# Help with Fundamental Frequency problem

1. Nov 21, 2004

### Purduenub03

In order to decrease the fundamental frequency of a guitar string by 2%, by what percentage should you reduce the tension?

I'm so lost can you point me in the right direction?

2. Nov 21, 2004

### swatikiss

just a guess - but why not look at the length of the string?

3. Nov 21, 2004

### Hurkyl

Staff Emeritus
Well, you have a problem involving frequency and tension... have you at least thought if you know any formulae that relate frequency and tension?

4. Nov 21, 2004

### Purduenub03

I think that F = 1/2L * sqrt(t/m)

I have no clue what to do though

5. Nov 21, 2004

### Hurkyl

Staff Emeritus
All right.

Have you yet assigned variables knowns and unknowns in this problem, and written down the information in the problem algebraically?

6. Nov 21, 2004

### Purduenub03

I'm guessing

.96F = 1/2L * sqrt(x*T/g)

?

7. Nov 21, 2004

### Hurkyl

Staff Emeritus
Nope, you're jumping too far ahead. When you're lost you need to look one step at a time, even if it seems like a trivial step!

8. Nov 21, 2004

### Purduenub03

.96 F = 1/2L * sqrt(t/m)

?

9. Nov 21, 2004

### Hurkyl

Staff Emeritus
Nope, think much more basic.

For instance, the problem is about changing the fundamental frequency of a string.

So, it would make sense to start off by defining, for instance:

$F_0$ is the frequency of the string before the change.
$F_1$ is the final frequency of the strign after the change.

(aside: in text, we usually write $F_0$ as F_0)

10. Nov 21, 2004

### Purduenub03

So F1 is .96 F0

11. Nov 21, 2004

### Hurkyl

Staff Emeritus
Don't say "is"! Say:

F_1 = .96 F_0

(BTW, it should be .98, unless you made a typo when you copied the problem)
(Yes, I'm being picky, but I've seen plenty of people confuse themselves by thinking in terms of "is" instead of thinking in terms of an equation)

Can you think of any other variables you might want to define for this problem?

Last edited: Nov 21, 2004
12. Nov 21, 2004

### Purduenub03

Wait......

since 1/2L is constant we can count that out........and t = mg so you can cancel out the m being left with

.98 = sqrt(g/1)

.98^2 = g/1

1-.98^2 = g

3.96%?

13. Nov 21, 2004

### Hurkyl

Staff Emeritus
Well, I think you have the right answer, so it's time for me to go to bed!

I still think your derivation looks confused though (I don't know if it's just the way you wrote it, or if you really do still have some confusion), and your formula looks different than what's in my physics book (I don't know if it's just different letters or not -- it's been a while since I've reviewed this).

Just FYI, what I was trying to get you to do was to say that you need to find looking for $T_1/T_0$ (if you defined $T$ to be the tension in the string), and the easiest to see method (though not the quickest) for finding this was to solve your formula for T and do some substitutions.