Answer to Question #293803 in Physics for Lug

Question

A stretched wire emits a fundamental note of 256Hz keeping the stretching force constant and reducing the length of wire by 10cm, the frequency becomes 310Hz . Calculate the original length of the wire

Answer 1

How does frequency change with length and how does it depend on tension?

$f=\dfrac 1{2L}\sqrt\dfrac{T}{\mu}.$

For the condition given, we have

$f_1=\dfrac 1{2L_1}\sqrt\dfrac{T}{\mu},\\\space\\ f_2=\dfrac 1{2L_2}\sqrt\dfrac{T}{\mu},\\\space\\ \dfrac {f_1}{f_2}=\dfrac{L_2}{L_1},\\\space\\ \dfrac {f_1}{f_2}=\dfrac{(L_1-10)}{L_1},\\\space\\ L_1=\dfrac{10f_2}{f_2-f_1}=57\text{ cm}.$