Expression for a critically damped system

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The discussion revolves around deriving an expression for maximum displacement in a critically damped system given initial conditions of zero displacement and an initial velocity V. The user starts with the general solution for displacement and identifies that the amplitude A equals zero due to initial conditions. The challenge lies in determining the constant B, which is linked to the initial velocity. After differentiating the displacement equation and applying the initial conditions, the user arrives at the expression for maximum displacement as Vt exp(-rt/2m). The conversation highlights the difficulty of understanding the mathematics involved in damped harmonic motion.
hemetite
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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 displacement 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?
 
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hemetite said:
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 displacement 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?
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.
 
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

answer
x maximum= Vt exp (-rt/2m)

correct?
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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