Fundamental frequency and tension of a guitar string

[BlueSkyy]

Homework Statement

In order to decrease the fundamental frequency of a guitar string by 4%, by what percentage should you reduce the tension?

Homework Equations

f = sqrt [T/(m/L)] / 2L

I believe that is the equation that relates frequency to tension...

The Attempt at a Solution

I plugged in some theoretical values and got 76.96% which seemed wrong, and it is :)
How can I go about solving this problem?

 

[Vidatu]

$$f_1 = (1/(2L)) * \sqrt(F/\mu)$$

Your formula is correct.

First, do f and F vary directly or inversely with each other?

 

[BlueSkyy]

as L gets bigger, F gets bigger, right?

 

[Vidatu]

Yes, except in this case, they will be getting smaller.

What is the ratio between them?

 

[BlueSkyy]

if L doubles, than T is quadrupled, right?

 

[BlueSkyy]

but this has nothing to do with the frequency, since L stays the same...

 

[Vidatu]

Whoops, I misread what you typed. We don't care whether the tension and length vary directly or inversely. We want how frequency and tension are related. The rest of the equation isn't important, since this is just asking for a relative number.

$$f \alpha \sqrt(F)$$

 

[BlueSkyy]

so the frequency is proportional to the square of the tension?

 

[Vidatu]

Yep. So, putting what we know together;
$$0.96f \alpha \sqrt(xF)$$
You need to find x.

 

[BlueSkyy]

no idea. this is kinda where i got stuck...

 

[Vidatu]

No problem. As $$f \alpha \sqrt(F)$$, it makes sense that $$0.96 \alpha \sqrt(x)$$, right? Use $$0.96 = \sqrt(x)$$ to solve for x. This will give you a decimal value, which you multiply by 100 to turn into a percent. This is the percent of the original length needed to change the frequency by 4%, so to get the answer, you subtract it from 100%.

% to decrease = 100% - (100x)

 

[BlueSkyy]

okay, that's what i was thinking, but i didn't know if i could use an equals sign since we were working with a proportion :) thank you!

 

[Vidatu]

Glad to be of help :)