What is the relationship between string tension and frequency on a ukulele?

[Drizzy]

Homework Statement

You have a ukulele and you change the strings tension so that it becomes tighter. First the frequency was 431 Hz and after we tightened the string the frequency. how much bigger does the force have t be to get that frequency?

Homework Equations

PA*V^2 = F (P = density, A = area)
V=f*lambda

The Attempt at a Solution

F₂ = PA V²=PAf₂² * lambda²
F₁ = PA V²=PAf₁² * lambda²
Then i divided F₂ by F₁ and I got:
f₂²/f₁²
which equals to 1,06. My answer is 6 precent bigger.

So my questions are: can I assume that lambda is going to remain the same? Is my solution correct?

 

[haruspex]

If lambda here represents the wavelength of the wave in the string, yes you can assume it remains the same. The string length did not change, and we are only interested in the fundamental.
You seem to have omitted the new frequency from the problem statement.

 

[Drizzy]

Sorry, the new frequency is 444Hz. Why can we assume that? That was my initial thought but then I thought wouldn't the wavelength differ if the string is tighter?

 

[haruspex]

Sorry, the new frequency is 444Hz. Why can we assume that? That was my initial thought but then I thought wouldn't the wavelength differ if the string is tighter?

What is the equation relating the wavelength of the n^th harmonic to the length of the string?

 

[Drizzy]

it is l=n*lambda/2

 

[haruspex]

it is l=n*lambda/2

Right. So if we are only interested in the fundamental, that fixes n as 1. And the length l of the string does not change. So what does that tell you about lambda?