- #1
brunettegurl
- 138
- 0
Homework Statement
A stretched wire vibrates in its fundamental mode at a frequency of 384 Hz. What would be the fundamental frequency if the wire were half as long, its diameter were doubled, and its tension were increased five-fold?
Homework Equations
F= [tex]\frac{1}{2L}[/tex][tex]\sqrt[]{\frac{T}{\mu}}[/tex]
The Attempt at a Solution
ok i know that this F= [tex]\frac{1}{2L}[/tex][tex]\sqrt[]{\frac{T}{\mu}}[/tex] is the formula for the one where F1= 384 Hz
i know that length is cut in half and that tension increases by 5and when the diameter is doubled that means mass is quadrupled my problem is that i don't know how to show that wrt [tex]\mu[/tex]
for the second frequency would my equation look like
F2= [tex]\frac{1}{2(L/2)}[/tex][tex]\sqrt[]{\frac{5*T}{4*\mu}}[/tex]
and then if we cancel off the 2 in the l would it look like F= [tex]\frac{1}{L}[/tex][tex]\sqrt[]{\frac{5*T}{\mu}}[/tex] *[tex]\frac{1}{2}[/tex]