Fundamental Frequency Question

[dantechiesa]

Homework Statement

A stretched wire vibrates in its first normal mode at a frequency of 369Hz. What would be the fundamental frequency if the wire were one third as long, its diameter were tripled, and its tension were increased two-fold?

Homework Equations

f = 1/2L * squareroot(Ft/u)

The Attempt at a Solution

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Can someone explain what I've done wrong?

369 = 1/2L * Sroot(Ft/u)

L is 1/3, so 1/2L becomes 3/(2L)

The Ft is doubled, but the u needs to be broken down further.
m/l = u
l is 1/3
m = dv
d stays the same, so v is the relevant one.
V = pi r²h
r is tripled, so the V increases by a factor of 9. However, since the h (aka l) also changes by a third. Total change of V is x3.

so, 3*m / (l/3)
All together,
Sroot ( 2Ft / (9u)

However, 369 * (3/2) * Sqroot (2/9) is wrong. ( = 260.922402258)

Thanks.

 

[mjc123]

You had a factor of 1/2L. Now you have a factor of 3/2L. So you don't multiply by 3/2.

 

[Delta2]

Diameter is tripled

In the formula for V , r is the radius , so since diameter is tripled, radius goes up by x1.5.

 

[mjc123]

Radius and diameter are in proportion. If one is tripled, so is the other.

 

[dantechiesa]

You had a factor of 1/2L. Now you have a factor of 3/2L. So you don't multiply by 3/2.

Oh would you multiply by 3 only? Since the 1/2 was already apart of the equation?

ALso, does the remaining work with the square root seem right?

Thanks.

 

[mjc123]

Yes

 

[dantechiesa]

Yes

Thanks a lot!