Expression for a critically damped system

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SUMMARY

The discussion focuses on deriving the expression for maximum displacement in a critically damped system subjected to an initial velocity V. The user starts with the equation x = A exp(-rt/2m) + B*t* exp(-rt/2m) and determines that A equals zero due to initial displacement being zero. The differentiation of the displacement equation is performed using the product rule, leading to the conclusion that the maximum displacement can be expressed as x maximum = Vt exp(-rt/2m). This derivation confirms the correct application of mathematical principles in the context of simple harmonic motion (SHM) with damping.

PREREQUISITES
  • Understanding of critically damped systems in physics
  • Familiarity with simple harmonic motion (SHM) concepts
  • Knowledge of differentiation techniques, including the product rule
  • Basic understanding of exponential functions and their applications in motion equations
NEXT STEPS
  • Study the principles of critically damped systems in more detail
  • Learn about the mathematical modeling of simple harmonic motion (SHM)
  • Explore advanced differentiation techniques, especially in the context of physics
  • Investigate the implications of damping on system behavior in engineering applications
USEFUL FOR

Students studying physics, particularly those focusing on mechanics and dynamics, as well as educators and professionals involved in teaching or applying concepts of simple harmonic motion and damping in real-world scenarios.

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?
 

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