Waves question, finding new frequency, please help

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  • #1
Ush
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Homework Statement



A stretched wire vibrates in its fundamental mode at a frequency of 375 Hz. What would be the fundamental frequency if the wire were one third as long, its diameter were tripled, and its tension were increased five-fold?


Homework Equations



V = F*w = sqrt(T/u)

mass of cylinder = Volume x Density
m = πr2Lp

where
F = Frequency
w = wavelength,
T = tension
u = linear mass density
m = mass
L = length
p = density

The Attempt at a Solution



m = πr2Lp
m/L = πr2p
1. u = πr2p

F*w = sqrt(T/u)
F*2L = sqrt(T/u)
F24L2 = T/u
2. F2 = T/(u4L2)

sub in u from 1. into 2.

F2 = T/(πr2p4L2)

Question says..
T x 5
L x 1/3
D x 3 => r x 6

F22 / F12 = [5T/(π(6r)2p4(1/3L)2)] / [T/(πr2p4L2)]

F22 / F12 = 5 / 62(1/3)2

F2 / F1 = sqrt(5/4)
F2 = sqrt(5/4)*F1

...This gave me the wrong answer,
Could someone please tell me where I went wrong? or where I missed something?
Thank you
 

Answers and Replies

  • #2
Doc Al
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You're almost right. Keep it simple. Express the fundamental frequency in terms of the speed and length. Then you can figure out how the frequency changes by understanding how the speed and length changes. (By what factor does each variable change?)
 
  • #3
Ush
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thank you for the reply,

but I don't understand how I can express it in terms of velocity, don't I need to change V into sqrt(T/pπr2) because we're changing diameter?
 
  • #4
Ush
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Hm, I tried that, but it doesn't seem to give me the correct answer, =/ it gave me a oddly high frequency of 7546.5Hz
 
  • #5
Ush
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I believe there may be something wrong with your reasoning for calculating linear mass density?
 
  • #6
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I believe there may be something wrong with your reasoning for calculating linear mass density?

Yeah, I fixed it in my example. I was probably looking at a center of mass problem earlier and went with mass per unit area instead of mass per unit length. Hopefully the problem is cleared up now.
 
  • #7
Doc Al
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but I don't understand how I can express it in terms of velocity, don't I need to change V into sqrt(T/pπr2) because we're changing diameter?
My suggestion was to do it in steps. Much easier (to me) and less chance for error.

The fundamental frequency in terms of speed and length is given by: f = v/(2L).

How does the speed change? How does the length change?
 
  • #8
Ush
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AtticusFinch,
your method gave me a frequency of 279.508Hz, unfortunately this is the wrong answer,

This may be because Volume of a cylinder doesn't vary with D3; It varies with D2, or r2, depending on the volume equation used..; but even when I tried the same reasoning, with D2, it still gave me the wrong answer, (484.12Hz)

---

Doc Al,
I know the length is reduced to a 1/3 of original
Fnew = Vnew / (2/3)L

I don't understand how I can calculate how speed changes? Other then trying => speed varies with sqrt(T/u) ,Which isn't really working at the moment =/
 
  • #9
Doc Al
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45,075
1,382
Doc Al,
I know the length is reduced to a 1/3 of original
Fnew = Vnew / (2/3)L
Express that as:
L' = L/3

I don't understand how I can calculate how speed changes? Other then trying => speed varies with sqrt(T/u) ,Which isn't really working at the moment
Why isn't it working? Use v = sqrt(T/μ). How does T change? How does μ change?
 
  • #10
Ush
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Express that as:
L' = L/3

can I express it as,
Wold = Wnew/3 ?

Tension changes by a factor of 5,

the equation for mass of a cylinder is
m = πr2pL

therefore u (m/L) is u = πr2p

Diameter changes by a factor of 3
and since volume changes by r2
wouldn't that mean u changes by a factor of 36?

or am I complicating it again?
 
  • #11
Doc Al
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45,075
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can I express it as,
Wold = Wnew/3 ?
Sure. But let's keep L' = L/3.

Tension changes by a factor of 5,
OK. T' = 5T.

the equation for mass of a cylinder is
m = πr2pL

therefore u (m/L) is u = πr2p

Diameter changes by a factor of 3
and since volume changes by r2
wouldn't that mean u changes by a factor of 36?
How did you get 36? Since r' = 3r, r² increases by a factor of 3².

(Throughout this I assume that the new wire is made from the same material as the original wire, thus it has the same density.)
 
  • #12
Ush
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But 1 diameter = 2 radius?
so shouldn't an increase in diameter by 3, increase radius by 6?
and 62 => 36?
 
  • #13
Doc Al
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But 1 diameter = 2 radius?
so shouldn't an increase in diameter by 3, increase radius by 6?
and 62 => 36?
Nope. Both the radius and diameter are proportional to each other. If the area is proportional to r² (as in A = πr²), then it's also proportional to d² (as in A = π(d/2)² = (π/4)d²).
 
  • #14
Ush
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hm, then u would increase by a factor of 9 correct?

also, do we have to consider change in length when we're finding u?

since u = m/L
and L decreases to 1/3Lo

would that make unew = 3 uold

-> b/c
mold = πd2pL/4

mnew = π(3d)2p(1/3)L/4

3u = π(3d)2p /4
3u = π9d2p /4

divide by 3 again goes back to

u = π3d2p /4

so.. u increases by a factor of 3? :S
 
  • #15
Doc Al
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hm, then u would increase by a factor of 9 correct?
Right.

also, do we have to consider change in length when we're finding u?
No. μ is the mass per unit length, so it does not depend on the length.

since u = m/L
and L decreases to 1/3Lo

would that make unew = 3 uold
That would be true if you kept the same mass and just squished it into 1/3 of its original length. But that's not what we're doing here. We're just using a thicker piece of wire.
 
  • #16
Ush
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ah, okay

so that would mean..

T' = 5T
u' = 9u
L' = L/3

Fold*2L = sqrt(T/u)

Fnew*(2/3)L = sqrt(5T/9u)

Fnew / Fold = sqrt5
Fnew = sqrt5*Fold ?

is this correct?
 
  • #17
Doc Al
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Looks good to me.
 
  • #18
Ush
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thank you so much =]
 
  • #19
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Whoops yeah I saw that I was thinking in terms of a sphere not a wire. But you got it now so I just took out the example :)
 

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