What Is the Fundamental Frequency of a Wire Under Modified Conditions?

[hemetite]

Homework Statement

A stretched wire vibrates in its fundamental mode at a frequency of 400Hz. What would be the fundamental frequency if the wire were half as long, with twice the diameter and with four times the tension?

Homework Equations

fn = nV / 2L where n is the harmonics number. f1 is the fundmental frequency.

i think i should be using
V= Sqrt(T/u)

The Attempt at a Solution

how do i start? this is what i got...

since it gave 400Hz... then i think i should use...

fn = nV/2L -----> 400Hz = V/2L

therefore V= 400 * 2L

using V= Sqrt ( T/u) ---> 400*2L= Sqrt( T/U) ----

i am stuck here...i have a hunch that i need to find something it term of something that play back the two equation..afterwhich it will cancel out something..then i can get the value...dunno where to go from here...any hel thanks??

 

[tiny-tim]

dimensions

A stretched wire vibrates in its fundamental mode at a frequency of 400Hz. What would be the fundamental frequency if the wire were half as long, with twice the diameter and with four times the tension?

Hi hemetite!

This is a dimensions question …

write out the formula for the fundamental frequency, f₁ …

then see whether f₁ is proportional to T, to 1/T, to √T, or what?

same for L and d.

 

[hemetite]

k...let me struggle..abit...i seems to smell something already...hopefully it is not burnt... :)

physic is just so challenging... what a rush...!

 

[hemetite]

here what i get...

these are the formulas...

for fundamental frequency...n=1

so,

f= V/2L --------- 1

lamba = 2L ------- 2

v= lambda * f -----3

v= sqrt (T/ (m/L) ------4

---------------------------------------

using 3 ... v= (2L) 400Hz
so if we wire of 1/2 as long so
v= (1/2) (2L) 400Hz = 400L

put these into 1

f= V/2L = 400L/2L =200Hz

these is 1 of the answer...i have to get for 2 more right for the change for D and T?

...i see the question like 3 condition combine into 1?...or..3 mini question or 1 big question?