- #1
Sparky_
- 227
- 5
Greetings,
Regarding a mass on a spring – I know the classic differential equation is
[tex] m \frac {dx^2(t)}{dt^2} + B \frac {dx(t)}{dt} + kx = f(t) [/tex]
F(t) = outside force applied
B = damping coefficient
“X” is in the vertical direction and +x direction is down.
In reading I have seen that there are some problems involving the “bar” the spring is attached oscillating.
For example – let the bar oscillate according to h(t) where h(t) is the value above or below the original position.
I’ve seen the new differential equation:
[tex] m \frac {dx^2(t)}{dt^2} = -k(x-h) - B \frac {dx(t)} {dt} [/tex]
I don’t quite see why “h” is considered in -k(x-h).
Intuitively, I want to say … –k(h)
I’m thinking h is a value on the x-axis.
Is this a coordinate system problem also?
Can you help clear this up?
Thanks
-Sparky_
Regarding a mass on a spring – I know the classic differential equation is
[tex] m \frac {dx^2(t)}{dt^2} + B \frac {dx(t)}{dt} + kx = f(t) [/tex]
F(t) = outside force applied
B = damping coefficient
“X” is in the vertical direction and +x direction is down.
In reading I have seen that there are some problems involving the “bar” the spring is attached oscillating.
For example – let the bar oscillate according to h(t) where h(t) is the value above or below the original position.
I’ve seen the new differential equation:
[tex] m \frac {dx^2(t)}{dt^2} = -k(x-h) - B \frac {dx(t)} {dt} [/tex]
I don’t quite see why “h” is considered in -k(x-h).
Intuitively, I want to say … –k(h)
I’m thinking h is a value on the x-axis.
Is this a coordinate system problem also?
Can you help clear this up?
Thanks
-Sparky_
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