Wave Length that gives minimum velocity?

Join the discussion
Registration is free. Ask a follow-up in this thread, or start your own.
7 replies · 6K views
01010011
Messages
48
Reaction score
0

Homework Statement


The velocity of a wave of length L in deep water is v = K square root of (L/C + C/L)
where K and C are known positive constants. What is the length of the wave that gives the minimum velocity?

Homework Equations


Possibly a(t) = v'(t) = s"(t)

The Attempt at a Solution


I don't know how to work the question but here is my best guess...
Can I just cancel L with L and C with C (inside the square root) and be left with v = K.
Next, to find the length, can I just find the antiderivative of the velocity, like this: v(t) = Kt + C.

How do I workout this question?
 
Physics news on Phys.org
Hi 01010011! Welcome to PF! :smile:

The question is asking for the minimum of (L/C + C/L), where C is a constant. :wink:
 
tiny-tim said:
Hi 01010011! Welcome to PF! :smile:

The question is asking for the minimum of (L/C + C/L), where C is a constant. :wink:

Thanks for the welcome. Well I was thinking the answer was 0 because the Ls and Cs cancel out each other, but I am not sure
 
They don't cancel (why do you think they would? :confused:)

Try differentiating. :smile:
 
tiny-tim said:
They don't cancel (why do you think they would? :confused:)

Try differentiating. :smile:

Differentiating v = K square root of (L/C + C/L)? I thought I could just cancel the Ls and Cs. Here is another attempt:

v = k square root (L/C + C/L)

v(t) = s'(t)

v(t) = k square root [(L*C^-1) + (C*L^-1)] + C

v(t) = k * [(L*C^-1) + (C*L^-1)] ^ (1/2) + C

s(t) = k * {[(L*C^-1) + (C*L^-1)] ^ (3/2)} / 3/2 + C + D

This looks like madness. I am sure this is not correct.
What steps (and why) should I take to answer questions like this?
 
tiny-tim said:
Hi 01010011! Welcome to PF! :smile:

The question is asking for the minimum of (L/C + C/L), where C is a constant. :wink:

Right, since C is a constant, the smallest value C can be is 0, So...
 
01010011 said:
Differentiating v = K square root of (L/C + C/L)? I thought I could just cancel the Ls and Cs. Here is another attempt:

v = k square root (L/C + C/L)

v(t) = s'(t)

v(t) = k square root [(L*C^-1) + (C*L^-1)] + C

v(t) = k * [(L*C^-1) + (C*L^-1)] ^ (1/2) + C

s(t) = k * {[(L*C^-1) + (C*L^-1)] ^ (3/2)} / 3/2 + C + D

This looks like madness. I am sure this is not correct.
What steps (and why) should I take to answer questions like this?


Nooo, this is a mess. :redface:

i] s has nothing to do with it, the fact that v is a velocity has nothing to do with it, all you have to do is find the minimum of (L/C + C/L), where C is a constant

ii] why are you trying to minimise the square-root? the square-root is a minimum if and only if the whole thing is a minimum, so minimise the whole thing, it's easier! :wink:

iii] you haven't used the chain rule at all … look it up in your book :smile:

01010011 said:
Right, since C is a constant, the smallest value C can be is 0, So...

erm :redface: … that doesn't even make sense, does it?

get some sleep! :zzz:​
 
tiny-tim said:
Nooo, this is a mess. :redface:

i] s has nothing to do with it, the fact that v is a velocity has nothing to do with it, all you have to do is find the minimum of (L/C + C/L), where C is a constant

ii] why are you trying to minimise the square-root? the square-root is a minimum if and only if the whole thing is a minimum, so minimise the whole thing, it's easier! :wink:

iii] you haven't used the chain rule at all … look it up in your book :smile:



erm :redface: … that doesn't even make sense, does it?

get some sleep! :zzz:​

Ok, I got some much needed sleep lol!

Alright, let me try again:

v = K square root of (L/C + C/L)
dy/dx?
Let U = square root of (L/C + C/L)
V = KU
V = 1KU^(1-1)
V = K

hmmm...