# HELP! Calling Calculus Gurus...

• Oct 30th 2008, 07:21 PM
makgis
Hi there,

I have the following equation to determine the performance properties of a damper (shock absorber).

d(t) = -5e^-5t cos(10t)

u = -5 x 10^-5t
du/dx = 25 x 10^-5t

v = cos(10t)
dv/dx = -10 sin(10t)

Using the given Product rule equation, is applied to the first derivative:

d(uv)/dx = u/dv/dx + v/du/dx

Therefore
d(uv)/dx = 25 x 10^-5t x cos(10t) + -10 sin(10t) x -5 x 10^-5t

= 25 x 10^-5t (cos(10t) + 2 sin(10t))
0 = cos(10t) + 2 sin(10t)
0 = tan(10t) = -1/2

I have attached the model showing measured and modelled deflection for a given damper (shock).

Now I need to answer the following questions; can anyone help....

a) Can anyone explain how I determine the velocity of the rod as it passes the rest position? (please refer to attached graph).

b) how far the rod moves below the rest position after first rebound (dm)? (please refer to attached graph).

• Oct 31st 2008, 11:44 AM
anywho
I'm no guru but...

a) Can anyone explain how I determine the velocity of the rod as it passes the rest position? (please refer to attached graph).

The equation for velocity is simply the first derivative - evaluated it at the value for time when the rod passes the rest position.

b) how far the rod moves below the rest position after first rebound (dm)? (please refer to attached graph).

Graph it and find the min at that point? Evaluate the function at that time to find the postition of the piston?
• Oct 31st 2008, 01:00 PM
HallsofIvy
Quote:

Originally Posted by makgis
Hi there,

I have the following equation to determine the performance properties of a damper (shock absorber).

d(t) = -5e-5t cos(10t)

\$\displaystyle -5e^{-5t} cos(10t)\$?

Quote:

u = -5 x 10-5t
? If you are doing this as "u(t)v(t)" then \$\displaystyle u= -5 e^{-5t}\$
where did you get the "10- 5t"?

Quote:

v = cos(10t)

Using the given Product rule equation, is applied to the first derivative:
d(uv)/dx = u/dv/dx + v/du/dx

du/dx = 25 x 10-5t
dv/dx = -10 sin(10t)

Therefore d(uv)/dx

= 25 x 10-5t x cos(10t) + -10 sin(10t) x -5 x 10-5t
= 25 x 10-5t (cos( + 2 sin(10t))
0 = cos(10t) + 2 sin(10t)
0 = tan(10t) = -1/2

Therefore given that y = tan(10t) we can determine the maximum positive rebound height at the first rebound, by using an approximate x value (time) read from the figure 2; x = 0.27.

y = tan(10t)
y = tan(10 x 0.27) + π/2
y = 1.098

I have attached the model showing measured and modelled deflection for a given damper (shock).

Now I need to answer the following questions; can anyone help....

a) Can anyone explain how I determine the velocity of the rod as it passes the rest position? (please refer to attached graph).

b) how far the rod moves below the rest position after first rebound (dm)? (please refer to attached graph).

• Oct 31st 2008, 04:37 PM
makgis
Quote:

Originally Posted by HallsofIvy
\$\displaystyle -5e^{-5t} cos(10t)\$?

? If you are doing this as "u(t)v(t)" then \$\displaystyle u= -5 e^{-5t}\$
where did you get the "10- 5t"?

Hi yes should be d(t) = -5 x 10^-5t cos(10t)
u = -5 x 10^-5t
v = cos(10t)

cheers.