allowed to reach equilibrium. It is then displaced $1$ meter downward and released. If the weight experiences a damping force in Newtons equal to $64$ times the velocity at every point, find the equation of motion.
$y(t)= ?$
where $t$ is time and $y(t)$ is displacement in time.
So I found that the characteristic equation would be;
$$5r^2 + 64r + 320 = 0$$
and I solved that equation for its roots which are:
$r = -(32/5) + (24/5)i$ and $-(32/5) - (24/5)i$
So the general equation would be:
$$c_1e^{-32t/5} cos(24t/5) + c_2e^{-32t/5} sin(24t/5) = y(t)$$
The only initial condition I can pull from the problem is $y(0) = 1$ which tells me that $c_1=1$ but I do not know how to find $c_2$.
Are there more initial conditions in the problem or is my method wrong?
calculus ordinary-differential-equations
edited Dec 13 '18 at 2:48
Andrei
12.4k21128
asked Dec 13 '18 at 2:31
S. SnakeS. Snake
485
$endgroup$
add a comment |
$begingroup$
A spring with a spring constant $k$ of $320$ Newtons per meter is loaded with a $5$-kg weight and allowed to reach equilibrium. It is then displaced $1$ meter downward and released. If the weight experiences a damping force in Newtons equal to $64$ times the velocity at every point, find the equation of motion.
$y(t)= ?$
where $t$ is time and $y(t)$ is displacement in time.
So I found that the characteristic equation would be;
$$5r^2 + 64r + 320 = 0$$
and I solved that equation for its roots which are:
$r = -(32/5) + (24/5)i$ and $-(32/5) - (24/5)i$
So the general equation would be:
$$c_1e^{-32t/5} cos(24t/5) + c_2e^{-32t/5} sin(24t/5) = y(t)$$
The only initial condition I can pull from the problem is $y(0) = 1$ which tells me that $c_1=1$ but I do not know how to find $c_2$.
Are there more initial conditions in the problem or is my method wrong?
calculus ordinary-differential-equations
edited Dec 13 '18 at 2:48
Andrei
12.4k21128
asked Dec 13 '18 at 2:31
S. SnakeS. Snake
485
$endgroup$
A spring with a spring constant $k$ of $320$ Newtons per meter is loaded with a $5$-kg weight and allowed to reach equilibrium. It is then displaced $1$ meter downward and released. If the weight experiences a damping force in Newtons equal to $64$ times the velocity at every point, find the equation of motion.
$y(t)= ?$
where $t$ is time and $y(t)$ is displacement in time.
So I found that the characteristic equation would be;
$$5r^2 + 64r + 320 = 0$$
and I solved that equation for its roots which are:
$r = -(32/5) + (24/5)i$ and $-(32/5) - (24/5)i$
So the general equation would be:
$$c_1e^{-32t/5} cos(24t/5) + c_2e^{-32t/5} sin(24t/5) = y(t)$$
The only initial condition I can pull from the problem is $y(0) = 1$ which tells me that $c_1=1$ but I do not know how to find $c_2$.
Are there more initial conditions in the problem or is my method wrong?
calculus ordinary-differential-equations
calculus ordinary-differential-equations
edited Dec 13 '18 at 2:48
Andrei
12.4k21128
asked Dec 13 '18 at 2:31
S. SnakeS. Snake
485
edited Dec 13 '18 at 2:48
Andrei
12.4k21128
asked Dec 13 '18 at 2:31
S. SnakeS. Snake
485
edited Dec 13 '18 at 2:48
Andrei
12.4k21128
edited Dec 13 '18 at 2:48
Andrei
12.4k21128
edited Dec 13 '18 at 2:48
Andrei
12.4k21128
12.4k21128
asked Dec 13 '18 at 2:31
S. SnakeS. Snake
485
asked Dec 13 '18 at 2:31
S. SnakeS. Snake
485
asked Dec 13 '18 at 2:31
S. SnakeS. Snake
485
485
add a comment |
add a comment |
1 Answer
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Take the derivative, to calculate the velocity as a function of time. The weight is released from rest, so $y'(t=0)=0$
answered Dec 13 '18 at 2:43
AndreiAndrei
12.4k21128
$endgroup$
add a comment |
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1 Answer
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$begingroup$
Take the derivative, to calculate the velocity as a function of time. The weight is released from rest, so $y'(t=0)=0$
answered Dec 13 '18 at 2:43
AndreiAndrei
12.4k21128
$endgroup$
add a comment |
$begingroup$
Take the derivative, to calculate the velocity as a function of time. The weight is released from rest, so $y'(t=0)=0$
answered Dec 13 '18 at 2:43
AndreiAndrei
12.4k21128
$endgroup$
add a comment |
$begingroup$
Take the derivative, to calculate the velocity as a function of time. The weight is released from rest, so $y'(t=0)=0$
answered Dec 13 '18 at 2:43
AndreiAndrei
12.4k21128
$endgroup$
Take the derivative, to calculate the velocity as a function of time. The weight is released from rest, so $y'(t=0)=0$
answered Dec 13 '18 at 2:43
AndreiAndrei
12.4k21128
answered Dec 13 '18 at 2:43
AndreiAndrei
12.4k21128
answered Dec 13 '18 at 2:43
AndreiAndrei
12.4k21128
answered Dec 13 '18 at 2:43
AndreiAndrei
12.4k21128
12.4k21128
add a comment |
add a comment |
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