Damped Simple Harmonic Motion

  • Thread starter roam
  • Start date
  • #1
1,267
11

Homework Statement



http://img534.imageshack.us/img534/6503/questiono.jpg [Broken]

Homework Equations



The Attempt at a Solution



(a) I'm not sure how to approach this part. I appreciate any help to get me started.

(b) I know that the period of oscillation is given by

[itex]T'=\frac{2 \pi}{\omega '}[/itex]

I can use this later on to express my results in terms of the period T'. The one term in the given expression for x(t) that can turn negative as time increases would be the δ in the cosine. So how do I evaluate the right value for the phase delta?
 
Last edited by a moderator:

Answers and Replies

  • #2
1,024
32
for first one ,differentiate x(t) wrt time and put velocity so obtained=0 at t=0,perhaps that will do.
 
  • #3
1,267
11
for first one ,differentiate x(t) wrt time and put velocity so obtained=0 at t=0,perhaps that will do.

Thanks. I'm still a bit confused. When I differentiated the expression with respect to t I got:

[itex]x'(t)=\frac{\left( \frac{A_0}{\cos (\delta)}\right) e^{-t/\tau} (\tau \omega ' \sin(\omega' t + \delta)+\cos(\omega' t + \delta))}{\tau}[/itex]

I've found x'(0):

[itex]x'(0)=\frac{\left( \frac{A_0}{\cos (\delta)}\right) (\tau \omega ' \sin( \delta)+\cos(\delta))}{\tau}[/itex]


But I have to show that δ is given by:

[itex]\tan \delta = - \frac{1}{\omega ' \tau}[/itex]

So I'm not sure, how can do I use the expression for x'(0) to show this?
 
  • #4
1,267
11
Did you mean I write down the velocity as:

[itex]v(t)=\omega ' . \frac{A_0}{\cos (\delta)} e^{-t/\tau} \cos (\omega ' t + \delta)[/itex]

[itex]\therefore \ v(0) =A_0 \omega '[/itex]

So, how does this help? :confused:
 
  • #5
1,024
32
Thanks. I'm still a bit confused. When I differentiated the expression with respect to t I got:

[itex]x'(t)=\frac{\left( \frac{A_0}{\cos (\delta)}\right) e^{-t/\tau} (\tau \omega ' \sin(\omega' t + \delta)+\cos(\omega' t + \delta))}{\tau}[/itex]

I've found x'(0):

[itex]x'(0)=\frac{\left( \frac{A_0}{\cos (\delta)}\right) (\tau \omega ' \sin( \delta)+\cos(\delta))}{\tau}[/itex]


But I have to show that δ is given by:

[itex]\tan \delta = - \frac{1}{\omega ' \tau}[/itex]

So I'm not sure, how can do I use the expression for x'(0) to show this?

ok.you have done it.equate it to zero(because velocity is initially zero) don't you get what is required.
 
  • #6
1,267
11
ok.you have done it.equate it to zero(because velocity is initially zero) don't you get what is required.

I'm a bit confused. What do I have to do once I equate that expression to zero?
 
  • #7
1,024
32
by equating you will get,
w'(tau)sin(delta)=-cos(delta)
which gives
tan(delta)=-1/w'(tau)
 
  • #8
1,267
11
Oh oops I didn't see that part. Thank you so much for the help I really appreciate that!!

Any ideas how to approach the second part?
 
  • #9
1,024
32
the best way to it will be to draw(or see elsewhere) the graph and judge when x(t) will be zero for first time or just put it equal to zero and find t for first time.the graph is cosine curve ,shifted leftward and with decreasing amplitude.max. negative displacement will occur after first cycle means after when it crosses first zero and see wikipedia for other assistance.
 
Last edited:
  • #10
1,267
11
If I want to calculate it. We have by setting it equal to zero:

x(0) = (A0/cos) cos (δ) = 0

So what value do I use for δ? Do I have to use tan δ = 1/ω'τ and then substitute ω'=2π/T'?
 
  • #11
1,024
32
I am not saying to put t=0 and then equate x=0.you have to find t by equating it.since
e(-t/tau) can not be zero.cosine part will be zero find when for the first time this will be zero.
cos(w't+delta)=0,and choose cos(pi/2) from here you will get t for first zero.max. negative displacement can be obtained at T'/2 time from the beginning means after
-delta/w'.can you tell the answer so that I can check it.
 

Related Threads on Damped Simple Harmonic Motion

  • Last Post
Replies
3
Views
2K
  • Last Post
Replies
2
Views
6K
Replies
4
Views
8K
  • Last Post
Replies
2
Views
732
Replies
8
Views
3K
Replies
1
Views
1K
  • Last Post
Replies
4
Views
401
  • Last Post
Replies
0
Views
4K
  • Last Post
Replies
3
Views
828
  • Last Post
Replies
2
Views
10K
Top