# Homework Help: Expression for a critically damped system

1. Sep 1, 2008

### hemetite

man..i have 5 questions on SHM with damping..and it so difficult..it seem that the book have little coverage on this..

Qn1
A system, which is critically damped, has zero displacement at time t=0 and receive an impulse which gives it an intially velocity V. Obtain an expression for the maximum displacment in the subsequent motion.

i will be using

x=A exp (-rt/2m) + B*t* exp (-rt/2m) --------------> equation 1

I know that A is the amplitude...but what is B?

at t=0, displacement = 0
substitute the values in equation 1 will get
A=0.

next, it said that it has an initial velocity = V at t=0.

so i think i need to express x into velocity...so i need to differentiate dx/dt

x(t)=A exp (-rt/2m) + B*t* exp (-rt/2m)

dx/dt = (A) [-r/2m exp (-rt/2m)] + ( i don't know how to differentiate here...can help me..?)

i think after i differentiate the dx/dt, that will be
V= something...then i will put t= 0 and x= 0

all the values i will get will be put back to equation 1...

what i am tryging to do correct?

2. Sep 1, 2008

### Hootenanny

Staff Emeritus
You're on the right lines and what you have done thus far is correct. To differentiate the second term in your expression for displacement, you need to use the product rule.

3. Sep 1, 2008

### hemetite

been six years i touch math...so sorry..

here i try again...
dx/dt = (A) [-r/2m exp (-rt/2m)] + B + [-r/2m exp (-rt/2m)]

therefore

V= (A) [-r/2m exp (-rt/2m)] + B + [-r/2m exp (-rt/2m)]

substitute t=0, A= 0, B=V

into
x=A exp (-rt/2m) + B*t* exp (-rt/2m)

therefore