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Solution to a differential equation of motion

0

$begingroup$

Can someone please check my working for the following question?

"A particle moves along the x-axis under the action of a conservative force F1(x) with potential energy V(x) = $k(x-l)^2/2$ and a force of $F2(dot x)=vdot x$ , where k, l,v are positive constants, v < $2sqrt km$"

a) find the differential equation of motion

b) find the general solution to the equation

---

a) m$ddot x$= $dv/dx$ + v$dot x$

[V(x)]' = k(x-l)

m$ddot x$ = k(x-l) + v$dot x$

m$ddot x$- kx+ kl- v$dot x$=0

$ddot x$ - $(k/m)$x+ $(k/m)$l- (v/m)$dot x$ = 0

b)
$ddot x$ - $(k/m)$x+ $(k/m)$l- (vm)/$dot x$ = 0

let $xi$ = x-l

$ddotxi$ - $(v/m)dot xi$ -$(k/m)xi$ = 0

let $omega^2 = (k/m)$ and $delta= v/m$

$ddotxi$ - $dot xi$$delta -$$omega^2xi$ = 0

Giving us the characteristic equation:
$lambda^2$ - $delta$$lambda$ - $omega^2$ = 0

Roots: $lambda_1 and lambda_2$

= [$delta pm$ $sqrt (-(delta^2)-4(-omega^2)$]/2

= [$delta pm$ $sqrt ((delta^2)+ 4omega^2$]/2

$b^2-4ac>0$, so roots are real

General solution is:

$xi$(t) = $c_1$e^($lambda_1t$) + $c_1$e^($lambda_2t$)

x(t) = l + $c_1$e^($lambda_1t$) + $c_1$e^($lambda_2t$)

ordinary-differential-equations proof-verification physics

share|cite|improve this question

edited Mar 25 at 21:52

Lee David Chung Lin

4,50841342

asked Mar 25 at 19:40

vjhyvjhvjhvjhgvjhyvjhvjhvjhg

12

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Welcome to MSE. Please edit and use MathJax to properly format math expressions.
  $endgroup$
  – Lee David Chung Lin
  Mar 25 at 21:51
  

  
  
  
  

add a comment | 

0

$begingroup$

Can someone please check my working for the following question?

"A particle moves along the x-axis under the action of a conservative force F1(x) with potential energy V(x) = $k(x-l)^2/2$ and a force of $F2(dot x)=vdot x$ , where k, l,v are positive constants, v < $2sqrt km$"

a) find the differential equation of motion

b) find the general solution to the equation

---

a) m$ddot x$= $dv/dx$ + v$dot x$

[V(x)]' = k(x-l)

m$ddot x$ = k(x-l) + v$dot x$

m$ddot x$- kx+ kl- v$dot x$=0

$ddot x$ - $(k/m)$x+ $(k/m)$l- (v/m)$dot x$ = 0

b)
$ddot x$ - $(k/m)$x+ $(k/m)$l- (vm)/$dot x$ = 0

let $xi$ = x-l

$ddotxi$ - $(v/m)dot xi$ -$(k/m)xi$ = 0

let $omega^2 = (k/m)$ and $delta= v/m$

$ddotxi$ - $dot xi$$delta -$$omega^2xi$ = 0

Giving us the characteristic equation:
$lambda^2$ - $delta$$lambda$ - $omega^2$ = 0

Roots: $lambda_1 and lambda_2$

= [$delta pm$ $sqrt (-(delta^2)-4(-omega^2)$]/2

= [$delta pm$ $sqrt ((delta^2)+ 4omega^2$]/2

$b^2-4ac>0$, so roots are real

General solution is:

$xi$(t) = $c_1$e^($lambda_1t$) + $c_1$e^($lambda_2t$)

x(t) = l + $c_1$e^($lambda_1t$) + $c_1$e^($lambda_2t$)

ordinary-differential-equations proof-verification physics

share|cite|improve this question

edited Mar 25 at 21:52

Lee David Chung Lin

4,50841342

asked Mar 25 at 19:40

vjhyvjhvjhvjhgvjhyvjhvjhvjhg

12

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Welcome to MSE. Please edit and use MathJax to properly format math expressions.
  $endgroup$
  – Lee David Chung Lin
  Mar 25 at 21:51
  

  
  
  
  

add a comment | 

$begingroup$

Can someone please check my working for the following question?

"A particle moves along the x-axis under the action of a conservative force F1(x) with potential energy V(x) = $k(x-l)^2/2$ and a force of $F2(dot x)=vdot x$ , where k, l,v are positive constants, v < $2sqrt km$"

a) find the differential equation of motion

b) find the general solution to the equation

---

a) m$ddot x$= $dv/dx$ + v$dot x$

[V(x)]' = k(x-l)

m$ddot x$ = k(x-l) + v$dot x$

m$ddot x$- kx+ kl- v$dot x$=0

$ddot x$ - $(k/m)$x+ $(k/m)$l- (v/m)$dot x$ = 0

b)
$ddot x$ - $(k/m)$x+ $(k/m)$l- (vm)/$dot x$ = 0

let $xi$ = x-l

$ddotxi$ - $(v/m)dot xi$ -$(k/m)xi$ = 0

let $omega^2 = (k/m)$ and $delta= v/m$

$ddotxi$ - $dot xi$$delta -$$omega^2xi$ = 0

Giving us the characteristic equation:
$lambda^2$ - $delta$$lambda$ - $omega^2$ = 0

Roots: $lambda_1 and lambda_2$

= [$delta pm$ $sqrt (-(delta^2)-4(-omega^2)$]/2

= [$delta pm$ $sqrt ((delta^2)+ 4omega^2$]/2

$b^2-4ac>0$, so roots are real

General solution is:

$xi$(t) = $c_1$e^($lambda_1t$) + $c_1$e^($lambda_2t$)

x(t) = l + $c_1$e^($lambda_1t$) + $c_1$e^($lambda_2t$)

ordinary-differential-equations proof-verification physics

share|cite|improve this question

edited Mar 25 at 21:52

Lee David Chung Lin

4,50841342

asked Mar 25 at 19:40

vjhyvjhvjhvjhgvjhyvjhvjhvjhg

12

$endgroup$

share|cite|improve this question

edited Mar 25 at 21:52

Lee David Chung Lin

4,50841342

asked Mar 25 at 19:40

vjhyvjhvjhvjhgvjhyvjhvjhvjhg

12

share|cite|improve this question

edited Mar 25 at 21:52

Lee David Chung Lin

4,50841342

asked Mar 25 at 19:40

vjhyvjhvjhvjhgvjhyvjhvjhvjhg

12

edited Mar 25 at 21:52

Lee David Chung Lin

4,50841342

edited Mar 25 at 21:52

Lee David Chung Lin

4,50841342

asked Mar 25 at 19:40

vjhyvjhvjhvjhgvjhyvjhvjhvjhg

12

asked Mar 25 at 19:40

vjhyvjhvjhvjhgvjhyvjhvjhvjhg

12