- 194

- 1

## Main Question or Discussion Point

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_

Last edited: